Rajasthan Board RBSE Solutions for Class 10 Science Chapter 12 Electricity Textbook Exercise Questions and Answers.

## RBSE Class 10 Science Solutions Chapter 12 Electricity

### RBSE Class 10 Science Chapter 12 Electricity InText Questions and Answers

Page 200.

Question 1.

What does an electric circuit mean?

Answer:

An electric circuit consists of closed path in which different components like cells, ammeter, voltmeter, resistance etc. are attached.

Question 2.

Define the unit of current.

Answer:

Ampere is the unit of current ampere is defined as the flow of 1 coulomb of charge for one second.

i.e \(I=\frac{Q}{t}\)

If Q = 1C,

t = Is,

then I = 1A.

Question 3.

Calculate the number of electrons constituting one coulomb of charge.

Answer:

Charge on 1 electron = 1.6 x 10 ^{-19}C

1 Coulomb = \(\frac{1}{1.6 \times 10^{-19}}\) electrons 1.6 x to

= 6.2 x 10^{18} electrons.

Page 202.

Question 1.

Name a device that helps to maintain a potential difference across a conductor.

Answer:

A cell or battery helps to maintain a potential difference across conductor.

Question 2.

What is meant by saying that the potential difference between two points is 1V?

Answer:

Potential difference between two points in a current carrying conductor will be IV when 1 Joule of work is done to move a charge of 1 coulomb from one point to other.

∴ \(1 \text { Volt }=\frac{1 \text { Joule }}{1 \text { Coulomb }}\)

Question 3.

How much energy is given to each coulomb of charge passing through a 6V battery?

Answer:

Given: V = 6V

Q = 1C

Potential difference between two points (V)

= \(\frac{Work done (W)}{Charge (Q)}\)

\(V =\frac{W}{Q}\)

From this formula, the energy given to each coulomb of charge will be:

W = V x Q

Energy (E) = V x Q

= 6V x 1C = 6 Joule.

Page 209.

Question 1.

On what factors does the resistance of a conductor depend?

Answer:

Resistance of a conductor depends on (i) length, (ii) area of cross section, (iii) nature of material, (iv) temperature.

Question 2.

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:

As we know that resistance produces obstruction in the path of current which is inversely proportional to the area of cross – section of the conductor (R α \(\frac{1}{A}\)) So a thick conductor offers lesser resistance than a thin wire.

Question 3.

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:

According to Ohm’s law, V α I. As the value of potential difference decreases to half of its former value, the value of current also changes to half of former value.

Question 4.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:

Generally coils of electric toasters and electric irons are made of nichrome. Nichrome is an alloy of nickel (60%), chromium (12%), manganese (2%) and iron (26%). The use of this metal is because

- Resistivity of an alloy is generally higher than that of pure metal.
- They have high melting point.
- On heating ( 800°C), they does not oxidise easily.

Question 5.

Use the data in Table 12.2 to answer the following questions:

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Answer:

(a) Iron is a better conductor as it has low resistivity.

(b) Silver is the best conductor.

Page 213.

Question 1.

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, an 8Ω resistor, and a 12D resistor, and a plug key, all connected in series.

Question 2.

Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

The circuit diagram shown ahead is equivalent resistance of the circuit

R_{s} = R_{1} + R_{2} + R_{3}

5 + 8 + 12 = 25Ω

Page 216.

Question 1.

Judge the equivalent resistance when the following are connected in parallel: (a) 1Ω and 10^{6}Ω, (b) 1Ω. and 10^{3}Ω, and 10^{6}Ω.

Answer:

Question 2.

An electric lamp of 100Ω, a toaster of resistance 50Ω, and a water filter of resistance 500Ω. are connected in parallel to a 220V source. What is the resistance of an electric iron connected to the same sources that takes as much current as all three appliances, and what is the current through it?

Answer:

In parallel combination resistance of electric iron will be equal to the resultant of these three appliances connected in parallel.

∴ \(\frac{1}{R_{p}}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}\)

\(\frac{1}{R_{p}}=\frac{5+10+1}{500}\)

⇒ R_{p} = \(\frac{500}{16}\) = 31.25

Now current through it

I = \(\frac{V}{R_{p}}\) = \(\frac{220}{31.25}\) = 7.04A

Question 3.

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

1. In parallel combination the potential difference across each device remains the same where as in series connections it is distributed.

