# NCERT Solutions for Class 10 Science Chapter 12 Electricity

In Text Questions

Page No: 200

Q 1: What does an electric circuit mean?

The electric circuit is the continuous path which allows electric current or electrons to flow through it. An electric circuit consists of connecting wires, other elements (like electric bulb, resistances, key etc.) and a battery.

Q 2: Define the unit of current.

The SI unit of electric current is Ampere, commonly denoted by letter A. 1 Ampere is defined as – When an amount of charge equivalent to 1 coulomb flows through a cross-section of a conductor in 1 second.

Q 3: Calculate the number of electrons constituting one coulomb of charge.

We know that, the charge of an electron, e = 1.6 × 10-19 Coulomb

Therefore, 1.6 × 10-19 C of charge is equivalent to = 1 electron

By unitary method,

If the charge is 1C then no of electrons,

Thus,

6.25 ×1018 electrons constitute 1 coulomb of charge.

In Text Questions

Page No: 202

Q 1: Name a device that helps to maintain a potential difference across a conductor.

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

Q 2: What is meant by saying that the potential difference between two points is 1 V?

The potential difference between two points is said to be 1 volt (1V) if the amount of work done in moving a charge of 1 Coulomb (1C) between the two points is equivalent to 1 Joule (1J).

Q 3: How much energy is given to each coulomb of charge passing through a 6 V battery?

The charge passing through the battery is 1 coulomb. The potential difference is 6V. We have to find out the energy which is equal to the work done in moving the charge through 6V. Now,

Since the work done on each coulomb of charge is 6 joules. Therefore, the energy given to each coulomb of charge is also 6 joules.

In Text Questions

Page No: 209

Q 1: On what factors does the resistance of a conductor depend?

The electrical resistance of a conductor depends on the factors listed below:

1) Nature of the material of the conductor – e.g. A wire made of copper would be a better conductor than a wire made of iron.

2) Length of the conductor – e.g. A shorter wire would be a better conductor than a longer wire.

3) Area of cross-section of the conductor – e.g. A thick copper wire is a better conductor than a thin copper wire.

4) Temperature of the conductor – resistance goes up with increase in temperature.

Q 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?

The current will flow more easily through a thick wire than a thin wire of the same material when connected to the same source. This is mainly because the resistance of a wire is inversely proportional to the square of its diameter. And, a thick wire has greater diameter than a thin wire. Therefore, thick wire will offer less resistance to the current flow and thin wire will offer more resistance to the current flow. Thus, current flows easily through thick wire.

Q 3: Let the resistance of an electrical component remain 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?

We know, From Ohm’s Law: V = IR,

Since the resistance is constant and the potential difference is decreased to half of the former value. Thus, the current through the component will also decrease to half of its former value.

Q 4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

The coils (or heating elements) of toasters and electric irons are made of an alloy rather than a pure metal due to the following two reasons:

1) The resistivity of an alloy is much higher than that of a pure metal, and

2) An alloy does not undergo oxidation easily even at high temperature

Q 5: Use the data in Table on page 23 of this book to answer the following:

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

(b) Which material is the best conductor?

(a) The electrical resistivity of iron is 10.0 × 10-8 Ωm. And, the electrical resistivity of mercury is 94.0 × 10-8 Ωm.

Since the resistivity of mercury is greater than that of iron, it will offer more hindrance to current flow as compared to iron. Therefore, iron is a better conductor than mercury.

(b) Silver metal has the lowest resistivity nearly about 1.60 × 10-8 Ωm. Therefore, silver metal is the best conductor of electricity.

In Text Questions

Page No: 213

Q 1: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.

In this problem, we have 3 cells of voltage 2 V each. Thus, the total voltage of the three cells connected together to form a battery will be 3 × 2V = 6V. The circuit also consists 3 resistors of 5Ω, 8Ω, 12Ω respectively and a plug key, all connected in series as shown in the figure given below:

Q 2: Redraw the circuit of Question 1 putting 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 reading in the ammeter and the voltmeter?

We redraw the circuit by including an ammeter circuit and a voltmeter across the 12 Ω resistor, as shown in figure. Please note that the ammeter is always connected n series with the circuit but the voltmeter has been put in parallel with the 12 Ω resistor

We will now calculate the current reading in the ammeter and potential difference reading in the voltmeter:

(i) Calculation of current flowing in the circuit. The three resistors of 5 Ω, 8 Ω and 12 Ω are connected in series. So, net resistance in series is the sum of all the resistances.

