Class 12

RBSE Solutions for Class 12 Physics Chapter 6 Electric Circuit

March 8, 2021 by Prasanna Leave a Comment

RBSE Solutions for Class 12 Physics Chapter 6 Electric Circuit are part of RBSE Solutions for Class 12 Physics. Here we have given Rajasthan Board RBSE Solutions for Class 12 Physics Chapter 6 Electric Circuit.

Rajasthan Board RBSE Class 12 Physics Chapter 6 Electric Circuit

RBSE Class 12 Physics Chapter 6 Text Book Exercise with Answers

RBSE Class 12 Physics Chapter 6 Multiple Choice Type Questions

Question 1.
Kirchhoffs first and second laws are based on :
(a) Charge and Energy Conservation Laws
(b) Current and Energy Conservation Laws
(c) Mass and Charge Conservation Laws
(d) None of these
Answer:
(a) Charge and Energy Conservation Laws
Kirchhoffs first law is based on law of conservation of charge and second law is based on law of conservation of energy.

Question 2.
The potential difference between the points a and b in given figure will be :

(a) R₁ – R₂
(b) R₂ – R₁
(c) [latex]\frac{R_{1} R_{2}}{R_{1}+R_{2}}[/latex]
(d) Zero
Answer:
(a) R₁ – R₂
I₁ = I ₂ = 1 amp
∴ Potential at a, V_a = I₁ × R₁
= 1 × R₁ = R₁
Potential at b, V_b = I₂R₂ = 1 × R₂ = R₂
V_a – V_b = R₁ – R₂

Question 3.
The value of I in a given figure will be :

(a) 6 A
(b) 11 A
(c) 7 A
(d) 5 A
Answer:
(c) 7 A
From first law of Kirchhoff,
ΣI = 0
5 + 2 + 4 – 4 – I = 0
I = 7 A

Question 4.
In a Wheatstone bridge, the position of battery and galvanometer is interchanged, then the new balanced position will be :
(a) Unchanged
(b) Will changed
(c) Can’t say anything
(d) May change or may not be, it depends upon the resistance of galvanometer and battery
Answer:
(a) Unchanged
There will be no effect

Question 5.
The potential difference between the points a and b in a given figure is :

Answer:
(d)
In the circuit [latex]\frac{P}{Q}=\frac{R}{S}[/latex]
∴ [latex]\frac{8}{4}=\frac{6}{3}[/latex] = true
So the circuit work as Wheatstone Bridge

Question 6.
The current flow in a given circuit will be :

(a) 2.5 A
(b) 0.75 A
(c) 0.5 A
(d) 0.25 A
Answer:
(d) 0.25 A
From Kirchhoff’s second law
ΣE = IR
2 – 4 = -2I – 5I – I
-2 = -8I
I = [latex]\frac{2}{8}\frac{1}{4}=[/latex] = 0.25 A

Question 7.
Potentiometer is a type of instrument by which potential difference can be measured; then its resistance will be ;
(a) Zero
(b) Infinite
(c) Uncertain
(d) Depends on external resistance
Answer:
(b) Infinite
Effective resistance of potentiometer is infinite.

Question 8.
The physical quantity which can not be measured with the help of potentiometer is :
(a) EMF of cell.
(b) Inductance and capacitance
(c) Resistance
(d) Electric current
Answer:
(b) Inductance and capacitance
Capacitance and self inductance are not measured with the help of potentiometer.

Question 9.
In the circuit zero deflection is shown in figure in meter bridge experiment. Find out the resistance R:

(a) 220 Ω
(b) 110 Ω
(c) 55 Ω
(d) 13.75 Ω
Answer:
(a) 220 Ω

Question 10.
Temperature coefficient of resistance of potentiometer wire should be :
(a) High
(b) Low
(c) Negligible
(d) Infinite
Answer:
(c) Negligible
The specific resistance and the coefficient of linear expansion of potentiometer wire must be negligible.

Question 11.
Formula of internal resistance of primary cell in the form of balancing length is {l₁and l₂ are the balancing lengths in open and closed circuit respectively} :

Answer:
(a)
r = [latex]\left(\frac{l_{1}-l_{2}}{l_{2}}\right)[/latex]R

Question 12.
Cell of emf e is balanced at length L in potentiometer experiment. Now other cell having emf e connects parallel to the first cell. Now the new balancing length will be :
(a) 2L
(b) L
(c) [latex]\frac{L}{2}[/latex]
(d) [latex]\frac{L}{4}[/latex]
Answer:
(b) L
In parallel combination there will be no effect on potential difference. So there will be no effect on balance length.

Question 13.
The standard cell of 1.1 V is balanced on 2.20 m length in potentiometer. Potential difference across any resistance is balanced at 95 cm and voltmeter shows 0.5 V reading then error in the reading of voltmeter will be :
(a) +0.025 V
(b) +0.525 V
(c) -0.025 V
(d) -0.525 V
Answer:
(a) +0.025 V
Potential drop measured by potentiometer
V₂ = [latex]\frac{E}{L}[/latex] × R₂
= [latex]\frac{1.1}{2.20} \times \frac{95}{100}[/latex] = 0.475V
Potential difference measured by voltmeter = 0.5 volt
Error in the reading of voltmeter
ΔV = V – V’
ΔV = 0.500 – 0.475
= + 0.025 volt

RBSE Class 12 Physics Chapter 6 Very Short Answer Type Questions

Question 1.
Write the mathematical form of Kirchhoff’s junction law.
Answer:
The mathematical form of Kirchhoff’s junction law, ΣI = 0

Question 2.
Kirchhoff’s voltage law is based on which conservation law?
Answer:
Law of conservation of energy.

Question 3.
Write the condition of the balanced state of the Wheatstone bridge.
Answer:
When the Wheatstone bridge is balanced, then the ratio of the resistances of ratio arms are equal.
[latex]\frac{P}{Q}=\frac{R}{S}[/latex]

Question 4.
What is the principle of meter bridge?
Answer:
Meter bridge is based on the principle of Wheatstone bridge.

Question 5.
Why the potential gradient of the wire of potentiometer is based on the temperature?
Answer:
When the temperature of the potentiometer wire increases its resistance also increases so that its potential gradient affected.

Question 6.
If the emf of unknown cell in secondary circuit is greater than the emf of primary circuit of potentiometer then what will happen?
Answer:
Null point will not be found on potentiometer wire.

Question 7.
Write down the definition of the potential gradient.
Answer:
Potential drop per unit length of the wire of potentiometer is known as potential gradient of potentiometer.

Question 8.
Why the cross-sectional area of the wire should be uniform in a potentiometer?
Answer:
So that the potential gradient remains constant at all positions.

Question 9.
Which cell can be used for standardization of the potentiometer other than Daniel cell?
Answer:
Cadmium cell or Leclanche cell. ’

Question 10.
How can the sensitivity of the potentiometer be increased?
Answer.
Potential gradient k = [latex]\frac{V_{A B}}{l_{A B}}[/latex]
So, the sensitivity of potentiometer can be increased by increasing the length of the wire of potentiometer.

Question 11.
Length of the potentiometer wire is 10 m. A standard cell of 1.1 V is balanced at 8.8 m length. How much maximum potential difference can be measured with the help of this potentiometer?
Answer:
Potential gradient (k) = [latex]\frac{V}{l}=\frac{1.1 V}{8.8}[/latex]
= 0.125 V/ m
Maximum potential difference
V = kl = 0.125 × 10 = 12.5 volt

Question 12.
Why copper wire can not be used in potentiometer?
Answer:
Temperature coefficient of resistance is large and specific resistance is less for the copper wire.

Question 13.
The potential gradient of the wire of potentiometer is 0.3 V/m. In the calibration of ammeter, potential difference across 1 Ω coil is balanced at 1.5 m length. If the reading of ammeter is 0.28A then find out the error in the reading of ammeter.
Answer:
Virtual current is measured by Ammeter I = 0.28 A

RBSE Class 12 Physics Chapter 6 Short Answer Type Questions

Question 1.
Write the statement of Junction and Loop Law of Kirchhoff s.
Answer:
“In an electric circuit, the algebraic sum of current at any junction point must be zero.”
i. e., ΣI = 0
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kirchhoff s First Law or Junction Law.

It is based on the law of Conservation of Charge. When currents in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.

In figure (6.1) at junction O,
I₁ + I₂ – I₃ – I₄ + I₅ = 0
or I₁ + I₂ + I₅ = I₃ + I₄ ………………. (1)
In figure (6.1) the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.

Kirchhoff’s Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., ΣV = 0 …………… (1)
The algebraic sum of the emf’s in any loop of a circuit is equal to the sum of the voltages across the resistance.
ΣE = ΣIR ………….. (2)
Sign convention for applying Loop Rule
1. The product of current and resistance (IR) is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance (IR) is taken as negative if the resistor is traversed in the opposite direction of assumed current.

In loop we can move from A to B then the product of IR take as positive V_AB = +IR. [In the direction of current]
In a loop we can move from B to A then the product of IR take as negative.
∴ V_BA = -IR [Opposite direction of current]

2. When we move from B to A then E is taken positive (+). When we move from AtoB, then E is taken negative (-).

Loop law can be understood by the figure 6.4.
In the given figure for point d by the Junction Law, current flowing through the resistance R% will be
I₃ = I₁ + I₂ …………… (3)
In the loop adcba by the Kirchhoff s Loop Law.
I₁R₁ – I₂R₂ = ε₁ – ε₂
or I₂R₂ = I₂R₂ = ε₂ – ε₁ ………… (4)
In the loop defcd by the Loop Law
I₃R₃ + I₂R₂ = ε₂ …………….. (5)
By solving equations (3), (4) and (5) we can find out electric currents in different branches as well as we can find potential differences across the resistance.

Question 2.
Write the process of finding unknown resistance by meter bridge and deduce the necessary mathematical formula.
Answer:
Meter Bridge
It is the simplest practical application of the Wheatstone bridge that is used to measure an unknown resistance.

Construction : It consists of usually one meter long Manganin or constant on wire of uniform cross-sectional area, stretched along a meter scale fixed over a wooden board and with its two ends soldered to two L- shaped thick copper strips M and N. Between these two copper strips, another copper strip I is fixed so as to provide two gaps ab and a₁b₁. A resistance box R.B. is connected in the gap ab and the R.B. G unknown resistance S is connected to the gap ab. and the unknown resistance S is connected to the gap a₁b₁. A source of emf e is connected across AC. A movable jockey and a galvanometer are connected at BD as shown in the figure.

Working : After taking out a suitable resistance R from the resistance box, we touch the jockey at point A and C and observe opposite deflection in galvanometer. If galvanometer shows deflection in one direction then we change the value of known resistance until galvanometer shows opposite deflection.