2. If any appliance is switched ‘ON’ or ‘OFF’ it does not affect the working of other devices.

Question 4.

How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?

Answer:

(a) To get a total resistance of 4Ω from three resistors 2Ω, 3Ω and 6Ω, then resistors 3Ω and 6Ω should be connected in parallel with 2Ω in series with them.

Now, equivalent resistance of 3Ω and 6Ω

\(\frac{1}{\mathrm{R}_{\mathrm{eq}}} = \frac{1}{3}+\frac{1}{6}\)

\(\frac{1}{\mathrm{R}_{\mathrm{eq}}} = \frac{6+3}{18} = \frac{9}{18} = \frac{1}{2}\)

⇒ \(\mathbf{R}_{\mathrm{eq}}\) = 2Ω

Now, 2Ω resistance is in series with \(\mathbf{R}_{\mathrm{eq}}\) = 2Ω, then total resistance becomes

\(\mathrm{R}_{\text {total }}\) = 2 + 2 + 2 = 4Ω

(b) The resistances should be connected in parallel

\(\frac{1}{R} = \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

⇒ \(\frac{1}{R} = \frac{3+2+1}{6}\)

⇒ \(\frac{1}{R} = \frac{6}{6}\)

⇒ R = 1Ω

Question 5.

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, 24Ω?

Answer:

(a) The highest total resistance obtained when resistances are connected in series.

R = R_{1} + R_{2} + R_{3} + R_{4}

= 4 + 8 + 12 + 24

= 48Ω

(b) The lowest total resistance obtained when resistances are connected in parallel.

⇒ \(\frac{1}{R} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}\)

⇒ \(\frac{1}{\mathrm{R}}\) = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}\)

⇒ \(\frac{1}{\mathrm{R}}\) = \(\frac{6+3+2+1}{24}\)

⇒ R = \(\frac{24}{12}\) = 2Ω

Page 218.

Question 1.

Why does the cord of an electric heater not glow while the heating element.

Answer:

The coil of the electric heater is made of alloy nichrome. The resistivity of nichrome is very high. So when the current is flowing, the heating element glow. While the cord of the electric heater is made of copper or aluminium, the resistance of which is very low so it does not glow.

Question 2.

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50V does ?

Answer:

V = 50 volt

t = 1hr = 60 x 60 sec.

Current flowing

\(\text { (I) } = \frac{Q}{t}\)

\(I=\frac{96000}{3600}=26.66 \mathrm{~A}\)

Amount of heat generated by current I in time t

H + VIt

H = 50 x 26.66 x 3600H = 4.8 x 10^{6} Joule.

Question 3.

An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30s.

Answer:

Given

R = 20Ω

I = 5A

t = 30s

By H = I^{2}Rt

= (5)^{2} x 20 x 30 = 15000 Joule

Hence H = 1.5 x 10^{4} Joule

Page 220.

Question 1.

What determines the rate at which energy is delivered by a current?

Answer:

Electric power determines the rate at which energy is delivered by a current.

Question 2.

An electric motor takes 5A from a 220V line. Determine the power of the motor and the energy consumed in 2h.

Answer:

Given

I = 5A

V = 220V

as, P = VI

= 220 x 5A = 1100 W

Energy consumed in 2 hours

= 1100W x 2h

= 2200Wh

= 2200Watt x 60 x 60 secs.

= 7920000 Joule = 7.92 x 10^{6} Joule.

### RBSE Class 10 Science Chapter 12 Electricity Textbook Questions and Answers

Question 1.

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’ then the ratio R/R’ is.

(a) 1/25

(b) 1/5

(c) 5

(d) 25

Answer:

(d) 25

Question 2.

Which of the following terms does not represent electrical power in a circuit?

(a) I^{2}R

(b) IR^{2}

(c) VI

(d) V^{2}/R

Answer:

(b) IR^{2}

Question 3.

An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be.

(a) 100W

(b) 70W

(c) 50W

(d) 25W

Answer:

(d) 25W

Question 4.

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be.

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer:

(c) 1 : 4

Question 5.

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

A voltmeter is always connected parallel in the circuit to measure the potential difference between two points.

Question 6.

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^{-8}Ω m. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?

