Class 12

RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.1

September 1, 2021 by Fazal Leave a Comment

Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.1

Inverse Function Calculator tool gives the inverse function of the input in no time. Check the Calculator to get inverse function.

Question 1.
For which value of x, matrix
is singular ?
Solution:

⇒ 1 (- 6 – 2) + 2 (-3 – x) + 3(2 – 2x) = 0
⇒ -8 – 6 – 2x + 6 – 6x = 0
⇒ -8 = 8
⇒ x = $\frac { 8 }{ -8 }$ = -1
Hence, x = -1

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Question 2.
If matrix , then find adj.A, Also prove that, A(adj.A) = |A| I₃ = (adj.A)A.
Solution:

Matrix made from adjoint of matrix A,

Question 3.
Find the inverse matrix of the following matrix:

Solution:
(i) Let

= 1(1 + 3) – 2(-1 + 2) + 5(3 + 2)
= 4 – 2 + 25
|A| = 27 ≠ 0
So. A^-1 exists.
Cofactors of matrix A,

Matrix made from adjoint of matrix A,

Cofactors of matrix A,

Matrix made from adjoint of matrix A,

Cofactors of matrix A.

Matrix made from adjoint of matrix A,

Question 4.
If matrix

then find A^-1 and prove that:
(i) A^-1A= I3
(ii) A^-1 = F(-α)
(iii) A(adj.A) = |A|I = (adj A).A
Solution:

Then, |A| = cos α (cos α – 0) + sin α (sin α – 0) + 0(0 – 0)
= cos² α + sin² α
|A|= 1 ≠ 0 So,
A^-1 exists.
Cofactors of matrix A,

Matrix made from adjoint of matrix A

So, A(adj.A) = |A|I = (adj.A) A Hence Proved.

Question 5.
, then prove that : A^-1 = A^T.
Solution:
Let

Question 6.
If matrix , then prove that A^-1 = A³.
Solution:
Given,

So, A^-1 exists.
On finding adjoint of matrix A,
a₁₁ = – 1, a₁₂ = – 2, a₂₁ = 1, a₂₂ = 1
Matrix formed by adjoint of A,

From (i) and (ii),
A^-1 = A³
Hence proved.

Question 7.

then find (AB)^-1.
Solution:
Given

Then, |A| = 5(3 – 4) – 0(2 – 2) + 4(4 – 3)
= – 5 – 0 + 4
|A| = -1 ≠ 0
So, A^-1 exists.
On finding adjoint of matrix A,

Matrix formed by adjoint of A

Question 8.
If
Solution:
Given

Question 9.
Show prove that matrix satisfies equation A² – 6A + 17I = O. Thus find A^-1.
Solution:
Given

Question 10.
If matrix
, then show that A² + 4A – 42I = O. Hence A².
Solution:
Given

So, given matrix satisfies A² + 4A – 42I = O
Now A² + 4A – 42I = O
⇒ A² + 4A = 42I
⇒ A^-1 (A² + 4A) = 42A^-1.I
⇒ A^-1.A² + 4A^-1.A = 42A^-1.I
⇒ A + 4I = 42A^-1
(∵ A^-1.A = I and A^-1.I = A^-1)

RBSE Solutions for Class 12 Maths

RBSE Solutions for Class 12 Physics Chapter 5 Electric Current

July 12, 2021 by Prasanna Leave a Comment

RBSE Solutions for Class 12 Physics Chapter 5 Electric Current are part of RBSE Solutions for Class 5 Electric Current. Here we have given Rajasthan Board RBSE Solutions for Class 12 Physics Chapter 5 Electric Current.