Now the jockey is moved along the wire AC till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge. If P and Q are the resistances of the parts AB and BC of the wire, then for the balanced condition of the bridge we have
[latex]\frac{P}{Q}=\frac{R}{S}[/latex] ……………….. (1)
If resistance of unit length of the wire is [latex]\sigma \frac{\mathrm{ohm}}{\mathrm{cm}}[/latex] and the balance point occurs at l cm from A then the length of part BC will be (100 – l) cm.
P = Resistance of wire of AB part = σl …………. (2)
Q = Resistance of wire of BC part = σ (100 – l) ……………… (3)
In equation (1) putting the value P and Q

Knowing l and R, unknown resistance can be determined.
The formula of determination of specific resistance of unknown resistannce (S)
We know that
Unknown resistance (S) = [latex]\rho \frac{l^{\prime}}{A}[/latex]
Where l’ = Length of unknown resistance
A = Area of cross section = πr²
From equation (4)

Limitations of Meter Bridge :

Copper strips have their own resistances but in the calculation of meter bridge formula we neglect this resistance so that our results are errorful. For minimising this error, at time of performing experiment we should calculate unknown resistance by interchanging the position of unknown resistance and known resistance. In this process we take average value of obesrvations by which error can be minimised.
The resistances of end points affect the sensitivity of meter bridge. To nulify this effect Carey-Foster s bridge is used.
Current should not flow for long time through the wire otherwise wire will heat up and the resistance of wire will be changed.
Jockey should not slide tightly on resistance wire otherwise it’s cross-section area will change. That is why the resistance of the wire will be changed.

Question 3.
What is Wheatstone bridge? Deduce the balanced condition of it with the help of Kirchhoff’s law.
Answer:
Construction : The construction of Wheatstone’s bridge is shown in following diagram (figure 6.5) in which a quadrilateral ABCD is prepared by

connecting four resistances P, Q, R and S. A cell is connected between the points A and C through key K₁ which is known as battery key and a galvanometer between the point B and D through another key K₂ which is known as galvanometer key. If on pressing the key K₁ and K₂, there is no deflection in galvanometer, then the bridge is said to be balanced condition. In this situation the relation between all the resistances is given by :
[latex]\frac{P}{Q}=\frac{R}{S}[/latex]

Principle of Wheatstone Bridge and Condition of Balance
When battery key K₁ is pressed, then main current I starts flowing in the circuit. At junction A this current splits in two parts I₁ and I₂ as shown in figure 6.6. These currents I₂ and I₂ again obtain two paths at junctions B and D respectively. Now following three situations are possible :

(i) When V_B > V_D. then current I₂ passes as such through resistance S but current I₁ splits in two parts at junction B, one part I_g passes through galvanometer showing deflection in one direction while rest current (I₁ – I_g) passes through Q.

(ii) When V_B < V_D. then opposite situation to previous situation (i) is observed i.e. .current I₂ splits in two parts at D. Now the current I_g produces deflection in galvanometer in opposite direction to previous deflection.

(iii) When V_B = V_D, then no current passes through galvanometer i.e. galvanometer shows no deflection because now I _g = 0. This situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called ‘Bridge circuit’ Thus in balanced condition.

Therefore, “it is clear that in balanced situation of the bridge, the ratio of resistances of any two corresponding arms of quadrilateral ABCD is equal to ratio of two remaining corresponding arms”.
From equation (3).
S = [latex]\frac{Q}{P} \times R[/latex]
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below :
(i) P and Q are called ratio arms.
(ii) R is called variable resistance arm.
(iii) S is called unknown resistance arm.

Question 4.
What is potential gradient? On which factors potential gradient depends?
Answer:
Principle of Potentiometer
Potentiometer : Potentiometer is a apparatus with help of which the e.m.f. of a cell and potential difference can be measured accurately.

Principle : Suppose AB is a resistance of uniform area of cross section and length LAB- If a standard cell of emf E and negligible internal resistance is connected across the ends of wire A and B with the help of connection wires of negligible resistance, then the potential difference across the wire will be E. Therefore potential gradient.
k = [latex]\frac{E}{L_{A B}}[/latex] ……………. (1)
If we consider a point C at a distance l from A, then
V_AC = kl ………….. (2)
Now if a cell of unknown emf E’ is connected between A and C as shown in the figure, then the deflection in galvanometer will depend on the fact that whether V_AC > E’ or E’ > V_AC. When V_AC = E’, then galvanometer will show no deflection. This fact can be understood well with the help of following equivalent circuit between A and C. This is null position galvanometer.

E’ = V_AC
or E’ = kl ………………… (3)
This is the principle of potentiometer.
If E’ > E, then galvanometer will show the deflection in one direction only. In this situation the measurement of E’ is not possible.
If the resistance of potentiometer wire is RAB and the current is I then from the ohm’s law
E = IR_AB ……………… (4)
From the equation (1)
k = [latex]\frac{I R_{A B}}{L_{A B}}[/latex]
Let [latex]\frac{R_{A B}}{L_{A B}}[/latex] = Density of wire is represented by the ρ
∴ k = ρI ……………… (5)
The circuit diagram for potential gradient

The potential gradient

Value of potential gradient under different conditions : .
Case-1 : When only the battery of negligible internal resistance is connected to potentiometer.
∴ r = 0 and R_h = 0
From the equation (8)

Case-2: When the battery of internal resistance (r) is connected to the potentiometer

Case-3: When a some other resistance or rheostat of resistance (Rh) is connected in series with the battery in the circuit of potentiometer. Then the potential gradient is calculated by equation (8).
k = [latex]\left(\frac{E}{R_{A B}+r+R_{h}}\right) \times \frac{R_{A B}}{L_{A B}}[/latex]

Factors effecting the value of potential gradient:
From the equation (5)

Question 5.
What is standardisation of potentiometer? Write the process of it and draw the necessary diagram.
Answer:
Standardisation of Potentiometer
For the potentiometer process of finding practical value of potential gradient is called standardisation of potentiometer.

For this process, a standard cell of known emf is required which is connected in secondary circuit as shown in figure (6.12).

Standard cell is a type of cell whose emf remains contant for long time. Cadmium cell or Daniel cell is used as standard cell. Its emf is 1.0186 V and 1.08 V at 20° C respectively.

For finding potential gradient from standard cell, we slide jockey on wire and find out the balancing length where zero deflection is obtained in galvanometer. If the emf of standard cell is es then from the principle of potentiometer.
ε_s = kxl₀
k = [latex]\frac{\varepsilon_{s}}{l_{0}}[/latex] ……………… (1)

Question 6.
What is the sensitivity of potentiometer, how can it be increased?
Answer:
Sensitivity of Potentiometer
A potentiometer is sensitive if :

It is capable of measuring very small potential difference.
It shows a significant change in balancing length for a small change in the potential difference being measured. The sensitivity of a potentiometer depends upon the potential gradient along the wire. Smaller the potential gradient, greater will be the sensitivity of the Potentiometer.

The sensitivity of potentiometer can be increased by reducing the potential gradient. This can be done in two ways :

For a given potential difference, the sensivity can be increased by the increasing the length of the potentiometer wire.
For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of a rheostat. When in primary circuit current decreases then potential difference across the wire decreases so that without changing in primary circuit for reducing potential gradient the total length of the wire increases.

Graph between V and L
∵ The slope of graphical line

Question 7.
For comparing the emfs of the two primary cells by potentiometer draw the circuit diagram and deduce the formula.
Answer:
Comparison of Electromotive Forces of Two Primary Cells
Suppose we have to compare e.m.f. of two primary cells E₁ and E₂ with the help of potentiometer. For this purpose we prepare the circuit diagram as figure 6.15.

First of all we press key K₁ and find out the balancing length l₁ for cell E₁. If k be the potential gradient of potentiometer wire, then
E₁ = kl₁ …………….. (1)

Now the key K₁ is removed and K₂ is fitted, then if l₂ be the new balancing length, then
E₂ = kl₂ …………….. (2)
From equations (1) and (2), we have
[latex]\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}[/latex] ……………. (3)

Question 8.
Standard cell of 1.2 V is balanced at the length of 2.40 m in potentiometer. Find out the balancing length for the potential difference across 3.5 Ω resistance where 0.2 A current is flowing through it, also find out potential gradient.
[x = 0.5V/m, l = 1.40 m]
Answer:

Question 9.
The real value of emf of cell or potential difference across resistance can not be measured by voltmeter, why? How real value can be measured by potentiometer?
Answer:
Potentiometer is a device which does not draw any current from the circuit and still measures the potential difference.

Question 10.
Why should we get the null point in the middle of the meter bridge wire? Explain.
Answer:
The metre bridge is most sensitive when the four resistances forming the Wheatstone’s bridge are equal. This is possible only if the balance point is somewhere near the middle of the wire.

Question 11.
Why should the current be not passed through potentiometer wire for a long time?
Answer:
This will heat up the potentiometer wire and will change its resistance. Potential drop per unit length of the wire will be also changed.

Question 12.
Why the value of current remain constant in primary circuit? Explain.
Answer:
On changing the current, the potential gradient will change and due to this error occurs is measurement.

Question 13.
Write the two precautions for using potentiometer.
Answer:
Precautions with Potentiometer

emf of the cell connecting in primary circuit must be more than or equal to the emf of the cell of secondary circuit otherwise zero deflection can not be obtained.
All the high potential points or positive terminals should be connected at A
Balancing length should be calculated from A
Area of cross-section of the wire should be uniform otherwise potential gradient will not be constant.
Current should not be passing through potentiometer wire for a long time otherwise this will heat up the wire and will change its resistance and hence potential gradient will also changed.

Question 14.
What is the calibration of voltmeter by potentiometer. Draw the necessary circuit diagram.
Answer:
Calibration of Voltmeter
Observations measured from voltmeter is errorful due to different reasons :
(i) In built mechanical error,
(ii) Parallex error,
(iii) Unsymmetricity in the spring of voltmeter etc. Potentiometer can increase precise value of potential difference.

Checking the errors of observations of voltmeter by potentiometer is known as Calibration of Voltmeter.

In figure 6.17 necessary circuit diagram is shown for calibration of voltmeter. Primary circuit is usually connected as section before.

In secondary circuit positive terminal of standard cell is connected at point A (high potential end) while negative terminal is connected with the terminal 1 of two way key. In this circuit other cell ε, rheostat Rh₂, key K₂ and resistance box (R.B.) is connected in series as shown in figure. The high potential end of R.B. is connected at A while lower potential end is connected at the terminal 3 of two way key. The voltmeter which is to be calibrated should be connected across the ends of R.B. The middle terminal of two way key is connected with the one end of galvanometer (G) and other end of (G) is connected with jockey.