Total resistance =5Ω+8Ω+12Ω=25Ω

Potential difference, V = 6 V

And, Current, I =?

We know,

Since the current in the circuit is 0.24 A, therefore, the ammeter will show reading of 0.24 A.

(ii) Calculation of potential difference across the 12 Ω resistor. From (i) we have calculated that a current of 0.24 A flows in the circuit. Since, in series same current flows through the whole circuit. The same current of 0.24 A also flows through the 12 Ω.

Now, for the 12 Ω resistor:

Current, I = 0.24A

Resistance, R = 12 Ω

And, Potential difference, V =?

We know that,

Thus, the potential difference across the 12 Ω resistor is 2.88V. So, the voltmeter will show a reading of 2.88 V.

In Text Questions

Page No: 216

Q 1: Judge the equivalent resistance when the following are connected in parallel:

(a) 1 Ω and 106Ω

(b) 1Ω, 103Ω and 106Ω

(a) The equivalent resistance of two resistances 1 Ω and 106 Ω connected in parallel will be less than 1 Ω. This is because when a number of resistances are connected in parallel, then their equivalent resistance is less than the smallest individual resistance. Here, since the smallest resistance is 1 Ω. Therefore, the net resistance will be less than 1Ω.

(b) The equivalent resistance of three resistances 1Ω, 103Ω and 106Ω connected in parallel will be less than 1 Ω because the smallest resistance is 1 Ω.

Q 2: An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Potential difference of the source, V = 220 V
These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.

7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V= 7.04 A
Let R’ be the resistance of the electric iron. According to Ohm’s law,
V= IR’

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

Q 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Advantages of connecting electrical devices in parallel:
(i) When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that of battery which is not possible in series connection.
(ii) Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, the same current flows through all devices, irrespective of their resistances.
(iii) If one appliance fails to work, others will continue to work properly. If they are connected in parallel.

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

There are three resistors of resistances 2Ω, 3Ω, and 6Ω respectively.
(a) The following circuit diagram shows the connection of the three resistors.

Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω
Hence the total resistance of the circuit is 4 Ω.

(b) The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in series.
Therefore, their equivalent resistance will be given as

Therefore, the total resistance of the circuit is 1 Ω.

Q 5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively.
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

Therefore, 2 Ω is the lowest total resistance.

Intext Questions

Page No: 218
Q 1: Why does the cord of an electric heater not glow while the heating element does?

The cord of an electric heater is made up of metallic wire such as copper or aluminium which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also, heat produced ‘H’ is
H = I2Rt
Thus, for the same current H ∝ R, more heat is produced by heating element as it has more resistance, and it glows.

Q 2: Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

Given Charge, Q = 96000C
Time, t= 1hr = 60 × 60= 3600s
Potential difference, V= 50volts
Now we know that H= VIt
So we have to calculate I first
As I= Q/t
∴ I = 96000/3600 = 80/3 A

Therefore, the heat generated is 4.8 × 106 J.

Q 3: An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30s.

The amount of heat (H) produced is given by the joule’s law of heating as

H= Vlt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 × 20 = 100 V
H= 100 × 5 × 30 = 1.5 × 104 J.
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.

Intext Questions

Page No: 220
Q 1: What determines the rate at which energy is delivered by a current?

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

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

Power (P) is given by the expression, P = VI
Where,
Voltage, V = 220 V
Current, I = 5 A
P= 220 × 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 × 60 × 60 = 7200 s
∴ P = 1100 × 7200 = 7.92 × 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 × 106 J

Exercises

Page No: 221

Q 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

(D) 25

The resistance of wire is R. This wire is cut into five equal pieces as a result resistance of each piece of wire will be R/5. Now, when these five pieces are connected in parallel, the equivalent resistance for the combination is R’. We can write, (Assume that each piece of wire is named R1, R2, R3, R4, R5)

Q 2: Which of the following terms does not represent electrical power in a circuit?

A. PR

B. IR2

C. VI

D. V2/R

(b) IR2

The electrical power is not represented as IR2

Q 3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

A. 100 W

B.75W

C.50 W

D.25W

(d) 25 W

The bulb parameters are as given below:

Power, P = 100W

Potential difference, V = 220V

Then Resistance, R =?

Now,

This resistance of the bulb will remain constant i.e. 484 Ω

Now, when we operate this bulb on 110 V, then,

Power, P =?