Rajasthan Board RBSE Class 12 Physics Chapter 5 Electric Current

RBSE Class 12 Physics Chapter 5 Text Book Exercise with Answers

RBSE Class 12 Physics Chapter 5 Multiple Choice Type Questions

Question 1.
The product of resistivity and conductivity of conductor depends upon :
(a) the cross-sectional area
(b) the temperature
(c) the length
(d) none of these
Answer:
(d) none of these

Question 2.
Two wires of same shape whose resistivity ρ₁and ρ₂ are connected in series. The equivalent resistivity of the combination will be :
(a) $\sqrt{p_{1} \rho_{2}}$
(b) 2 (ρ₁ + ρ₂)
(c) $\frac{\rho_{1}+\rho_{2}}{2}$
(d) ρ₁ + ρ₂
Answer:
(c) $\frac{\rho_{1}+\rho_{2}}{2}$
Resistance in series

Question 3.
A conducting resistance is connected to the battery and temperature of conductor decreases by the process of cooling then the value of current will be :
(a) increased
(b) decreased
(c) remain constant
(d) zero
Answer:
(a) increased
The resistance of the conducting wire will decrease according to equation R_t = R₀ (1 ± α Δt) due to which current will increase according to Ohm’s law (I = $\frac{V}{R}$ )

Question 4.
A cell of e.m.f. 2.1 V gives a current of 0.2 A. This current is passing through the resistance of 10 Ω. The internal resistance of cell is :
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 0 Ω
Answer:
(b) 0.5 Ω

Question 5.
The current I and voltage V curves for a given metallic conductor at two different temperatures T₁ and T₂ are shown in the figure then :

(a) T₁ = T₂
(b) T₁ > T₂
(c) T₁ < T₂
(d) None of these
Answer:
(c) T₁ < T₂
Gradient of graph,
tanθ = $\frac{1}{R}=\frac{I}{V}$
But, according to equation R_t = R₀ (1 α Δt)
∴ The resistance R₂ > R₁
∴ T₂ > T₁

Question 6.
The electric power is supplied through the copper wires from a one city to another city which is 150 km apart. If the terminal voltage and average resistance of per kilometer are 8 volt and 0.5 Ω respectively then the power loss in a wire is :
(a) 19.2 W
(b) 19.2 kW
(c) 19.2 J
(d) 12.2 kW
Answer:
(b) 19.2 kW
Total resistance of wire = 0.5 × 150 km = 75 Ω
Potential drop on wire = 8 × 150 = 1200 volt
Power loss m wire (P) = $\frac{V^{2}}{R}=\frac{1200 \times 1200}{75}$
= 19200 watts = 19.2 kW

Question 7.
Five resistances of R Ω were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is :

Answer:
(b)
Equivalent resistance,

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Question 8.
Drift velocity (v_d) depends upon the electric field, in which the following dependence of drift velocity on the electric field obey the ohm’s law is :
(a) v_d ∝ E²
(b) v_d ∝ E
(c) v_d ∝ E^1/2
(d) v_d = constant
Answer:
(b) v_d ∝ E

Question 9.
A carbon resistance has a colour sequence of bands as blue, yellow, red and silver. The value of resistance is :
(a) 64 × 10² Ω
(b) (64 × 10² ± 10%) Ω
(c) 642 × 10⁴Ω
(d) (26 × 10³ ± 5%) Ω
Answer:
(b) (64 × 10² ± 10%) Ω

Question 10.
When the wire is connected with a battery is heated up due to electric current which quantity do not vary :
(a) Drift velocity
(b) Resistivity
(c) Resistance
(d) Number of free electrons
Answer:
(d) Number of free electrons
Number of free electrons does not depend on temperature.

RBSE Class 12 Physics Chapter 5 Very Short Answer Type Questions

Question 1.
Calculate the resistance of resistor by V-I curve

Answer:

Question 2.
Write the S.I. unit of current density?
Answer:
Current density, J = $\frac{I}{A}$ = Ampere/metre²

Question 3.
Write the relation between conductivity and current density?
Answer:
Conductivity $\vec{J}=\sigma \vec{E}$

Question 4.
Write the two examples of non-ohmic resistance?
Answer:
Diode and electrolytes.

Question 5.
Write the dependency of resistivity of metals on temperature?
Answer:
Resistivity of a conductor depends upon temperature as
ρ = ρ₀(l + α Δt)

Question 6.
Write the name of two substances of which when its temperature increases its resistivity decreases.
Answer:
The resistivity of semiconductor decreases with increase in its temperature. So the examples are Germanium and Silicon.