Working : Firstly we completed the primary circuit and insert the plug between terminal 1 and 2 and by the help of jockey finding balancing length l₀for the standard cell ε_s
[latex]\varepsilon_{s}=k l_{0} \quad \text { or } \quad k=\frac{\varepsilon_{s}}{l_{0}}[/latex] …………….. (1)
k is potential gradient.
By the standardisation of cell, then find out k.
Now eliminate the plug between 1 and 2 and insert it between 2 and 3. Now key K₂ be closed and suitable resistance out from R.B. By the help of Rh₂ necessary current pass through the resistance and takes the deflection in voltmeter, which is V. It is errorful observation, corresponding to V. Find out the real value of potential difference by potentiometer. Find the balancing length l₂. Now by the principle of potentiometer the real value of potential difference
V’ = kl₂
V’ = [latex]\left(\frac{\varepsilon_{s}}{l_{0}}\right)[/latex]l₂ ………… (2)
Error in the observation of voltmeter
ΔV = V – V’
By the help of R.B. and Rh₂, observe different values of voltmeter and for it find corresponding values of potential difference by potentiometer. Now taking difference of V (observations of voltmeter) and V’ (potential difference by potentiometer) we can find error in it
ΔV = V – V’
Calibration Curve : A graph is plotted between the readings of voltmeter and the corresponding errors. This graph is called calibration curve of the voltmeter. The graph has been no definite shape. The any graph of calibration is shown in figure 6.18.

Question 15.
Draw the necessary diagram for measuring small resistance by potentiometer?
Answer.
Measurement of Small Resistance
In figure 6.16 necessary circuit is shown.
Primary circuit is as usual. In secondary circuit, unknown small resistance r, known resistance R, rheostat Rh, key K₂ and emf ε’ all are connected in series. Higher potential end of R, connects at point A which has higher potential, while lower potential end of R and r connects with the ends 1 and 3 of two way key. Terminal 2 of two way key connects with galvanometer end. Galvanometer also connects with jockey.

Working : Key K₁ is closed, key K₂ is also closed, now plug is inserted between terminal 1 and 2

of two way key. Now find out potential difference across resistance R by potentiometer. If I current flowing through secondary circuit, potential difference V across the ends of resistance R and the balancing length is l₁ for V.
By the principle of potentiometer
V = kl₁ ………….. (1)
but V = IR (by Ohm’s law)
IR = kl₁ ………… (2)
Eliminate the plug between 1 and 2 and insert it between 2 and 3. Now R and r connect in series. Current I remains same in this circuit also now the potential difference across the ends of the (R + r) is V₁ and balancing length for it is l₂.

Putting the values l₁, l₂ and R, then internal resistance of primary cell (r) can be determined.

RBSE Class 12 Physics Chapter 6 Long Answer Type Questions

Question 1.
Give statements of Kirchhoff s junction and loop laws. Find out the balance condition of Wheatstone’s bridge with the help of these laws.
Answer:
“In an electric circuit, the algebraic sum of current at any junction point must be zero.”
i. e., ΣI = 0
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kirchhoff s First Law or Junction Law.

Kirchhoff’s Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., ΣV = 0 …………… (1)
The algebraic sum of the emfs in any loop of a circuit is equal to the sum of the voltages across the resistances.
ΣE = ΣIR ………….. (2)
Sign convention for applying Loop Rule
1. The product of current and resistance (IR) is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance (IR) is taken as negative if the resistor is traversed in the opposite direction of assumed current.

In loop we can move from A to B then the product of IR take as positive V_AB = +IR. [In the direction of current]
In a loop we can move from B to A then the product of IR take as negative.
∴ V_BA = -IR [Opposite direction of current]

2. When we move from B to A then E is taken positive (+). When we move from AtoB, then Eis taken negative (-).

Loop law can be understood by the figure 6.4.
In the given figure for point d by the Junction Law, current flowing through the resistance R% will be
I₃ = I₁ + I₂ …………… (3)
In the loop adcba by the Kirchhoff s Loop Law.
I₁R₁ – I₂R₂ = ε₁ – ε₂
or I₂R₂ = I₂R₂ = ε₂ – ε₁ ………… (4)
In the loop defcd by the Loop Law
I₃R₃ + I₂R₂ = ε₂ …………….. (5)
By solving equations (3), (4) and (5) we can find out electric currents in different branches as well as we can find potential differences across the resistances.

Construction : The construction of Wheatstone’s bridge is shown in following diagram (figure 6.5) in which a quadrilateral ABCD is prepared by

connecting four resistances P, Q, R and S. A cell is connected between the points A and C through key K₁ which is known as battery key and a galvanometer between the point B and D through another key K₂ which is known as galvanometer key. If on pressing the key K₁ and K₂, there is no deflection in galvanometer, then the bridge is said to be balanced condition. In this situation the relation between all the resistances is given by :
[latex]\frac{P}{Q}=\frac{R}{S}[/latex]

(iii) When V_B = V_D , then no current passes through galvanometer i.e.. galvanometer shows no deflection because now I _g = 0. This situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called ‘Bridge circuit’ Thus in balanced condition.

Therefore, “it is clear that in balanced situation of the bridge, ratio of resistances of any two corresponding arms of quadrilateral ABCD is equal to ratio of two remaining corresponding arms”.
From equation (3).
S = [latex]\frac{Q}{P} \times R[/latex]
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below :
(i) P and Q are called ratio arms.
(ii) R is called variable resistance arm.
(iii) S is called unknown resistance arm.

Question 2.
What is the meter bridge? On which principle does it work? Explaining its construction obtain the expression for determining unknown resistance with the help of meter bridge. Give necessary diagram also.
Answer:
Meter Bridge
It is the simplest practical application of the Wheatstone bridge that is used to measure an unknown resistance.

Working : After taking out a suitable resistance R from the resistance box, we touches the jockey at point A and C and observe opposite deflection in galvanometer. If galvanometer shows deflection in one direction then we change the value of known resistance until galvanometer shows opposite deflection.

Copper strips have their own resistances but in the calculation of meter bridge formula we neglect this resistance so that our results are errorful. For minimising this error, at time of performing experiment we should calculate unknown resistance by interchanging the position of unknown resistance and known resistance. In this process we take average value of obesrvations by which error can be minimised.
The resistances of endpoints affect the sensitivity of the meter bridge. To nulify, this effect Carey-Foster s bridge is used.
Current should not flow for long time through the wire otherwise wire will heat up and the resistance of wire will be changed.
Jockey should not slide tightly on resistance wire otherwise it’s cross-section area will change. That is why the resistance of wire will be changed.

Question 3.
What do you understand by internal resistance of a cell? Establish the relation for determining the internal resistance of a cell by potentiometer.
Answer.
Determination of Internal Resistance of Primary Cell
To determine the internal resistance of a cell, we prepare a circuit as shown in the figure 6.15. Primary circuit is prepared by connecting a battery, a key K and a rheostat in series with potentiometer wire AB. In secondary circuit the cell E of unknown internal resistance r and a resistance box are connected with outer terminals of a two way key. The middle terminal of this key is connected with galvanometer and second terminal of galvanometer to jockey which can move on potentiometer wire.The whole practical is conducted in following steps :

(i) First of all the key K₁ is closed then the cell E is only is working condition and no current flow through the external resistance in a secondary circuit. So, potentiometer measured the emf of cell. Let the balance point is obtained at the length l₁ from the point A If k be the potential gradient.
∴ emf of cell = Potential gradient × Length
E = kl₁ …………… (1)
(ii) Now key K₂ also is pressed and some appropriate resistance R is applied by resistance box. In this situation the balance point is obtained at length of l₂ corresponding to terminal voltage V, therefore
V = kl₂
Since the internal resistance is obtained by following relation :

Substituting the values of l₁, l₂ and R we can get the internal resistance (r) of the cell.

Question 4.
What is meaning of calibration of voltmeter and ammeter. Explain the method of calibration of voltmeter and ammeter with the help of potentiometer. Make the required circuit diagram and draw the calibration curve.
Answer:
Calibration of Voltmeter
Observations measured from voltmeter is errorful due to different reasons :
(i) In built mechanical error, (ii) Parallex error,
(iii) Unsymmetricity in the spring of voltmeter etc. Potentiometer can increase precise value of potential difference.

Checking the errors of observations of voltmeter by potentiometer is known as Calibration of Voltmeter.

In figure 6.17 necessary circuit diagram is shown for calibration of voltmeter. Primary circuit is usually connected as section before.

Calibration of Ammeter
Checking of the errors of observations of electric current by ammeter by potentiometer is known as calibration of ammeter.

Circuit : For calibration of ammeter necessary circuit diagram is shown in figure 6.19.

Primary circuit is usually connected. In secondary circuit positive terminal of standard cell ε_s is connected at point A (high potential end) while negative terminal is connected with the terminal 1 of two-way key. In this circuit other cell ε₁ rheostat Rh₂, key K₂ and 1 Ω coil and ammeter are connected in series as shown in figure.

Working : Firstly we complete the primary circuit and insert the plug between terminal 1 and 2 and by the help of jockey find balancing length l₀ for the standard cell ε_s
ε_s = kl₀
or k = [latex]\frac{\varepsilon_{s}}{l_{0}}[/latex] ………………… (1)
k is potential gradient. By the standardisation of cell find out the k. Now eliminating the plug between the terminals 1 and 2 and inserting it between 2 and 3. Now key K₂ is closed. By the help of Rh₂ necessary current pass through the 1 Ω coil and takes the deflection in ammeter, which is I. It is errorful observation.

By Ohm’s law, current flowing through 1 Ω resistance is equal to the potential difference across the ends of the 1 Ω coil. Obtain balancing length l₂ for the potential difference across the ends of the 1 Ω coil. Obtain balancing length l₂ for the potential difference V’ which is across the ends of the coil.
V’ = kl₂ but V’ = I’R, R = 1 Ω
I’ = kl₂
I’ = [latex]\left(\frac{\varepsilon_{s}}{l_{0}}\right)[/latex] l₂ …(2)
I’ is the real value of current which is measured by the potentiometer. Now obtain the error in current observed by ammeter as ΔI.
ΔI = I – I’
For different readings of ammeter obtain corresponding real current by potentiometer then the corresponding error will be I – I’. Plotting the curve between I – I’ and I which is called calibrated curve of ammeter. Zig-zag curve is obtained as shown in figure 6.20.
We can find real reading by calibrated curve

Question 5.
What is potentiometer? Explain its principle. Describe the method of determining a small resistance by potentiometer. Make necessary diagram also.
Answer:
Principle of Potentiometer
Potentiometer : Potentiometer is a apparatus with help of which the e.m.f. of a cell and potential difference can be measured accurately.