Answer:

Given, diameter of the wire

D = 0.5 mm = 0.5 x 10^{3} m

Radius = \(\frac{0.5}{2} \mathrm{~mm}=2.5 \times 10^{-4} \mathrm{~m}\)

resistivity of copper.

p = 1.6 x 10^{6} ohm – cm

= 1.6 x 10^{8} ohm – m

Resistance required,

R = 10 ohm

As \(\mathrm{R}=\rho \frac{l}{\mathrm{~A}}\)

⇒ \( l =\frac{\mathrm{RA}}{\rho}\)

On putting the values

= \(\frac{10 \times 3.14 \times\left(2.5 \times 10^{-4}\right)^{2}}{1.6 \times 10^{-8}}\)

= \(\frac{10 \times 3.14 \times 25}{4 \times 1.6}\)

l = 122. 72 m

If diameter is doubled, then the diameter will be (2 x 0.5 mm) 1 mm or 0.001 m. So, using formula

\(\mathrm{R}=\rho \frac{l}{\mathrm{~A}}\)

On putting the value.

\(R=1.6 \times 10^{-8} \times \frac{122.72}{3.14 \times \frac{0.001}{2}}\)

= 250.2 x 10 ^{2} = 2.5Ω

Therefore, when the diameter is doubled. The resistance with remained one – fourth (1/4 R).

Question 7.

The values of current flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below.

I Amperes | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

V Volts | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

Plot a graph between V and I and calculate the resistance of that resistor.

Answer:

The resistance of the resistor will be equal to the slope of the graph.

Question 8.

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer:

Given

I = 2.5 mA = 2.5 x 10^{3}A

V = 12V

Using Ohm’s law, resistance

\(\mathrm{R} = \frac{\mathrm{V}}{\mathrm{I}} = \frac{12}{2.5 \times 10^{-3}} = \frac{120 \times 10^{3}}{25}\)

⇒ \(\mathrm{R} = \frac{120 \times 1000}{25} = 120 \times 40\)

= 4800Ω

⇒ R = 4.8 x 10^{3}Ω

Question 9.

A battery of 9V is connected in series with resistors of 0.2Ω, 0.3Ω, 0.4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ω resistor?

Answer:

Given that, all the resistances are in series, therefore equivalent resistance of the circuit (R) is

R = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}

= 0.2 + 03 + 0.4 + 0.5 + 12

= 13.4Ω

From ohm’s Law

Current through the circuit

\( I = \frac{V}{R} = \frac{9}{13.4} = 0.67 \mathrm{~A}\)

Since, same current flows through all the resistances in series combination therefore, current through 12Ω resistor.

= 0.67Ω

Question 10.

How many 176Ω resistors (in parallel) are required to carry 5A on a 220V line?

Answer:

Lets the number of resistor required is x

In parallel, the resistance is

\(\frac{1}{R} = x \times \frac{1}{176}\)

\(\mathrm{R} =\frac{176}{x}\)

From ohm’ law

\(R = \frac{V}{I}\) [v = 220v I = 5A ]

On putting value

\(\frac{176}{x} = \frac{220}{5}\)

\(x = \frac{176 \times 5}{220}\)

x = 4

Question 11.

Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.

Answer:

(i) In order to get a resistance of 9Ω. from three resistors, each of resistance 6Ω, we connect two resistors (i.e., of 6Ω each) in parallel and with their parallel combination, we connect the third resistor in series. This scheme is shown in figure.

Now 3Ω resistance is in series with 6Ω resistance

∴ \(\mathrm{R}_{\mathrm{eq}} = 3 \Omega+6 \Omega = 9 \Omega\)

(ii) In order to get a resistance of 4Ω from three resistors, two of resistance 6Ω are connected in series to get a resistance of 12Ω Now 12Ω and 6Ω resistances connected in parallel. The scheme is shown in figure.

Question 12.

Several electric bulbs designed to be used on a 220V electric supply line,are rated 10W. How many lamps can be connected in parallel with each other across the two wires of 220V line if the maximum allowable current is 5A?

Answer:

Let n bulbs are connected in parallel to get the power as

P = n x Power of one bulb = n x 10W

∴ P = 10nW

Given

V = 220V, I = 5A

Then P = VI

⇒ 10nW = 220V x 5A

⇒ \(n = \frac{220 \times 5}{10}\) = 110

∴ 110 bulbs can be connected in parallel.

Question 13.

A hot plate of an electric oven connected to a 220V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Given that, resistance of each coil = 24Ω

(i) If each of the resistance is used separately then current.

\(\mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}} = \frac{220}{24} = 9.2 \mathrm{~A}\)

(ii) if both resistance are connected in series then equivalent resistance.