Question 7.
Find the value of current which is flowing through the bulb of 40 W 220 V.
Answer:

RBSE Class 12 Physics Chapter 5 Short Answer Type Questions

Question 1.
What will be the value of charge when an electric current flows through a conductor?
Answer:
On flowing a electric current in conductor it does not charged. So total charge across it will be zero.

Question 2.
In the given material the resistivity of single metal are ρ₁ and ρ₂ (Ω × m), find the ratio of ρ₁ and ρ₂.
Answer:
Resistivity of a conductor does not depend on length and area of cross section of the material.
∴ ρ₁ : ρ₁ = 1 : 1

Question 3.
Two identical cells having equal emf and negligible internal resistance are combined in parallel combination. Find out the electric current in resistance R:

Answer:
In parallel combination the potential difference will be equal.
V = E
∴ Current, I = $\frac{V}{R}$
∴ I = $\frac{E}{R}$

Question 4.
What is the difference between emf and terminal potential difference?
Answer:
See the article no. 5.10.
Electromotive Force (or e.m.f.) of a Cell
“The energy given by the cell to a unit charge for flowing through the circuit i.e., the work done in flowing the unit charge through the complete circuit by the cell is called the electromotive force or e.m.f. of the cell.”
It is represented by E. Therefore
E = $\frac{W}{q}$ ……………. (1)
Where W is the work done by the cell in flowing charge q throughout the circuit.
∴ Unit of E = — $\frac{\mathrm{J}}{\mathrm{C}}$ = JC^-1 or volt.
If q = 1C, W = 1J, then E = 1JC^-1
i. e., “If the work done in flowing unit charge is lJ then the e.m.f. of the cell is 1 JC^-1 or 1 volt.”
The value of e.m.f. for one cell is constant and different for different cells. For example, e.m.f. of voltaic cell is 1.08 volt, for daniell cell it is 1.12 volt and for dry cell it is 1.5 volt.
Important Points
1. Although we call e.m.f. as force but actually it is energy which is given by the cell to unit charge for flowing through the circuit.
2. If external resistance is connected between the terminals of the cell the the potential difference between the terminals is ‘Terminal voltage’ of the cell.

3. When there is no external resistance connected between the terminals the potential difference is equal to the e.m.f. of the cell.

Terminal Potential Difference
“Work done by the cell in flowing a unit positive charge from one terminal to other terminal through external resistance in a circuit is called the terminal potential difference or terminal voltage of the cell.”
Therefore terminal voltage
V = $\frac{W_{\mathrm{ext}}}{q}$ …………… (2)
Where W_ext is the work done by the cell in flowing q charge through external circuit.
If R be external resistance in the circuit and i the flowing current, then
V = $\frac{W_{\mathrm{ext}}}{q}$ = iR (according to Ohm’s law)
Similarly if r be the internal resistance of the cell, then the potential drop i.e., the energy consumed due to internal resistance r,
v = ir
According to definition of e.m.f.
E = Work done (external work + internal work) in flowing unit charge.

i.e., “in open circuit, the terminal potential difference of the cell is called the e.m.f. of the cell.”
From equation (4),

Therefore, internal resistance of a cell can be obtained with the help of equations (5) and (6).

Question 5.
Write the definition of drift velocity.
Answer:
Drift Velocity (v_d)
When a potential difference (V) is applied across the conductor of length (l), then an electric field $\overrightarrow{(E)}$ develops in the conductor $\left(E=\frac{V}{l}\right)$ Due to this field each free electron of the conductor experiences an electric force $\vec{F}=-e \vec{E}$ towards the positive end of the conductor and hence it starts accelerated motion $\left(\vec{a}=\frac{\vec{F}}{m}\right)$ towards the positive end. During its accelerated motion it collides with the other electrons and positive ions of the conductor. Therefore its velocity always remains changing. This motion of electron is known as ‘Drift motion’ and the average velocity between two successive collisions is known as ‘Drift velocity.’ It is denoted by v_d.