Principle : Suppose AB is a resistance of uniform area of cross section and length LAB- If a standard cell of emf E and negligible internal resistance is connected across the ends of wire A and B with the help of connection wires of negligible resistance, then the potential difference across the wire will be E. Therefore potential gradient.
k = [latex]\frac{E}{L_{A B}}[/latex] ……………. (1)
If we consider a point C at a distance l from A, then
V_AC = kl ………….. (2)
Now if a cell of unknown emf E’ is connected between A and C as shown in the figure, then the deflection in galvanometer will depend on the fact that whether V_AC > E’ or E’ > V_AC. When V_AC = E’, then galvanometer will show no deflection. This fact can be understood well with the help of following equivalent circuit between A and C. This is null position galvanometer.

E’ = V_AC
or E’ = kl ………………… (3)
This is the principle of potentiometer.
If E’ > E, then galvanometer will show the deflection in one direction only. In this situation the measurement of E’ is not possible.
If the resistance of potentiometer wire is RAB and the current is I then from the ohm’s law
E = IR_AB ……………… (4)
From the equation (1)
k = [latex]\frac{I R_{A B}}{L_{A B}}[/latex]
Let [latex]\frac{R_{A B}}{L_{A B}}[/latex] = Density of wire is represented by the ρ
∴ k = ρI ……………… (5)
The circuit diagram for potential gradient

The potential gradient

Value of potential gradient under different conditions : .
Case-1 : When only the battery of negligible internal resistance is connected to potentiometer.
∴ r = 0 and R_h = 0
From the equation (8)

Case-2: When the battery of internal resistance (r) is connected to the potentiometer

Case-3: When a some other resistance or rheostat of resistance (Rh) is connected in series with the battery in the circuit of potentiometer. Then the potential gradient is calculated by equation (8).
k = [latex]=\left(\frac{E}{R_{A B}+r+R_{h}}\right) \times \frac{R_{A B}}{L_{A B}}[/latex]

Factors effecting the value of potential gradient:
From the equation (5)

Measurement of Small Resistance
In figure 6.16 necessary circuit is shown.
Primary circuit is as usual. In secondary circuit, unknown small resistance r, known resistance R, rheostat Rh, key K₂ and emf ε’ all are connected in series. Higher potential end of R, connects at point A which has higher potential, while lower potential end of R and r connects with the ends 1 and 3 of two way key. Terminal 2 of two way key connects with galvanometer end. Galvanometer also connects with jockey.

RBSE Class 12 Physics Chapter 6 Numerical Questions

Question 1.
Find out equivalent resistance between a and b.

Solution:
Arranging the circuit :

This circuit follows a Wheatstone bridge so arm cd does not work.

Adding resistances of path acb in series
R₂ = R + R = 2R
Adding resistances of path adb in series
R₂ = R + R = 2R
Adding R₁ and R₂ in parallel
[latex]R^{\prime}=\frac{R_{1} \times R_{2}}{R_{1}+R_{2}}=\frac{2 R \times 2 R}{4 R}=R \Omega[/latex]

Question 2.
Meter bridge is in balanced state shown in figure. Resistance of per unit length of the meter bridge wire is 1 Ω/cm. Find unknown resistance X and find electric current flowing through it.

Solution:
According to condition of metre bridge

Question 3.
Resistances of four sides of Wheatstone bridge is AB = 100 Ω, BC = 10 Ω, CD = 5 Ω and DA = 60 Ω. Galvanometer of 15 Ω resistance is connected between BD. Find the current flowing through the galvanometer. Potential difference between A and C is 10 V.

Solution:
For closed mesh ABDA

Question 4.
What should be the resistance R shown in figure so that galvanometer shows zero deflection?

Solution:

This circuit represents Wheatstone bridge.
∴ Resistance in arm AS will not work.
∴ Current flowing through R = 0
Hence, for the any value of resistance R, the current flowing by the ammeter will be zero.

Question 5.
Length of potentiometer wire is L. In primary circuit battery of 2.5 V and 10 ft resistance is combined in series. Balance length
is [latex]\frac{L}{2}[/latex] for emf 1.0 Ω. If the resistance of primary circuit becomes double then what is the new balance length?
Solution:
Length of wire = L Potential (V) = 2.5 (V)
Let a resistance of R₁ ohm is placed in series with battery, then current

When we double the resistance R₁, connected in series with potentiometer then total resistance (R’) = R + 2R₁
R’ = 10 + 5 = 15 Ω
In this condition if current in potentiometer is i, then

Question 6.
The Wheatstone bridge is shown in figure. What should be the value of X so that Wheatstone bridge will be balanced?

Solution:
Equivalent resistance of X Ω and 100 Ω in parallel

Question 7.
Standard cell of 1.1 V emf is balanced at 0.88 m length of the potentiometer wire. Potential difference across 1 Ω resistance is balanced at 0.20 m length of the wire of potentiometer. If the reading of ammeter is 0.20 A, then find out the error in the measurement of ammeter.
Solution:
From principle of potentiometer,

Reading in ammeter (I) = 0.20 amp
Error in the measurement of current (ΔI) = I – I’
= 0.20 – 0.25 = -0.05 amp

Question 8.
In the experiment of potentiometer, for 1.25 V emf balance length is obtained at 4.25 m. For an another cell it is obtained at 6.80 m. Find out the emf of another cell.
Solution:
From potentiometer condition

Question 9.
The resistance of 10 m potentiometer wire is 1 Ω/m. Battery accumulator of 2.2 V and negligible internal resistance and high resistance connecting in series with the wire. For achieving 2.2 mV/m potential gradient how much high resistance will have to take?
Solution:
Length of potentiometer wire
L_AB = 10m
Density of resistance of wire = 1 Ω/m
Resistance of potentiometer

Question 10.
In potentiometer experiment two cell of emf of ε₁ and ε₂ combined in series (ε₁> ε₂) for it balance length obtained at 60 cm. If the polarity of the cell having lower emf is then the balance length achieves at 20 cm for the combination. Find out the ratio of emf’s ε₁ and ε₂
Solution:
Let gradient of potentiometer is k

RBSE Solutions for Class 12 Physics

RBSE Solutions for Class 12 Physics Chapter 3 Electric Potential

February 6, 2021 by Prasanna Leave a Comment

You can Download RBSE Solutions for Class 12 Physics Chapter 3 Electric Potential Guide Pdf, help you to revise the complete Syllabus and score more marks in your examinations.

Rajasthan Board RBSE Class 12 Physics Chapter 3 Electric Potential

RBSE Class 12 Physics Chapter 3 Text Book Exercise with Answers

RBSE Class 12 Physics Chapter 3 Multiple Choice Type Questions

RBSE Solutions For Class 12 Physics Chapter 3 Question 1.
Electric potential at any point due to a point charge is 300 volt and intensity of electric field is 50 V/m. The distance of charge of that point is.
(a) 9 m
(b) 15 m
(c) 6 m
(d) 3 m
Answer:
(c) 6 m
Electric field intensity E = [latex]\frac{V}{d}[/latex]

RBSE Solutions For Class 12 Physics Question 2.
The charges are placed at corners of a square as shown in figure. Let the electric field and electric potential at its centre be [latex]\vec{E}[/latex] and V respectively. If the charges at A and B are displaced with respect to the charges at C and D, then :
(a) V changes and [latex]\vec{E}[/latex] remains unchanged
(b) Both [latex]\vec{E}[/latex] and V changes
(c) Both [latex]\vec{E}[/latex] and V remains unchanged
(d) [latex]\vec{E}[/latex] changes and V remains unchanged
Answer:
(d) [latex]\vec{E}[/latex] changes and V remains unchanged
Electric field [latex]\vec{E}[/latex] is changed while electric potential (V) remains unchanged

RBSE Solution Class 12 Physics 3 Electric Potential Question 3.
The potential at a point in an electric field is 200 V, then the work done in bringing an electron to that point is :
(a) -3.2 × 10^-17J
(b) 200 J
(c) -200 J
(d) 100 J
Answer:
(a) -3.2 × 10^-17J
Work done W = qv
W = -1.6 × 10^-19 × 200
⇒ W = -3.2 × 10^-17 J

RBSE Solutions Physics Class 12 Ch 3 Electric Potential Question 4.
Two charged conducting sphere of radii r₁ and r₂ are at same potential, then the ratio of the surface charge densities will be :

Answer:
(a)

Class 12 Physics RBSE Solutions 3 Electric Potential Question 5.
A charge of 10 μC is situated at origin point on an X-Y coordinates. The potential difference between the points (α, 0) and [latex]\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)[/latex] will be (in volts)
(a) 9 × 10⁴
(b) Zero
(c) [latex]\frac { 9\times { 10 }^{ 4 } }{ \alpha } [/latex]
(d) [latex]\frac{9 \times 10^{4}}{\sqrt{2}}[/latex]
Answer:
(b) Zero

RBSE Solution Class 12th Physics 3 Electric Potential Question 6.
The electric potential at the surface of a charged conducting hollow sphere of radius 2 m is 500 volt. The potential at the distance of 1.5 m from the centre is :
(a) 375 V
(b) 250 V
(c) zero
(d) 500V
Answer:
(d) 500V
Electric potential inside and on surface of charged conducting sphere will be same.
Electric potential inside the hollow spherical conductor is equal to the potential at the surface of the conductor.
∴ V = 500 Volt

Physics Class 12 RBSE Solutions 3 Electric Potential Question 7.
The kinetic energy of a α-particle to move from the points where potential have 70 V to 50 V from rest is :
(a) 20 eV
(b) 40 eV
(c) 20 MeV
(d) 40 MeV
Answer:
(b) 40 eV
Work done W = qV
W = (2e)[V_B – V_A]
= (2e) [20] volt or
W = 40 eV

RBSE Solution Physics Class 12 Ch 3 Electric Potential Question 8.
The electric field intensity is zero at the place, then the variation of potential with distance in that field will be :
(a) V ∝ [latex]\frac{1}{r}[/latex]
(b) V ∝ [latex]\frac{1}{r^{2}}[/latex]
(c) V = zero
(d) V = constant
Answer:
(d) V = constant
Electric field intensity

12th Physics RBSE Solutions 3 Electric Potential Question 9.
Two charged conducting spheres of equal surface charge densities have radii R₁ and R₂ If the potential at the centre is V₁ and V₂ respectively, then [latex]\frac{V_{1}}{V_{2}}[/latex] is :

Answer:
(a)

RBSE Solutions For Class 12th Physics 3 Electric Potential Question 10.
The function of potential in an electric field is given by V = -5x + 3y + [latex]\sqrt{15 z}[/latex]. The electric field intensity at point (x, y, z) in S.I. system will be :
(a) [latex]3 \sqrt{2}[/latex]
(b) [latex]4 \sqrt{2}[/latex]
(c) [latex]5 \sqrt{2}[/latex]
(d) 7
Answer:
(d) 7

RBSE Solution Of Class 12th Physics 3 Electric Potential Question 11.
A unit charge is rotated around a charge of q in a circular path at the distance of r. Then, work done will be :
(a) zero
(b) [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} J[/latex]
(c) 2 πr J
(d) 2πrq J
Answer:
(a) zero
Circular path will act like a equipotential surface. So work done in moving unit charge will be zero.