R = R + R = 24 + 24 = 48Ω

then current I = \(\frac{\mathrm{V}}{\mathrm{R}} = \frac{220}{48} = 4.6 \mathrm{~A}\)

(iii) If both the resistances are connected in parallel then equivalent resistance

\(\mathrm{R}_{\mathrm{P}} = \frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}} = \frac{24 \times 24}{24+24}\) = 12Ω

then current = \(\mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}} = \frac{220}{12} = 18.3 \mathrm{~A}\)

Question 14.

Compare the power used in the 2Ω resistor in each of the following circuits : (i) a 6V battery in series with 1Ω and 2Ω resistors, and (ii) a 4V battery in parallel with 12Ω and 2Ω resistors.

Answer:

(i) In first case, when 6V battery is connected in series with 1Ω and 2Ω resistors, then current in the circuit,

\( I = \frac{6 V}{1 \Omega+2 \Omega} = \frac{6 V}{3 \Omega} = 2 \mathrm{~A}\)

power used in the 2Ω resistor

\(\mathbf{P}_{1} = \mathrm{I}^{2} \mathrm{R} = (2 \mathrm{~A})^{2} \times 2 \Omega = 8 \mathrm{~W}\)

(ii) In second case, when 4V battery is connected in parallel with 12Ω and 2Ω resistors,

\( \mathrm{P}_{2} = \frac{\mathrm{V}^{2}}{\mathrm{R}} = \frac{(4 \mathrm{~V})^{2}}{(2 \Omega)} = 8 \mathbf{W}\)

∴ \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} = \frac{8 \mathrm{~W}}{8 \mathrm{~W}} = 1\)

Question 15.

Two lamps, one rated 100W at 220V, and the other 60W at 220V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220V?

Answer:

Since both the lamps are connected in parallel, so, the potential difference between them will be same. The lamp with power of 100W, the current consumed by it is

P = V x I

I = P/V [ P = 100W, V = 220V ]

On putting the value,

\( I_{1} = \frac{100 \mathrm{~W}}{220 \mathrm{~V}} = \frac{100}{220} \mathrm{~A}\)

Similarly the current consumed by the lamp with power 60 W is

\( I_{2} = \frac{60 \mathrm{~W}}{220 \mathrm{~V}} = \frac{60}{220} \mathrm{~A}\)

The current flowing in the parallel combination I = I_{1} + I_{2}

\(\frac{100}{220}+\frac{60}{220} = 0.727 \mathbf{A}\)

Question 16.

Which uses more energy, a 250W TV set in 1hr, or a 1200W toaster in 10 minutes?

Answer:

Given

For TV P_{1} = 250W

t_{1} = 1 hour = 60 x 60 sec

t_{1} = 3600 sec

Energy consumed

H_{1} = P_{1} x t_{1} = 250W x 3600 s

= 900000J

= 9 x 105J

For toaster

P_{2} = 12W

t_{2} = 10 min

= 600s

∴ Energy consumed

H_{2} = P_{2} x t_{2} = 120 x 600

= 72000J

= 7.2 x 104J

Here it is clear that H_{1}> H_{2}

Therefore, the TV set consumed more energy.

Question 17.

An electric heater of resistance 8Ω draws 15A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

Given, R = 8Ω, I = 15A, t = 2hr

Rate at which heat is developed H = P = I_{2}

R = 15 x 8 = 1800W or 1800J/s

Question 18.

Explain the following

- Why is the tungsten used almost exclusively for filament of electric lamps?
- Why are the conductors of electric heating devices, such as bread – toasters and electric irons, made of an alloy rather than a pure metal?
- Why is the series arrangement not used for domestic circuits?
- How does the resistance of a wire vary with its area of cross – section?
- Why are copper and aluminium wires usually employed for electricity transmission

Answer:

- Due to high melting point of tungsten ( 3380° ), it becomes incandescent at 2400K.
- The resistivity of alloys are generally higher than that of pure metals of which they are made of.
- In series circuits, same current flows then if any one of the electric appliance fails or in off position, then all the other appliances stop working because same electric current is flowing through all of them.
- The resistace of a wire (R) varies inversely as its cross – sectional area (A) i.e., R α \(\frac{1}{\mathrm{~A}}\)
- Due to very low resistivity, copper and aluminium wires are generally used for electricity transmission.

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