“i.e., the maximum velocity achieved by the electrons due to imposed electric field with which they collide with other ions, is known as drift velocity.” Time taken in two successive collisions is called ‘Relaxation time’. For most of the conductors, the relaxation time is the order of 10^-14s. Just before colliding with a ion, the velocity is maximum and after collision for a moment the velocity becomes zero. Again electron gets accelerated by applied electric field and repeats the previous situation of colliding with ions of the conductor. Thus “the potential difference of battery does not provide accelerated motion to electrons but it can provide a small constant velocity along the length of the conductor, which is imposed upon the random motion of the electrons. This constant velocity of the electron is called drift velocity of electrons.” The order of this drift velocity about 10^-4 m/s.

Reason of small value of drift velocity : In following figure 5.4, the random motion of electron is shown by thick lines and in presence of external electric field by dotted lines. It is obvious from the figure that the electron moves with small drift velocity and due to this reason point X is shifted to X’. Thus due to electric field, the net displacement becomes XX’. Which has very small value. This is why the drift velocity is very small.

Relaxation time (τ) : The average time taken by electron in two successive collisions is known as ‘Relaxation time’ and it is denoted by τ. It is obtained by

Question 6.
When a wire of 80 resistance is bent in the form of a circle then find out the resistance between any two ends of the diameter.
Answer:
When resistor is folded into the circular form, then across diameter, for equivalent resistance, shape is divided in two parts, therefore resistance becomes half.

Question 7.
What will be the effect on resistance and resistivity on produce the distortion in the size of a matter?
Answer:
Resistance is inversely proportional to area of cross-section of the substance while there is no effect on resistivity.

Question 8.
Can the potential difference between the plates of a cell more than its electromotive force (emf)?
Answer:
Yes, when the cell is in condition of charging.
V_T = E + Ir

RBSE Class 12 Physics Chapter 5 Long Answer Type Questions

Question 1.
What is called drift velocity? Obtain the equation of Ohm’s law ( $\vec{J}=\sigma \vec{E}$ ) on the basis of drift velocity where parameters are in their usual meaning.
Answer:
When a potential difference (V) is applied across the conductor of length (l), then an electric field $\overrightarrow{(E)}$ develops in the conductor $\left(E=\frac{V}{l}\right)$ Due to this field each free electron of the conductor experiences an electric force [latex]\vec{F}=-e \vec{E}[/latex] towards the positive end of the conductor and hence it starts accelerated motion $\left(\vec{a}=\frac{\vec{F}}{m}\right)$ towards the positive end. During its accelerated motion it collides with the other electrons and positive ions of the conductor. Therefore its velocity always remains changing. This motion of electron is known as ‘Drift motion’ and the average velocity between two successive collisions is known as ‘Drift velocity.’ It is denoted by v_d.

The relation between drift velocity and potential difference,
v_d = $\frac{e \tau}{m} \frac{V}{l}$ …………… (1)
and relation between drift velocity and electric current,
v_d = $\frac{i}{A n e}$ ………….. (2)
On comparing equations (1) and (2) we have

or V = ρ $\frac{l}{A}$ i ………….. (4)
Here $\frac{m}{n e^{2} \tau}$ which is characteristic of the substance of the conductor and it is called ‘specific resistance’ or ‘resistivity’ of the substance. It is constant for one substance and different for different substances.

If physical conditions of the conductor do not change at constant temperature, then l and A will also remain constant. Therefore,
ρ $\frac{l}{A}$ = constant = R (Resistance of the conductor)
∴ From equation (4),
V = Ri …………….. (5)
or V ∝ i
i.e., “the potential difference developed across a conductor is directly proportional to current flowing through the conductor provided the physical conditions of the conductor remain unchanged.” This is Ohm’s law.
Vector form of Ohm’s law :
From equation (3),

Equation (7) is called microscopic form of Ohm’s law.
Specific Resistance. “Ratio of intensity of electric field (E) and current density (J) at any point inside the current carrying conductor, is called the specific resistance of the material of the conductor. It is represented by ρ.”
∴ ρ = $\frac{E}{J}$
If the length of the conductor be l and potential difference across it be V, then
E = $\frac{V}{l}$ and j = $\frac{i}{A}$
Where A is the area of cross-section of the conductor

If A = 1 m²; l = 1 m, then ρ = R
i.e., “the specific resistance of a material is equal to the resistance of material when the length and area of cross-section of material take as unity.”