RBSE Solutions For Class 12 Physics In Hindi 3 Electric Potential Question 12.
The electric potential energy of a system in bringing an electron towards another electron is :
(a) increases
(b) decreases
(c) remains the same
(d) becomes zero
Answer:
(a) increases
On bringing similar charges to near the electric potential energy increases.

RBSE Solutions 12th Physics 3 Electric Potential Question 13.
1000 small water drops, each of radius r and charge q, coalesce to form a single big drop. The potential of big drop, how much times the small drop is.:
(a) 1000
(b) 100
(c) 10
(d) 1
Answer:
(b) 100
V_large = n^2/3V_small
V_large = (10³)^2/3V_small
[∵ n = 1000, n = 10³]
∴ V_large = 10²V_small = 100V_small

Solution of Physics Class 12 RBSE 3 Electric Potential Question 14.
As shown in figure work done in bringing a coulomb charge from P to Q, due to the arranged charges will be :

(a) 10
(b) 5
(c) ∞
(d) zero
Answer:
(d) zero
W_PQ = qV₀ [V_Q – V_P]
W_PQ = 1 [0 – 0]
W_PQ = 0

Chapter 3 Physics Class 12 RBSE 3 Electric Potential Question 15.
64 identical mercury ball (each of potential 10 V) combine to form a big ball, then the potential at the surface of the big ball will be :
(a) 80 V
(b) 160 V
(c) 640 V
(d) 320 V
Answer:
(b) 160 V
V_large = n^2/3V_small
V_large = (64)^2/3 × 10
V_large = (4³)^2/3 × 10
V_large = 160 Volt

RBSE Class 12 Physics Chapter 3 Very Short Answer Type Questions

Class 12th Physics RBSE Solutions 3 Electric Potential Question 1.
Is the electric potential a scalar or vector quantity?
Answer:
Scalar quantity.

Physics Chapter 3 Class 12 RBSE 3 Electric Potential Question 2.
Define electric potential.
Answer:
The electric potential at a point is equal to the work done by external force in bringing the unit positive charge from infinity to a point inside the electric field without changing the kinetic energy.

12th Physics Chapter 3 RBSE 3 Electric Potential Question 3.
Can two equipotential surfaces intersect each other?
Answer:
No two equipotential surfaces or places can not intersect each other because then there would be two values of electric potential at the intersection point, which is impossible.

Physics Ch 3 Class 12 RBSE Question 4.
What would be the potential at infinity due to an isolated charge?
Answer:
V = [latex]\frac{k Q}{r}[/latex], r → ∞ ∴ V = 0

Electric Potential In Hindi Class 12 RBSE Question 5.
Can electric potential be zero at a point in a vacuum, the electric field at that point is not zero. Give example.
Answer:
Yes, it is possible.
Example :

Between line joining electric dipole
On equatorial line of an electric dipole.

Physics Class 12 Chapter 3 RBSE Question 6.
Can electric field at a point be zero, but the electric potential is non-zero. Give example.
Answer:
Yes, it is possible.
Example :

Inside uncharged spherical shell and charged conductor electric field is zero but electric potential cannot zero.
Between the line joining two similar charges of equal magnitude.

Chapter 3 Class 12 Physics RBSE Question 7.
How much work is done between the points at the distance 10 cm on an equipotential surface in taking the charge of 200 μC?
Answer:
W = q₀[V_B – V_A]
On equipotential surface,
V_B = V_A ⇒ W = 0

Question 8.
What would be the shape of the equipotential surfaces due to :
(a) A point charge
(b) A uniform electric field
Answer:
(a) Equipotential surface due to point charge is spherical.
(b)

Question 9.
What is the electric potential energy when an electric dipole is placed parallel to the electric field?
Answer:
U = -pE cosθ
θ = 0°, cos(0) = 1
U = -pE

Question 10.
The potential on charging the surface of a conducting sphere of radius 10 cm is 15 V. What is the potential at its centre?
Answer:
On surface and inside surface the electric potential will be same. Therefore at centre the electric potential will be 15 volt.

Question 11.
The potential on a uniformly charged non-conducting sphere of radius 5 cm is 10 V. What is the potential at its centre?
Answer:

Question 12.
The electric potential at a point (x, y, z) (all in metres) in vacuum is V = 2x² volt, Calculate electric field intensity at (1 m, 2 m, 3 m).
Answer:

Question 13.
Write the formula for potential energy for a system of two point charges.
Answer:

Question 14.
Write the formula for potential energy for a System of three point charges.
Answer:

Question 15.
Write the unit of potential gradient.
Answer:
Volt/metre.

Question 16.
How much work is done in taking an electron between two points that have potential difference of 20 V?
Answer:
W = q₀V = 1.6 × 10^-19 × 20 = 32 × 10^-19 J

Question 17.
The electric potential at a point in vacuum due to another point charge is 10 V. If a material of dielectric constant 2 is placed
around the point, then what electric potential?
Answer:

Question 18.
What is the work done in rotating an electric dipole in an external uniform electric field [latex]\vec{E}[/latex] from 0° to 180°?
Answer:
W = pE(cosθ₁ – cosθ₂)
θ₁ = 0°, θ₂ = 180°
W = pE( cosθ°-cos180°)
= pE [1 – (1)]
⇒ W = 2pE Joule

Question 19.
What is the electric potential of earth?
Answer:
Zero.

Question 20.
If potential function, V = (4x+ 3y) V, then what is the electric field intensity at point (2, 1) (all in metres)?
Answer:

RBSE Class 12 Physics Chapter 3 Short Answer Type Questions

Question 1.
What is electric potential? Write its formula and unit.
Answer:
Electric Potential
In fig. 3.1 it is shown that in electric field due to a configuration of charges, the work done in carrying a test charge (q₀) from point A to B, depends only upon the distance between A and B not on the path adopted.

“Electric potential is the factor which decides the direction of flow of electric charge.”
It is scalar quantity and it is denoted by V. As liquids flow from higher gravitation level to the lower level; heat flows from higher temperature to the lower temperature, similarly charge (positive) flows from higher electric potential to the lower potential. Negative charge flows from lower electric potential to the higher potential.

If V_A and V_B be the electric potentials at points A and B respectively, then the potential difference between points A and B is given below,
V_B – V_A = [latex]\frac{W_{A B}}{q_{0}}[/latex] ……………… (1)
Where, W_AB is the work done in carrying test charge + q₀ from A to B.
Here we assume that in this process, the test charge does not disturb the source of field, i.e., all the charges remain at their position. If for a displacement, the change in potential energy is U_B — U_A, then the potential difference between the points A and B can be given as

If q₀ = +1C, then V_B – V_A = W_AB

Question 2
Prove that the potential inside a charged spherical shell is equal to that on its surface.
Answer:
Inside the charged spherical shell (r < R)
According to definition of potential

Dependence of electric field upon distance is different e.g. outside spherical shell
E = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \hat{r}[/latex]
and inside the shell E = 0.
Therefore the integration can be written as under,

In other words, E = 0
By the relation of electric field and potential

Question 3.
What is equipotential surface? Draw the equipotential surface due to point charge.
Answer:
Equipotential Surface
The surface in an electric field where the value of electric potential is same at all the points on the surface is called equipotential surface. The potential difference between two points in an equipotential surface is zero. Thus, work done in moving the charge from one point to another in an equipotential surface is zero. Since the work done is zero when the direction of force (electric field) and displacement is normal. Thus, electric field is perpendicular to the plane in an equipotential surface.

It is to be noted for equipotential surface :
1. For a uniform electric field [latex]\vec{E}[/latex], the equipotential surface is normal to its field lines. According to figure 3.8, I (a) II and III are equipotential surfaces.

2. For an isolated point charge :
Potential at a distance r due to point charge +q,

If we draw a sphere of radius r surrounding the +q charge. Then, potential will be same at all points on the sphere i.e., equipotential surfaces are spherical whose charge is at the centre. For a positive point charge, as the radius of the spherical surface increases, the potential of spherical surface decreases.

3. Equipotential surface for an electric dipole : The equipotential surface for an electric dipole are shown dotted in figure 3.8 (c).

4. Equipotential surfaces for a pair of similar point charges : The equipotential surface is as shown in figure 3.8 (d)

Question 4.
Determine the electric potential energy for a system of three point charges.
Answer:
Electric Potential Energy of a System of Charges
Electric potential energy of a system of charges is equal to amount of work done in forming the system of charges by bringing them at their particular positions from infinity without any acceleration and against the electrostatic force. It is denoted by U.
U = W = qV(r)
(a) Electric potential energy of system of two charges :

Suppose two charges + q₁ and + q₂ are situated at a distance r. We have to find out the electrical potential energy of this system.

When the charge q₁ is brought from infinity to in its position, no work is done because there is no other charge to repel or attract it. The electric potential due to this charge q₁ at a distance r i.e., in the position of charge q₂,
V₁ = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r}[/latex]
Therefore the work done in bringing the charge q₂ from infinity to its own position i.e., the electric potential energy of the system.
W = U = V₁q₂
or U = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}[/latex] ……………. (1)
If both the charges are of same nature, the potential energy will be positive and for unlike charges it will be negative. Therefore during the process of evaluation of potential energy, the value of charge should be used with its sign.

(b) Electric potential energy of a system of three charges :

Let us first of all consider a system of three point charges q₁, q₂ and q₃. Such that q₂ and q₃ are initially at infinite distance from the charge (q₁) [In figure 3.18(a)], Work done to bringing the charge q₂ from infinity to point B,
W₁₂ = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{\overrightarrow{r_{12}}}[/latex] …………… (2)

Work done bringing charge q₃ (In the presence of charge q₁ and q₂) from infinity to point C

Now the total work done to bringing all the three charges from infinity to their respective position is given by

This work done is stored in the form of potential energy
∴ U = W = potential energy of three system of

(c) Electric potential energy due to four system of charges :
Suppose there are four charges in a system of charges, situated as shown in figure (3.18 (b)).

The total energy of this system can be obtained by adding the potential energies of each possible pair.

Similarly, the electrical potential energy of the system of n charges can be obtained. We can write the equivalent expression as under,

To obtain this sum, we have to use each pair of charge only once. Therefore the general formula for electrical potential energy will be as under

Here we have to multiply the expression by [latex]\frac{1}{2}[/latex] because in this expression each pair comes two times.

Question 5.
Why is static electric potential of the whole volume of the charged conductor equivalent to that on its surface?
Answer:
Electric field inside whole volume of charged conductor is zero.
∵ E = [latex]-\frac{d V}{d r}[/latex]
E = 0 ⇒ V = constant
(Electric potential is uniform in whole volume)
Inside conductor the work done in bringing any test charge from inside point to surface will be zero.
W = q₀(V_B – V_A)
V_B – V_A = 0
∴ V_A = V_B
Potential of both points will be equal.