Potential difference, V = 110V

And, resistance, R = 484 Ω

Q 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be:
A.1:2
B.2:1
C.1:4
D.4:1

(c)

The Joule heating is given by, H = i2Rt

Let, R be the resistance of the two wires.

The equivalent resistance of the series connection is RS = R + R = 2R

If V is the applied potential difference, then it is the voltage across the equivalent resistance.

Suppose the resistance of each one of the two wires is R.
Case 1: When both wires are connected in series. Then, net resistance will be R+ R=2R.

If V is the applied potential difference, then it is the voltage across the equivalent resistance.

The heat dissipated in time t is,

The equivalent resistance of the parallel connection is

V is the applied potential difference across this Rp.

The heat dissipated in time t is,

So, the ratio of heat produced is,

Note: H ∝ R also H ∝ i2 and H ∝ t. In this question, f is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H α R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated I in terms of voltage and resistance and used in the equation H = I2Rt to find the ratio.

Q 5: How is a voltmeter connected in the circuit to measure the potential difference between two points?

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

Q 6: A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10−8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Area of cross-section of the wire, A =π (d/2)2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that

If the diameter of the wire is doubled, new diameter = 2 × 0.5=1 mm = 0.001 m.

let new Resistance be R´

Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.

Q 7: The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below −

 A (amperes) 0.5 1 2,0 3 4 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. The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

 V (volts) 1.6 3.4 6.7 10.2 13.2 A (amperes) 0.5 1 2 3 4

The I-V characteristic of the given resistor is plotted in the following figure.

The slope of the line gives the value of resistance (R) as,

Resistance (R) of a resistor is given by Ohm’s law as, V= IR
⇒ R = V/I

Where,

Potential difference, V= 12 V

Current in the circuit, I= 2.5 mA = 2.5 × 10-3 A

Therefore, the resistance of the resistor is 4.8 kΩ

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

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

V= IR
I = (V/R)
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R.
R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential difference, V= 9 V
I = 9/13.4 = 0.671 A

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

Q 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as V= IR
R = V/I
where,
Supply voltage, V= 220 V

Current, I = 5 A Equivalent resistance of the combination = R, given as

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

Q 11: Show how you would connect three resistors, each of resistance 6 Ω so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

(i) When two 6 Ω resistances are in parallel and the third is in combination with this, the equivalent resistance will be 9Ω.

(ii) When two 6 Ω resistances are in series and the third is in parallel to them, then it will be 4Ω.

Q 12: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts

Because of R=V2/P

According to ohm’s law

V=IR

Let R is the Total Resistance of the circuit for x number of electric bulbs

Resistance of each electric bulb, R1 = 4840 Ω

∴ Number of electric bulbs connected in parallel is 110.

Q 13: A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Supply voltage, V= 220 V
Resistance of one coil, R= 24 Ω
(i) Coils are used separately
According to Ohm’s law,
V= I1R1 Where,
I1 is the current flowing through the coil
I1 = V/R1 = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law, V = I2R2 Where,
I2 is the current flowing through the series circuit
I2 = V/R2 = 220/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel
Total resistance, R3 is given as =

According to Ohm’s law,
V= I3R3
Where,
I3 is the current flowing through the circuit I3 = V/R3 = 220/12 = 18.33 A
Therefore, 18.33
A current will flow through the circuit when coils are connected in parallel.

Q 14: Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law,
V = IR Where,
I is the current through the circuit
I= 6/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A.
Power is given by the expression,
P= (I)2R = (2)2 × 2 = 8 W

(ii) Potential difference, V = 4 V 12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same.
Hence, the voltage across 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by P= V2/R = 42/2 = 8 W
Therefore, the power used by 2 Ω resistor is 8 W.

Q 15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,
Power = Voltage × Current
(Current = (Power/ Voltage)=(100/ 220}A)
Similarly, current drawn by the bulb of rating 60 W is given by,
Power = Voltage × Current
Current = (Power/Voltage) = 60/220 A

Hence, current drawn from the line

Q 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Energy consumed by an electrical appliance is given by the expression,

H= Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600 = 7.2×105 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Q 17: An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Rate of heat produced by a device is given by the expression for power as, P= I2R Where, Resistance of the electric heater, R= 8 Ω Current drawn, I = 15 A
P= (15)2 × 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.

Q 18: Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?