Specific resistance can also be obtained by following formula,
ρ = $\frac{m}{n e^{2} \tau}$ …………….. (9)
Where m is the mass of electron, n is the free electron density, e is the electronic charge and τ is the relaxation time.

Question 2.
Establish the relation between drift velocity and electric field. What is mobility? Describe the relation between mobility and velocity.
Answer:
Relation between Drift Velocity and Potential Difference
Suppose PQ is a conductor of length l and a potential difference V is applied across PQ as shown in the figure 5.5. i.e., its P end remains negative and Q end positive. Therefore an electric field is developed directing from Q to P and its intensity is given by
E = $\frac{V}{l}$ …………. (1)
Each free electron of the conductor experiences a force ( $\vec{F}$ ) towards the end Q,
∴ $\vec{F}=-e \vec{E}$

If m be the mass of the electron, then its acceleration

∵ The average thermal velocity of electrons is zero.
∴ u = 0
Therefore using first equation of motion (v = u + αt), we have

This is the relation between drift velocity and potential difference.

Mobility
In the expression of drift velocity (speed), the term $\left(\frac{e \tau}{m}\right)$ is constant because e and m are constant and τ is the characteristics of specific metal, so for specific metal at constant temperature τ is constant.
Constant term $\left(\frac{e \tau}{m}\right)$ for a given metal is called mobility (μ).

For specific metal, v_d ∝ E
Magnitude of drift velocity is directly proportional to the applied electric field

So the mobility of ions are defined as the ratio of drift velocity to the electric field.
Mobility is positive and its unit is $\frac{\mathrm{m} / \mathrm{s}}{\mathrm{volt} / \mathrm{m}}$ = m²s^-1volt^-1
When electric field is same for given metals
$\frac{\mu_{1}}{\mu_{2}}=\frac{v_{d_{1}}}{v_{d_{2}}}$
It means when in any metal drift speed is more than electrons, then its mobility will also high. We will discuss about mobility related to semiconductor in Chapter 16.
Relation between electric current and mobility for semiconductor : Conductivity in semiconductors is due to electrons and holes both.

Question 3.
Establish the relation between resistance and resistivity. How does resistivity depend on the temperature? Describe conductors, insulators and semiconductors on the basic of resistivity.
Answer:
Resistivity
On practical basis it is found that the resistance depends upon the length l and area of cross-section A of the resistor as under
R ∝ $\frac{l}{A}$
or R = ρ $\frac{l}{A}$
where ρ = specific resistance or resistivity of the material of the conductor. It is the characteristic of the conductor.
Special fact : Resistivity does not depend upon the dimensions of the conductor i. e., it does not depend upon the length and area of cross-section of the conductor.

Effect of temperature on resistivity of metallic conductors : Specific resistance is given by,
ρ = $\frac{m}{n e^{2} \tau}$ ……………. (1)
Where, m = mass of electron
e = charge of electron
n = number of free electrons per unit volume
τ = relaxation time
In equation, there is no effect of increase in temperature on m and e. If temperature rise is not much, then n will also remain unchanged. Now we have to study the effect of rise in temperature on relaxation time τ.
∵ τ = $\frac{\lambda}{v_{\mathrm{rms}}}$
Where λ is mean free path and v_rms is the root mean square velocity of electrons.