Question 6.
Establish the relation between electric potential and electric field intensity.
Answer:
Relation between Electric Field and Electric Potential
We have seen the relation between electric field and electric potential. With the help of this relation, we can evaluate potential difference between two points in known electric field.

In this section, our aim is to determine electric field for known potential function V.

For an electric field [latex]\vec{E}[/latex], the small element [latex]\vec{d} l[/latex] of displacement can be written in the form of derivative.
dV = [latex]-\vec{E} \cdot \overrightarrow{\mathrm{d} l}[/latex] = -Edl cosθ ………….. (1)
Where θ is the angle between electric field ([latex]\vec{E}[/latex]) and length ([latex]\vec{d} l[/latex])
[latex]-\frac{d V}{d l}[/latex] = Ecosθ

The term [latex]-\frac{d V}{d l}[/latex] represents decrease in potential with distance. It is clear from the above equation, that if the angle between [latex]\vec{d} l[/latex] and [latex]\vec{E}[/latex], θ = 0°, then the loss in potential with distance is maximum. In general form, the term is a scalar quantity but its maximum value [latex]\left(\frac{d V}{d l}\right)_{\max }[/latex] in a particular direction (θ = 0) i. e.. in the direction of electric field. Thus, the maximum rate of loss in potential can be considered to be a vector which is along [latex]\vec{E}[/latex]. In mathematical language, it is called potential gradient and it is represented as Grad V.

For an equipotential surface, the direction of Grad V is perpendicular to the surface. It can be understood with the help of figure 3.9. The equipotential surfaces are shown as S₁and S₂ on which the potential is V and V-dV respectively. The loss in potential in moving from point A on surface S₁ to points B and C is dV, but the rate of change in potential with distance is [latex]\frac{d V}{A B}[/latex] and [latex]\frac{d V}{A C}[/latex] respectivelv,
which are different from each other.
Since, AB < AC
Thus, [latex]\frac{d V}{A B}>\frac{d V}{A C}[/latex]
Now, surface AB is perpendicular to the surface. Thus, the rate of loss in potential is maximum in the direction normal to the surface.
Equation (1) can be written as :

where, E₁ = Ecos₁, component of [latex]\vec{E}[/latex] along [latex]\vec{d} l[/latex]. Please note that we have used the symbols of partial derivatives which tells that the above equation in a definite axis (here for l axis) sums up the change in potential and component of E along that axis. If we assume l axis to be X, Y and Z axes, then the components of [latex]\vec{E}[/latex] will be

This operator is called Del operator. With the help of equation (5), we can find out [latex]\vec{E}[/latex] when the potential function V(x, y, z) is known.

If the potential function is a function of radius r. Then radial electric field
E_r = [latex]-\frac{d V}{d r}[/latex] ……………… (7)

Question 7.
Obtain the formula for rotating the electric dipole in a uniform electric field.
Answer.
Work Done in Rotating of an Electric Dipole In a Uniform Electric Field
On placing an electric dipole in an uniform electric field, a torque is produced on the dipole which tries to align it in the direction of the electric field. Ultimately the dipole gets aligned with the field and attains equilibrium. Thus it is obvious that work will be done in rotating the dipole in uniform electric field.

The torque acting on an electric dipole when it is placed in a uniform electric field [latex]\vec{E}[/latex] making an angle θ with the field,
τ = pEsinθ ………(1)
Where p(= q × 2l) is electric dipole moment and E, the intensity of electric field.
Therefore work, done in giving an angular displacement dθ,
dW = τdθ
If the dipole is rotated from θ₁ orientation to θ₂,

Question 8.
Show that no work is done in taking a test charge from one point to another in an equipotential surface.
Answer.
The potential difference between any two points on an equipotential surface ΔV = 0,
Work done = q₀V = 0

Question 9.
What is meant by electric potential energy? Obtain the formula for potential energy for a system of charges.
Answer:
Electric Potential Energy of a System of Charges
Electric potential energy of a system of charges is equal to amount of work done in forming the system of charges by bringing them at their particular positions from infinity without any acceleration and against the electrostatic force. It is denoted by U.
U = W = qV(r)
(a) Electric potential energy of system of two charges :

Suppose two charges + q₁ and + q₂ are situated at a distance r. We have to find out the electrical potential energy of this system.

(c) Electric potential energy due to four system of charges :
Suppose there are four charges in a system of charges, situated as shown in figure (3.18 (b)).

Question 10.
Obtain the formula for electric potential energy of an electric dipole in an external electric field.
Answer:
Electric Potential Energy of an Electric Dipole in an External Electric Field
Electric potential energy of an electric dipole in an electric field is equal to work done in bringing an electric dipole from infinity to that field.

In figure 3.21, an electric dipole is brought from infinity to a uniform electric field [latex]\vec{E}[/latex] so that the dipole moment [latex]\vec{E}[/latex] is always along electric field [latex]\vec{E}[/latex]. The force on charge +q due to electric field [latex]\vec{E}[/latex], [latex]\vec{F}[/latex] = q [latex]\vec{E}[/latex] along the field and force on charge -q, [latex]\vec{F}[/latex] = -q [latex]\vec{E}[/latex] is opposite to the field. Thus, an external work is done for bringing charge q of dipole in electric field where as the electric field itself Work on charge -q. On coming from infinity to electric field, -q charge has to move 2a more distance that by q. Thus, the work done by the charge -q is more and negative. Thus, work done by the electric field,

W = Force on charge (-q) × Distance covered
W = -qE × 2a = -2 qaE
W = -pE ∵ p = 2qa
Thus, potential energy of electric dipole placed parallel to electric field [latex]\vec{E}[/latex].
U₁ = -pE ……………. (1)
Now, work done in rotating by angle θ from the parallel position of electric field [latex]\vec{E}[/latex].
U₂ = pE (1 – cosθ) ………….. (2)
Thus, potential energy of electric dipole at angle θ in electric field,
U = U₁+ U₂
U = -pE + pE (1 – cosθ)
U = -pE cosθ
In vector form, U = [latex]-\vec{p} \cdot \overrightarrow{\mathrm{E}}[/latex] …………. (3)
Equation (2) is the expression for potential energy of electric dipole.

Special Cases
(a) If the electric dipole is at 0° angle from the electric field, then the potential energy
U = -pE cosθ
U = -pE cos0°
i.e., U = -pE …………. (4)
(It is a condition of stable equilibrium)

(b) If electric dipole moment is perpendicular to the electric field, then the potential energy
U = -pEcos 90°
U = 0 ………….. (5)

(c) If the electric dipole moment is at 180° angle from the electric field, then the potential energy,
U = -pE cos 180°
U = pE …………….. (6)
(It is a condition of unstable equilibrium)

Case (a) is called the state of stable equilibrium becuase in this case, energy is minimum and case (c) is called the unstable equilibrium because in this case, the energy is maximum.

Question 11.
Write the formula for electric potential energy for two point charges q₁ and q₂ placed at displacement [latex]\overrightarrow{r_{1}}[/latex] and [latex]\overrightarrow{r_{2}}[/latex] respectively in a uniform external electric field.
Answer:
Electric Potential Energy of Charges in an External Electric Field
(i) Electric potential energy of a single charge in an external field : Let us consider an external electric field ([latex]\vec{E}[/latex]) have different values of electric potential at different points. Consider a point ‘ P’ at the distance (r) from the origin in this field having a electric potential [latex](V \vec{r})[/latex]. Then, work done to bringing a charge (q) from infinite to point ‘ P’ is given by q [latex](V \vec{r})[/latex]. This work is used as a potential energy of charge (q). Thus the potential energy of charge in external field.

(ii) Potential energy of a system of two charges in an external field : Let q₁ and q₂ be two charges placed at points P and Q having position vectors [latex]\overrightarrow{r_{1}}[/latex] and [latex]\overrightarrow{r_{2}}[/latex] respectively. In bringing (q₁) from infinity to point P, work done = [latex]q_{1} V \vec{r}[/latex], where [latex]V \vec{r}[/latex] is the potential at P due to external electric field.

To bringing (q₂) from infinity to point Q, work done = [latex]q_{2} V \overrightarrow{r_{2}}[/latex] where [latex]q_{2} V \overrightarrow{r_{2}}[/latex] is the potential at Q due to external electric field. If [latex]\left|\overrightarrow{r_{12}}\right|[/latex] is the distance between point P and point Q, then work done on q₂ against the
field due to q₁ = [latex]\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}[/latex]
∴ The total potential energy of two charges
[latex]U=q_{1}\left(V \vec{r}_{1}\right)+q_{2}\left(V \overrightarrow{r_{2}}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}[/latex]

Question 12.
Write any two properties of equipotential surface.
Answer:

No two equipotential surfaces or planes cut each other because then there would be two values of electric potential at the intersection point, which is not possible.
The surface of a conductor is always equipotential. Actually, the whole volume of a conductor is at equal potential.
Equipotential surfaces indicate regions of strong or weak electric fields. Using E = [latex]-\frac{d V}{d r}[/latex] or dr = [latex]-\frac{d V}{E}[/latex]
since dV (i.er potential difference) is constant on the equipotential surface, so dr ∝ [latex]\frac{1}{E}[/latex]
If E is strong, dr will be small and the separation of equipotential surfaces will small than the electric field is strong. The separation between equipotential surface is large than the electric field is weak.

Question 13.
Prove that the electric potential of point charge which has a dielectric medium around it, is 1/ε_r times lesser than that in vacuum.
Answer:
Electric potential at r distance from charge +q is
V = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}[/latex]
If a dielectric medium of er is surrounded by charge, then electric potential.

Question 14.
Prove that the electric potential at the centre of a uniformly charged non-conducting sphere is 1.5 times more than that on the surface.
Answer:
Electric Potential due to Charged Non-conducting Sphere
Consider a non-conducting sphere of radius R be charged by a charge q. The electric field intensity at the points outside the sphere, on the surface and inside the sphere is as follows :

The potential at a point in electric field intensity [latex]\vec{E}[/latex] is calculated by the help of following relation
[latex]-\int_{\infty}^{r} \vec{E} \cdot \vec{d} r[/latex]
The electric potential is calculated at the observation point in different conditions :
(a) At a point outside the non-conducting sphere

(b) At a point on the surface of non-conducting sphere
Putting r = R in equation (1),
V_S = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}[/latex] ……………. (2)

(c) At a point inside the non-conducting sphere (r < R)
For a point inside the non-conducting sphere, the electric field intensity E does not depend uniformly from infinity to r distance, i.e., from infinity to surface, it is inversely proportional to the square of the distance r and it is directly proportional to the distance from the surface to the centre r. Thus, integral is divided into two parts :

Thus, the electric potential at centre of a charged non-conducting sphere is 1.5 times that on its surface.
It is clear that the electric potential decreases with r² from centre to surface in a charged non-conducting sphere. After that, it decreases as per the law of r^-1 and becomes zero at infinity. It is shown in a graph in figure (3.16),

Question 15.
Two charges 10μC and 5μC are at lm apart. In order to make this distance 0.5 m, how much work would be done?
Answer:
q₁ = 10μC = 10 × 10^-6 C
q₂ = 5μC = 5 × 10^-6 C

Question 16.
Define electric potential difference. Distinguish between electric potential difference and electric potential.
Answer:
Electric Potential and Potential Difference
In fig. 3.1 it is shown that in electric field due to a configuration of charges, the work done in carrying a test charge (q₀) from point A to B, depends only upon the distance between A and B not on the path adopted.