When the temperature is increased, then the vibrational energy of atoms of the conductor increases and hence their amplitude of vibration increases and as result the value of mean free path λ decreases. Also due to increase in temperature the value of v_rms increases. Hence due to combined effect of decrease in λ and increase in v_rms, the relaxation time x will decrease. Thus it is clear from equation (1) the resistivity ρ will increase. Thus the specific resistance increases with increase in temperature. The change in resistivity (ρ) with rise in temperature is given by following relation,
ρ_t = ρ₀(1 + αt) ………….(2)
Where ρ_t = Specific resistance at t°C
ρ_o = specific resistance at 0°C
α = linear temperature coefficient of resistance The variation in resistivity of metallic conductors with temperature is shown in figure 5.11.

Temperature dependence of resistivity of semiconductor and insulator : Change in resistivity of semiconductor and insulator is different from change in resistivity of metallic conductors. For conductors the change in resistivity with temperature is linear but the change in resistivity of insulator
and semiconductor is represented by exponential curve as shown figure 5.13.

The mathematical variation of specific resistance with temperature
ρ_t = ρ₀(1 – α Δt)
Linear temperature coefficient of resistivity is negative for semiconductor and conductors.
Resistivity of a semiconductor varies with temperature according to following relation :
$\rho_{t}=\rho_{0} e^{-\left(\frac{E_{g}}{k T}\right)}$ …………. (3)
Where ρ_t = Resistivity at t°C
ρ₀ = Resistivity at 0°C
E_g = Energy gap
k = Boltzmann constant
T = (273 + t°C) K

Question 4.
Two cells of emf E₁ and E₂ and internal resistance r₁ and r₂ respectively are connected in parallel combination. Determine the equivalent emf and equivalent internal resistance of the combination.
If this combination is connected with external resistance R, then obtain the value of current flowing in this external resistance.
Answer:
Parallel Combination of Cells
(a) When e.m.f. and internal resistance is same for all cells : Suppose n cells of same e.m.f. E and internal resistance r are joined in parallel and this combination is connected with external resistance R as shown in the figure 5.19. All the cells are connected between junctions A and B. Any charge flowing in the circuit, passes only one cell. Therefore the equivalent e.m.f. = E = e.m.f. of one cell

The equivalent internal resistance

then the combination is significant.
(b) When the e.m.f. and internal resistance of the cells are different :
When the positive terminals of all cells are connected to one point and all their negative terminals to another point, the cells are said to be connected in parallel.

As shown in figure 5.20, suppose two cells of emf’s ε₁ and ε₂ and internal resistances r₁ and r₂ are connected in parallel between two points A and B.

The end points of A and B are connected with external resistance R.
If in each cell current is I₁ and I₂ respectively then total current in external resistance is
I = I₁ + I₂ ……………… (1)
If terminal potential between A and B is V.
For I cell V = ε₁ – I₁r₁ …………… (2)
For II cell V = ε₂ – I₁₂r₂ ………….. (3)
From above equations.

If equivalent e.m.f. is ε_eq and equivalent internal resistance is r_eq of combination then terminal potential difference will be
V = ε_eq – Ir_eq …………… (5)
Comparing equation (4) and (5),

By use of parallel combination it is clear :
(i) When ε₁ = ε₂ = ε and r₁ = r₂ = r means two cells of equal emf and equal internal resistance are connected in parallel then ε_eq = ε and r_eq = $\frac{r}{2}$

(ii) When n cells of equal emf and equal internal resistance are connected in parallel combination
than ε_eq = ε and r_eq = $\frac{r}{n}$ then electric, current in external resistance will be
I = $\frac{\varepsilon}{R+r / n}$ ………………….. (8)
(iii) We can display equations (6) and (7) in the below form

Similarly if n cells are combined in parallel system then,

RBSE Class 12 Physics Chapter 5 Numerical Questions

Question 1.
The length of cylindrical metal (copper) is 1 cm and radius is 2.0 mm. If 120 V potential difference is applied across the ends of the rod then find out the electric current flowing through it. (Resistivity of copper is 1.7 × 10^-8 Ω-m)
Solution:
Length (l) = 1 cm = 1 × 10^-2 m
Radius (r) = 2 mm = 2 × 10^-3 m
Area of cross section of wire
(A) = πr² = π ×(2 × 10^-3)² = 4π × 10^-6 m²
Potential across ends (V) = 120 V
Current (I) = ?
According to the Ohm’s law

Question 2.
Find out the equivalent resistance between a and b as shown in figure.

Solution:
Adding R₁ and R₂ in series
R’ = R₁ + R₂ = 6Ω
Adding R’ and R₉ in parallel

Question 3.
Find out the equivalent resistance between a and b of the circuit in which infinite resistances are connected.

Solution:
Let except first two resistances the equivalent resistance of all be x. Adding DG and EF resistance in parallel.

Question 4.
Three resistances of 1 Ω, 2 Ω and 3 Ω are connected in series combination. Find out equivalent resistance of the combination. If this combination is connected by the battery of 12 V e.m.f. and negligible internal resistance then find out the voltage across each ends of resistance.
Solution:
Connecting 1 Ω, 2 Ω and 3 Ω in series
R_equi = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6 Ω
Current flowing in circuit

Potential on 1 Ω resistance = V = IR = 2 × 1 = 2V
Potential on 2 Ω resistance V = IR = 2 × 2 = 4V
Potential on 6 Ω resistance (V) = IR = 2 × 3 = 6V

Question 5.
At room temperature (27°C) the resistance of thermal element is 100 Ω. If resistance of thermal element is 117 Ω, then find out the temperature of the element. Temperature coefficient of resistance of resistor is [1.70 × 10^-4 C^-1].
Solution:

Question 6.
A wire having length 15 m and 6 × 10^-7 m² cross-sectional area in which a negligible current is flowing, its resistance is 5 Ω. What is the resistivity of the substance at experimental temperature?
Solution:
Length of wire (l) = 15 m
Area of cross-section (A) = 6.0 × 10^-7 m²
Resistance (R) = 5.0Ω

Question 7.
The current of 0.5 Amp is flowing in a copper wire whose cross-sectional area is 1mm². If the number of free electrons is 8.5 × 10²² per cm³ in unit volume then find out the drift velocity of electrons?
Solution:
Area of cross section = 1 mm²
= 1 ×(10³)² = 1 × 10^-6 m²
Number of free electrons (n) = 8.5 × 10²² / cm³
= 8.5 × 10²⁸/m³
Current (I) = 0.5 amp

Question 8.
At what temperature the resistance of copper wire becomes doubled than the resistance of the wire at 0°C? Temperature coefficient of resistance is 4.0 × 10^-30°C^-1.
Solution:
Given : R_t = 2R,
R_t = R₀ (1 + α Δt)
2R₀ = R₀ (1 + α Δt)

Question 9.
E.m.f. of accumulator battery of car is 12 V and its internal resistance is 0.4 Ω, then find out the maximum current drawn from the battery.
Solution:
Electromotive force (E) = 12 V
Internal resistance (r) = 0.4 Ω

Question 10.
A coil which has a resistance of 4.2 Ω is immersed into water. If 2A current is flowing through it for 10 min then how much heat will produce in calorie?(</ = 4.2 J/cal)
Solution:
Given, R = 4.2 Ω, I = 2 Amp.
t = 10 minute = 10 × 60 = 600 sec
energy E = W = I²Rt
= (2)2 × 4.2 × 600 Joule
But W = JQ

Question 11.
The length of a cylindrical tube is / and its internal and external radius are a and b respectively. If resistivity of matter is ρ then find out the resistance between the ends of the tube.
Solution:
Resistance of cylinder, R = ρ $\frac{l}{A}$
Area of cross section for flow of current = External area – Internal area

= πa² – πb²
= π (a² – b²)
∴ R = $\frac{\rho l}{\pi\left(a^{2}-b^{2}\right)}$

Question 12.
In a house 4 bulbs of 100 W and 4 bulbs of 40 W glows daily for 4 hours and 6 hours respectively.
Two fans of 60 W running daily for 8 hours. Find out the consumption of electricity in one month of 30 days. If rate of consumption of electricity is ? 5 per unit.
Solution:
Energy consumed by 4 bulbs of 100 watt each.