“Electric potential is the factor which decides the direction of flow of electric charge.”
It is scalar quantity and it is denoted by V. As liquids flow from higher gravitation level to the lower level; heat flows from higher temperature to the lower temperature, similarly charge (positive) flows from higher electric potential to the lower potential. Negative charge flows from lower electric potential to the higher potential.

If V_A and V_B be the electric potentials at points A and B respectively, then the potential difference between points A and B is given below,
V_B – V_A = [latex]\frac{W_{A B}}{q_{0}}[/latex] ……………… (1)
Where, W_AB is the work done in carrying test charge + q₀ from A to B.
Here we assume that in this process, the test charge does not disturb the source of field, i.e., all the charges remain at their position. If for a displacement, the change in potential energy is U_B – U_A, then the potential difference between the points A and B can be given as

If q₀ = +1C, then V_B – V_A = W_AB

RBSE Class 12 Physics Chapter 3 Long Answer Type Questions

Question 1.
Deduce an expression for electric potential due to a charge at any point?
Answer:
Electric Potential due to a Point Charge
According to definition electric potential at a point is equal to the amount of work done in bringing unit positive test charge from infinity to that point in the electric field without any acceleration.

Suppose a point charge +q is placed at point O and potential due to this charge at point P situated at distance r from the charge, is to be determined. According to definition of potential, to find out the potential at P, we have to estimate the work done in bringing +1C charge from infinity to point P inside the electric field. For this purpose, let us consider another point P’ situated at distance x from O beyond P and + q₀ positive test charge is placed at P’, the electrostatic force acting on +q₀,
F = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q q_{0}}{x^{2}}[/latex]
Work done in giving a small displacement dx toward P,
dW = [latex]\vec{F} \cdot \vec{d} x[/latex] = Fdxcos 180° = Fdx(-1)
or dW = -F dx
Therefore work done in bring +q₀ charge from infinity to P

Where V → potential at P
If q is positive then potential V will also be positive and if q is negative, then potential will also be negative.”
From equation (1)
V ∝ [latex]\frac{1}{r}[/latex]
Electric field due to point charge
E = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}[/latex] ∴ E ∝ [latex]\frac{1}{r^{2}}[/latex]
If V and E both are plotted with distance r then the graphs will be as under figure [3.4(a)].
For negative charge V — r graph will be as under figure [3.4(b)].

Question 2.
Derive an expression for electric potential due to an electric dipole at any point (r, 0). Prove that the electric potential at point on axial is maximum and at equatorial is zero?
Answer:
Electric Potential due to an Electric Dipole at Any Point
An electric dipole AB is shown in figure 3.7. The charges at points A and B are -q and +q respectively and the distance between them is 2a. We have to determine the electric potential at point P from its centre O at distance r and the line OP makes an angle 0 with the axis of the dipole.

Now, to calculate (r₁ – r₂) and r₁r₂, we draw perpendiculars on AC and BD from points A and B on line OP respectively.

It is clear that potential due to dipole depends on r^-2 and not on r^-1 (as due to single charge).
Special Cases :

(i) If point P is on the axis when θ = 0°, cosθ = 1, then from equation (4),
V = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}[/latex]

(ii) If the point is normal, then θ = 90° , cosθ = 0, then from equation (4),
V = 0
In vector form, potential at point (r, θ) due to dipole,
[latex]V=\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{p} \vec{r}}{r^{3}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\vec{p} \vec{r}}{r^{2}}[/latex]

Question 3.
Derive an expression for potential due to charged spherical shell at following points (i) outer, (ii) at surface, (iii) inner point and also draw the graph for the variation of potential with the distance.
Answer:
Electric Potential due to Charged Spherical Shell
(a) At point P outside the shell (r > R)
When unit positive test charge is brought from infinity to point P, then the potential at P is equal to negative value of line integral of electric field between infinity and P.

Therefore, according to definition of potential

(b) On the surface of the charged spherical shell (r = R)
According to the definition of potential,

Where V is constant

(c) Inside the charged spherical shell (r < R)
According to the definition of potential

Dependence of electric field upon distance is different e.g. outside spherical shell
E = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \hat{r}[/latex]
and inside the shell E = 0.
Therefore the integration can be written as under,

Question 4.
Find the expression for electric potential due to an insulated charged sphere at outer point, surface and inner point?
Answer:
Electric Potential due to Charged Non-conducting Sphere
Consider a non-conducting sphere of radius R be charged by a charge q. The electric field intensity at the points outside the sphere, on the surface and inside the sphere is as follows :

The potential at a point in electric field intensity [latex]\vec{E}[/latex] is calculated by the help of following relation
[latex]-\int_{\infty}^{r} \vec{E} \cdot \vec{d} r[/latex]
The electric potential is calculated at the observation point in different conditions :
(a) At a point outside the non-conducting sphere

(b) At a point on the surface of non-conducting sphere
Putting r = R in equation (1),
V_S = [latex]\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}[/latex] ……………. (2)

Question 5.
Give the definition of electric potential energy. Find out the expression for electric potential energy of a dipole in uniform electric field. In which condition get the position of stable or unstable equilibrium
Answer:
Electric Potential Energy of an Electric Dipole in an External Electric Field
Electric potential energy of an electric dipole in an electric field is equal to work done in bringing an electric dipole from infinity to that field.

(b) If electric dipole moment is perpendicular to the electric field, then the potential energy
U = -pEcos 90°
U = 0 ………….. (5)

(c) If the electric dipole moment is at 180° angle from the electric field, then the potential energy,
U = -pE cos 180°
U = pE …………….. (6)
(It is a condition of unstable equilibrium)

Case (a) is called the state of stable equilibrium becuase in this case, energy is minimum and case (c) is called the unstable equilibrium because in this case, the energy is maximum.

RBSE Class 12 Physics Chapter 3 Numerical Questions

Question 1.
6 J of work is done to take a charge of 3 C between two points. Calculate the electric potential difference between these points.
Solution:
Potential difference between A and B

Question 2.
If the electric potential between two points A and B is 2 V and 4 V respectively. Then, how much work is done in bringing a point charge of 8 μC from point A to point B?
Solution:
q₀ = 8μC
q₀ = 8 × 10^-6C
Potential differnce between A and B
V_B – V_A = [latex]\frac{W_{A B}}{q_{0}}[/latex]
4 – 2 = [latex]\frac{W_{A B}}{8 \times 10^{-6}}[/latex]
W_AB = 2 × 8 × 10^-6
W_AB = 16 × 10^-6 J
W_AB = 1.6 × 10^-5 J

Question 3.
Four charges 100 μC, -50 μC, 20 μC and -60 μC are placed at the corners of a square of side [latex]\sqrt{2}[/latex] m. Calculate the electric potential at the centre of the square.
Solution:
According to figure
Diagonal of square

Question 4.
Two charges 3 × 10^-8C and -2 × 10^-8C are 15 cm apart. At what point on the line joining the charges, the electric potential is zero? Let the electric potential at infinity is zero.
Solution:
Potential at point P
V₁ + V₂ = 0

x = 0.45m = 45 cm

Question 5.
Each side of a square is 0.9 m long. The charges at its corners are -2 μC, + 3 μC, – 4 μC and +5μC. Determine the electric potential at the centre of the square.
Solution:

Question 6.
Charge of 5 μC is at each of the vertices of a regular hexagon of side 10 cm. Determine the electric potential at the centre of the regular hexagon.
Solution:
For Hexagonal shape
OA = OB – OC = OD = OE = OF = 10cm = 0.1 m
Potential at O

Question 7.
Charge of 2μC is placed at each corner of the square ABCD of side [latex]2 \sqrt{2}[/latex] cm. Calculate the electric potential at the centre of the square.
Solution:

Question 8.
The side of an equilateral triangle is 100 cm. The charges at its three corners are 1 μC, 2 μC and 3 μC respectively. Calculate the electric potential at a point (in the centre) equidistant from the three corners of the triangle.
Solution:
Given a = 100 cm
= 100 × 10^-2 m
Draw a ⊥OD and AB in ∆ ODB

Question 9.
The distance between the charges -1 μC and +1 μC of an electric dipole is 4 × 10^-14 m. Calculate the potential at an axial point 2 × 10^-6m from the centre of the dipole.
Solution:

Question 10.
(a) Calculate potential at a point at 9 cm distance due to charge 4 × 10^-7 C.
(b) Determine the work done in bringing a charge of 2 × 10^-9 C from infinity to that point. Does this work depend upon the path along which it was brought?
Solution:
(a) Potential at point P

(b) Work done in bringing charge 2 × 10^-9 C from infinity to point P
W = qV
= 2 10^-9 C × 4 × 10⁴ V
= 8 × 10⁵ J
This work done does not depend on path follows.

Question 11.
A charge of 30μC is at the reference point in the X-Y coordinate system. Calculate the potential difference between the points [latex]\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)[/latex] and (a, 0).

Solution:

V_B – V_A = [latex]\frac { kq }{ \alpha } -\frac { kq }{ \alpha } [/latex]
V_B – V_A = 0
Potential difference = 0

Question 12.
Three point charges -q, + q and +q are placed at the points (0, α), (0, 0) and (0, α) respectively in the x-y plane. Prove that the potential V at the distance r on the line making an angle 9 with the axis would be

Solution:
Here (0, α) and (0,- α) are position of charges – q and +q. They form an electric dipole. Electric potential is V₂ at point P due to this electric dipole.

Question 13.
How much work will be done in placing the charges +q, +2q, and +4q at the corners of the equilateral triangle of side α metre?

Solution:

Question 14.
Two charges 7 μC and -2μC are placed at (-9 cm, 0, 0) and (+9 cm, 0, 0) respectively. There is no external field on the system. Determine the electrostatic potential energy of this system.
(b) How much work will be done in keeping the two charges at infinite distance from each other?
Solution:
(a) q₁ = 7μC
q₁ = 7 × 10^-6 C
q₂ = -2pC
q₂ = -2 × 10^-6C
r = 18cm
r = 18 × 10^-2 m

(b) Potential energy at infinity
U₂ = 0
Potential energy at given position
U₁ = -0.7J
Work done in seperating the charges upto infinite
W = U₂ – U₁
⇒ W = 0 – (-0.7)
W = 0.7 J

Question 15.
The electric potential at a point (x, y) in an electric field is given by V = 6xy + y² – x² .
Calculate the electric field at that point.
Solution:

Question 16.
A charge of +15 μC is given to a hollow metallic sphere of radius 0.2 m. Determine;
(i) electric potential on the spherical surface
(ii) electric potential at the centre of the sphere
(iii) electric potential at 0.1 m from the centre of the sphere
(iv) electric potential at 0.3 m from the centre of the sphere.
Solution:
R = 0.2 m

V = 45 × 10⁴
V = 4.5 × 10⁵ volt

Question 17.
Three point charges +q, +2q and xq are placed at the corners of an equilateral triangle of side of length r. Calculate the value of x at which the potential energy of the system is zero?

Solution:
U = U₁₂ + U₁₃ + U₃

RBSE Solutions for Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise

February 5, 2021 by Fazal Leave a Comment

You can Download RBSE Solutions for Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise Guide Pdf, help you to revise the complete Syllabus and score more marks in your examinations.

Rajasthan Board RBSE Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise

Question 1.
, then find A.
Solution:

Question 2.
, then find (A – 2I) (A – 3I).
Solution:

Question 3.
, then find AB.
Solution:

Question 4.
, then find BA.
Solution:

Question 5.
, then find matrices A and B.
Solution:

Question 6.
, then find the value of x and y.
Solution:
On comparing,
x + 2 = -2 ∴ x = -4
– y – 2 = 5 ⇒ y = -7

Hence, x = -4, y = -7

Question 7.
Order of matrix A is 3 x 4 and B is a matrix, such that A^TB and AB^T defined, then write the order of B.
Solution:
∴ Order of A = 3 x 4
∴ Order of A^T = 4 x 3
But A^TB and AB^T is defined.
So, order of B is 3 x 4.

Question 8.
, is a symmetric matrix, then determine x.
Solution:
Given,

On comparing a_ij = a_ji
a₃₂ = a₂₃ ⇒ -x = 4
∴ x = -4

Question 9.
Construct a matrix of order 3 x 3, B = [b_ij], whose elements are b_ij= (i) (j).
Solution:
B₁₁ = 1 x 1 = 1
B₁₂ = 1 x 2 = 2
B₁₃ = 1 x 3 = 3

B₂₁ = 2 x 1 = 2
B₂₂ = 2 x 2 = 4
B₂₃ = 2 x 3 = 6

B₃₁ = 3 x 1 = 3
B₃₂ = 3 x 2 = 6
B₃₃ = 3 x 3 = 9

Question 10.

Solution:

Question 11.
Express matrix A as the sum of symmetric and skew-symmetric matrices, where
.
Solution:
Given,

Question 12.
then prove that :
(i) (A^T)^T = A
(ii) A + A^T is a symmetric matrix.
(iii) A – A^T is a skew-symmetric matrix.
(iv) AA^T and A^TA are symmetric matrix.
Solution:
(i) (A^T)^T = A

(ii) A + A^T is symmetric

So, A + A^T is symmetric matrix. Hence Proved.

So, A – A^T is skew symmetric matrix. Proved.

Here,
a₂₁ = a₁₂ = 0
a₃₁ = a₁₃ = 0
a₂₃ = a₃₂ = 6
So, AA^T is symmetric matrix.

Here,
a₁₂ = a₂₁ = 0
a₁₃ = a₃₁ = 0
a₃₂ = a₂₃ = 4
So, A^TA is symmetric matrix.

Question 13.
, and 3A – 2B + C is a null matrix, then determine matrix ‘C’.
Solution:

Question 14.
Construct a matrix B = [b_ij] of the order 2 x 3, whose elements are b_ij = (i +2j)²/2
Solution:
Given, B = [b_ij] whose elements are

Question 15.
, then find the element of 1^st row of ABC.
Solution:

So, element of 1^st row is 8.

Question 16.
, then find AA^T.
Solution:
Given,

Question 17.
, then find x.
Solution:

Question 18.
, then prove

Solution:
Given,

= (bc – ad)I₂ = R.H.S.
Hence Proved.

Question 19.
, then find the matrix form of the following (aA + bB) (aA – bB).
Solution:
Givn,

Question 20.
, then prove that (A – B)² ≠ A² – 2AB + B².
Solution:
Given,

From (i) and (ii),
(A – B)² + A² – 2AB + B²
Hence Proved.

Question 21.
, then find k, where A² = kA – 2IA₂ .
Solution:
Given,

On comparing corresponding element
From 3k – 2 = 1
3k = 3 ⇒ k = [latex]\frac { 3 }{ 3 }[/latex] ⇒ k = 1.

Question 22.
i = √-1 then prove that :
(1) A² = B² = C² = -I₂
(ii) AB = – BA = -C
Solution:

Question 23.
and f(A) = A² – 5A + 7I then find f(A).
Solution:

Question 24.
Prove that

Solution:

RBSE Solutions for Class 12 Maths

RBSE Class 12 Geography Notes

April 10, 2020 by Veer Leave a Comment

RBSE Class 12 Geography Notes Pdf download भूगोल in both Hindi Medium and English Medium are part of RBSE Class 12 Notes. Here we have given Rajasthan Board Books RBSE Class 12th Geography Notes Pdf Bhugol.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 12
Subject	Geography
Chapter	All Chapters
Exercise	Textbook & Additional
Category	RBSE Solutions

Rajasthan Board RBSE Class 12 Geography Notes in English Medium

We already provided RBSE Solutions for Class 12 Geography

Rajasthan Board RBSE Class 12 Geography Notes in Hindi Medium

खंड ‘अ’: मानव भूगोल के मूल तत्व

Chapter 1 मानव भूगोल: प्रकृति व विषय क्षेत्र
Chapter 2 विश्व की प्रमुख जनजातियाँ
Chapter 3 जनसंख्या: वितरण, घनत्व एवं वृद्धि
Chapter 4 विश्व: जनसंख्या संरचना
Chapter 5 जनसंख्या प्रवास एवं मानव विकास
Chapter 6 विश्व: मानव अधिवास
Chapter 7 मानव व्यवसाय: प्रमुख प्रकार
Chapter 8 प्राथमिक व्यवसाय
Chapter 9 द्वितीयक व्यवसाय
Chapter 10 विश्व: परिवहन एवं संचार
Chapter 11 विश्व: अंतर्राष्ट्रीय व्यापार
Chapter 12 पर्यावरणीय समस्याएँ एवं समाधान

खण्ड ‘ब’: भारत जनसंख्या एवं अर्थव्यवस्था

Chapter 13 भारत: जनसंख्या वितरण, घनत्व एवं वृद्धि
Chapter 14 भारत: जनसंख्या संरचना
Chapter 15 संसाधनों का वर्गीकरण, संरक्षण एवं पोषणीय विकास
Chapter 16 जैविक एवं अजैविक संसाधन
Chapter 17 ऊर्जा संसाधन
Chapter 18 कृषि
Chapter 19 उद्योग
Chapter 20 परिवहन
Chapter 21 भारत में नियोजन
Chapter 22 नियोजन एवं सतत् विकास
Chapter 23 सिंचाई एवं पेयजल
Chapter 24 राजस्थान: खनिज व उद्योग
Chapter 25 राजस्थान: जनसंख्या व जनजातियाँ

Pratical Geography भूगोल प्रायोगिक

Chapter 1 मानचित्र – वर्गीकरण और मानचित्रांकन
Chapter 2 आंकड़ों का एकत्रीकरण एवं विश्लेषण
Chapter 3 सांख्यिकीय आंकड़ों का निरूपण
Chapter 4 सुदूर संवेदन एवं भौगोलिक सूचना तन्त्र
Chapter 5 समपटल सर्वेक्षण
Chapter 6 क्षेत्रीय अध्ययन

We hope the given RBSE Class 12 Geography Notes Pdf download भूगोल in both Hindi Medium and English Medium will help you. If you have any query regarding Rajasthan Board Books RBSE Class 12th Geography Notes Pdf Bhugol, drop a comment below and we will get back to you at the earliest.

RBSE Class 12 Economics Notes

April 1, 2020 by Veer Leave a Comment

RBSE Class 12 Economics Notes Pdf download अर्थशास्त्र in both Hindi Medium and English Medium are part of RBSE Class 12 Notes. Here we have given Rajasthan Board Books RBSE Class 12th Economics Notes Pdf Arthashastra.

Rajasthan Board RBSE Class 12 Economics Notes in English Medium

RBSE Class 12 Economics Notes Microeconomics

RBSE Class 12 Economics Notes Macroeconomics

We already provided RBSE Solutions for Class 12 Economics

Rajasthan Board RBSE Class 12 Economics Notes in Hindi Medium

RBSE Class 12 Economics Notes व्यष्टि-अर्थशास्त्र

Chapter 1 अर्थशास्त्र का परिचय
Chapter 2 उपभोक्ता का संतुलन
Chapter 3 मांग की अवधारणा
Chapter 4 मांग की कीमत लोच
Chapter 5 पूर्ति की अवधारणा
Chapter 6 उत्पादन फलन
Chapter 7 उत्पादन की अवधारणा
Chapter 8 लागत की अवधारणा
Chapter 9 आगम की अवधारणा
Chapter 10 फर्म का संतुलन
Chapter 11 पूर्ण प्रतियोगी बाजार
Chapter 12 बाजार के अन्य स्वरूप
Chapter 13 व्यापार सन्तुलन

RBSE Class 12 Economics Notes समष्टि-अर्थशास्त्र

Chapter 14 राष्ट्रीय आय की मूल अवधारणाएँ
Chapter 15 राष्ट्रीय आय से सम्बन्धित समुच्चय
Chapter 16 राष्ट्रीय आय का मापन
Chapter 17 मुद्राः अर्थ, कार्य एवं महत्त्व
Chapter 18 व्यापारिक बैंकः अर्थ एवं कार्य
Chapter 19 केन्द्रीय बैंक: कार्य एवं साख नियन्त्रण
Chapter 20 उपभोग फलन, बचत फलन व निवेश फलन की अवधारणा
Chapter 21 आय-उत्पादन का निर्धारण
Chapter 22 अधिमाँग एवं न्यून माँग अवधारणा
Chapter 23 सरकारी बजट एवं अर्थव्यवस्था
Chapter 24 अन्तर्राष्ट्रीय व्यापार की अवधारणाएँ
Chapter 25 नकद विहीन लेनदेन

We hope the given RBSE Class 12 Economics Notes Pdf download अर्थशास्त्र in both Hindi Medium and English Medium will help you. If you have any query regarding Rajasthan Board Books RBSE Class 12th Economics Notes Pdf Arthashastra, drop a comment below and we will get back to you at the earliest.