Energy consumed by 2 fans of 60 watt each

Total consumed electric energy = 48 + 28.8 + 28.8 = 105.6 unit
Energy consumed = Electric energy × rate per unit
= 105.6 × 5 = 528.00 = ₹ 528

RBSE Solutions for Class 12 Physics

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) $\frac{R_{1} R_{2}}{R_{1}+R_{2}}$
(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 $\frac{P}{Q}=\frac{R}{S}$
∴ $\frac{8}{4}=\frac{6}{3}$ = 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 = $\frac{2}{8}\frac{1}{4}=$ = 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 = $\left(\frac{l_{1}-l_{2}}{l_{2}}\right)$ 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) $\frac{L}{2}$
(d) $\frac{L}{4}$
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 = $\frac{E}{L}$ × R₂
= $\frac{1.1}{2.20} \times \frac{95}{100}$ = 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.
$\frac{P}{Q}=\frac{R}{S}$

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 = $\frac{V_{A B}}{l_{A B}}$
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) = $\frac{V}{l}=\frac{1.1 V}{8.8}$
= 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
$\frac{P}{Q}=\frac{R}{S}$ ……………….. (1)
If resistance of unit length of the wire is $\sigma \frac{\mathrm{ohm}}{\mathrm{cm}}$ 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) = $\rho \frac{l^{\prime}}{A}$
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 :
$\frac{P}{Q}=\frac{R}{S}$

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 = $\frac{Q}{P} \times R$
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 = $\frac{E}{L_{A B}}$ ……………. (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 = $\frac{I R_{A B}}{L_{A B}}$
Let $\frac{R_{A B}}{L_{A B}}$ = 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 = $\left(\frac{E}{R_{A B}+r+R_{h}}\right) \times \frac{R_{A B}}{L_{A B}}$

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 = $\frac{\varepsilon_{s}}{l_{0}}$ ……………… (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
$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$ ……………. (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
$\varepsilon_{s}=k l_{0} \quad \text { or } \quad k=\frac{\varepsilon_{s}}{l_{0}}$ …………….. (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’ = $\left(\frac{\varepsilon_{s}}{l_{0}}\right)$ 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 :
$\frac{P}{Q}=\frac{R}{S}$

(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 = $\frac{Q}{P} \times R$
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 = $\frac{\varepsilon_{s}}{l_{0}}$ ………………… (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’ = $\left(\frac{\varepsilon_{s}}{l_{0}}\right)$ 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 = $\frac{E}{L_{A B}}$ ……………. (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 = $\frac{I R_{A B}}{L_{A B}}$
Let $\frac{R_{A B}}{L_{A B}}$ = 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 = $=\left(\frac{E}{R_{A B}+r+R_{h}}\right) \times \frac{R_{A B}}{L_{A B}}$

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
$R^{\prime}=\frac{R_{1} \times R_{2}}{R_{1}+R_{2}}=\frac{2 R \times 2 R}{4 R}=R \Omega$

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 $\frac{L}{2}$ 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 = $\frac{V}{d}$

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 $\vec{E}$ 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 $\vec{E}$ remains unchanged
(b) Both $\vec{E}$ and V changes
(c) Both $\vec{E}$ and V remains unchanged
(d) $\vec{E}$ changes and V remains unchanged
Answer:
(d) $\vec{E}$ changes and V remains unchanged
Electric field $\vec{E}$ 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 $\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)$ will be (in volts)
(a) 9 × 10⁴
(b) Zero
(c) $\frac { 9\times { 10 }^{ 4 } }{ \alpha }$
(d) $\frac{9 \times 10^{4}}{\sqrt{2}}$
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 ∝ $\frac{1}{r}$
(b) V ∝ $\frac{1}{r^{2}}$
(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 $\frac{V_{1}}{V_{2}}$ 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 + $\sqrt{15 z}$ . The electric field intensity at point (x, y, z) in S.I. system will be :
(a) $3 \sqrt{2}$
(b) $4 \sqrt{2}$
(c) $5 \sqrt{2}$
(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) $\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} J$
(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 = $\frac{k Q}{r}$ , 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: