Ch 11: Electricity (NCERT Solutions)

NCERT Solutions for Class 10 Science Ch 11: Electricity, is given below.

You will get solutions for;

• In-text questions as well as,
• Exercises questions.

So let’s dive into it!

In-text Questions Set-1

1) What does an electric circuit mean?

Solution:
An electric circuit is a continuous, closed path that allows electric current to flow. When the circuit is complete, the current can flow through it.

2) Define the unit of current.

Solution:
The SI unit of electric current is the ampere. A current of 1 ampere flows when 1 coulomb of charge moves through a conductor every second.

3) Calculate the number of electrons constituting one coulomb of charge.

Solution:
The value of the charge of an electron is 1.6 × 10^-19 C.

According to charge quantization,
Q = n × q_e, where n is the number of electrons and q_e is the charge of an electron.

Substituting the values in the above equation, the number of electrons in a coulomb of charge can be calculated as follows:

No. of electron in one coulomb of charge,
n = 1/(1.6 × 10^-19)
= 6.25 × 10¹⁸

In-text Questions Set-2

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

Solution:
A battery, which can have one or more electric cells, is a device that keeps a voltage difference across a conductor.

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

Solution:
When 1 joule of work is used to move a charge of 1 coulomb from one point to another, the potential difference between the two points is 1 volt.

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

Solution:
The potential difference between two points is given by the equation,

V = W/Q,
Where,
W is the work done in moving the charge from one point to another,
Q is the charge

From the above equation, we can find the energy given to each coulomb as:
W = V × Q

Substituting the values in the equation, we get
W = 6V × 1C = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.

In-text Questions Set-3

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

Solution:
The resistance of a conductor depends on:
– Its length
– Its cross-sectional area 
– The material it is made of 
– Its temperature

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?

Solution:
Resistance is given by the formula:
R = ρ l/A

Where:
ρ is the resistivity of the wire’s material,
l is the length of the wire,
A is the wire’s cross-sectional area.

This formula shows that resistance is inversely proportional to the cross-sectional area. This means a thinner wire has more resistance, while a thicker wire has less resistance.

Therefore, current flows more easily through a thick wire than a thin one.

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?

Solution:
According to the Ohm’s law;
V = IR

If the resistance remains constant, V is directly proportional to I. 
𝑉 ∝ 𝐼

Now, if the potential difference is reduced to half of its value, the current also becomes half of its original value. 

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

Solution:
Alloys have a higher melting point than pure metals because they resist melting more. This makes them suitable for heating appliances like electric toasters and irons, where high temperatures are needed.

5) Use the data in Table 11.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Solution:
(a) The resistivity of iron is 10.0 × 10⁻⁸ Ωm, while the resistivity of mercury is 94.0 × 10⁻⁸ Ωm.
Since mercury has higher resistivity than iron, iron is a better conductor.

(b) According to Table 11.2, silver has the lowest resistivity among the materials listed, making it the best conductor.

In-text Questions Set-4

In-text Questions Set-5

1) Judge the equivalent resistance when the following are connected in parallel:
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω, 10³ Ω and 10⁶ Ω

Solution:
When the resistances are joined in parallel, the resultant resistance in parallel arrangement is given by:
1/R = 1/R₁ + 1/R₂ + 1/R₃

(a) 1/R = 1/1 + 1/10⁶ = 1 + 10^-6
R = 1 Ω

(b) 1/R = 1/1 + 1/10³ + 1/10⁶ = 1 + 10^-3 + 10^-6
R = 1 Ω

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?

Solution:
Here, voltage (V) = 220 V
R₁ = 100 Ω
R₂ = 50 Ω and
R₃ = 500 Ω

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/100 + 1/50 + 1/500 = 16/500

R = 500/16 = 31.25 Ω

The resistance of electric iron, which draws as much current as all three appliances take together = R = 31.25 Ω.

Current passing through electric iron (I) = V/R = 220/31.25 = 7.04 A.

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

Solution:
When electrical devices are connected in parallel, they all receive the same voltage as the supply. This setup means there is no division of voltage among the devices. Additionally, connecting devices in parallel lowers the overall resistance of the circuit.

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

Solution:
(a) To obtain highest resistance, all the four resistances must be connected in series arrangement. In that case resultant;
R = R₁ + R₂ + R₃ + R₄
= 4 + 8 + 12 + 24
= 48 Ω

(b) To obtain lowest resistance, all the four resistance must be connected in parallel arrangement. 1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
= 1/4 + 1/8 + 1/12 + 1/24
= 1/2 Ω

So, R = 2 Ω

In-text Questions Set-6

1) Why does the cord of an electric heater not glow while the heating element does?

Solution:
When a heater and its cord are connected in series to a voltage source, they carry the same current. Since the cord’s resistance is much smaller than that of the heater element, it produces only a small amount of heat, while the heater element produces much more heat. As a result, the heater element glows, but the cord does not.

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

Solution:
Charge transferred (Q) = 96000 C,
time = 1 hour = 60 x 60 = 3600 s, and
potential difference (V) = 50 V.

Heat generated (H) = VIt
= V x Q
= 50 x 96000
= 4800000 J
= 4.8 x 10⁶ J.

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

Solution:
Resistance of electric iron (R) = 20 Ω,
Current (I) = 5 A, and
Time = 30 s

Heat generated (H) = I²Rt
= 5² x 20 x 30
= 15000 J

In-text Questions Set-7

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

Solution:
Electric power measures how quickly electrical energy is used by electrical devices. In other words, it shows the rate at which a device consumes energy when a current flows through it.

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.

Solution:
Current drawn by electric motor (I) = 5 A.
The line voltage V = 220 V,
Time (t) = 2 h.

Power of motor (P) = P = VI = 220 x 5 = 1100 W,

Now, the energy consumed (E) = Pt
So, E = 1100 W x 2 h
= 2200 Wh or 2.2 kWh.

Exercise Questions

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

Solution:
(d) 25

Explanation:
The resistance is divided into five equal parts, so each part has a resistance of R/5.

Since these parts are connected in parallel, we can find the equivalent resistance using the formula for parallel resistors.

1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R
= (5 + 5 + 5 + 5 + 5)/R
= 25/R

So, 1/R’ = 25/R
So, R/R’ = 25

2) Which of the following does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R

Solution:
(b) IR²

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) 75 W
(c) 50 W
(d) 25 W

Solution:
(d) 25 W

Explanation:
The energy consumed by the appliance is given by;
P = VI = V²/R

The resistance of the light bulb;
R = V²/P

Putting the values, we will get;
R = (220)²/100 = 484 Ω

Now, even if the supply voltage is reduced, the resistance will remain the same.

Hence, the power consumed can be calculated as;
P = V²/R

Again putting the value, we get;
P = (110)²/484 Ω = 25 W

Therefore, the power consumed for the electric bulb operating at 110 V is 25 W.

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

Solution:
(c) 1:4 

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

Solution:
A voltmeter is connected in parallel to the resistance where you want to measure the voltage.

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

Solution:
Diameter of wire (d) = 0.5 mm,
Resistivity (ρ) = 1.6 x 10^-8 Ωm,
Resistance (R) = 10 Ω.

R = ρL/A
L = πD²R/4ρ
= [22 x (5 x 10^-4)² x 10]/ [7 x 4 x 1.6 x 10^-8]
= 122.5 m

If the diameter is doubled, the new diameter, d = 1 mm = 0.001 m.

So the new resistance will be;
R = ρL/A
= (1.6 x 10^-8 x 122.5) /[π (0.001/2)²]
= 2.5 Ω

Now;
𝑅′/𝑅 = 2.5/10 = 1/4 
So, the new resistance will become 1/4 times the original resistance. 

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.

Solution:
Voltage of battery, V = 12 V
Current (I) = 2.5 mA = 2.5 x 10^-3 A

Resistance (R) = V/I = 12 V/ 2.5 x 10^-3 A = 4800 Ω.

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?

Solution:
Potential difference (V) = 9 V.

Total resistance (R) = R₁ + R₂ + R₃ + R₄ + R₅
= 0.2 +0.3 + 0.5 + 0.5 + 12
= 13.4 Ω

Current in the circuit (I) = V/R = 9 V / 13.4 Ω = 0.67 A.

In a series circuit, same current flows through all the resistance, hence the current of 0.67 A will flow through the 12 Ω resistor.

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

Solution:
Let a resistor of 176 Ω are joined in parallel.

Then their combined resistance (R);

1/R = 1/176 + 1/176 …… upto n times
= n/176
or
R = 176/n Ω

It is given that V = 220 V and I = 5 A

R = V/I

By putting the values, we will get,
176/n = 220/5
= 44 Ω

So, n = 176/44 = 4.
That means 4 resistors should be joined in parallel.

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?

Solution:
Each bulb is rated as 10 W, 220 V,
It draws a current (I) = P/V = 10 W/220 V
= 1/22 A.

Now the maximum allowable current is 5 A and all lamps are connected in parallel, hence maximum number of bulbs joined in parallel with each other = 5 x 22 = 110.

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?

Solution:
Potential difference (V) = 220 V.
Resistance of coil (A) = Resistance of coil (B) = 24 Ω

(i) When either coil is used separately, the current (I) = V/R = 220 V/ 24 Ω
= 9.2 A.

(ii) When two coils are used in series,
Total resistance (R)
= R₁ + R₂ = 24 + 24
= 48 Ω
Current flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.

(iii) When two coils are joined in parallel,
Total resistance (1/R) = 1/24 + 1/24
= 2/24,
So, R = 12 Ω.
Current (I) = V/R = 220 V / 12 Ω = 18.3 A.

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.

Solution:
(i) When a 6 V battery is in series with 1 Ω and 2 Ω resistors.
Total resistance (R) = 1 + 2 = 3 Ω.
Current (I) = 6 V/3 Ω = 2 A

Power used in 2 Ω resistor = I²R = 2² x 2 = 8 W

(ii) When a 4 V battery is in parallel with 12 Ω resistor and 2 Ω resistors,
Current flowing (I) in 2 Ω resistor = 4 V/ 2 Ω = 2 A

Power used in this 2 Ω resistor = I²R = 2² x 2 = 8 W

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?

Solution:
Current drawn by 1st lamp rated 100 W at 220 V = P/V = 100/ 220 = 5/11 A.

Current drawn by 2nd lamp rated 60 W at 220 V = 60/220 = 3/11 A.

In parallel arrangement the total current = I₁ + I₂ = 3/11 + 5/11 = 8/11 = 0.73 A.

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

Solution:
Energy used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.

Energy used by toaster of power 1200 W in 10 minute (10/60 h)
= P x t = 1200 W x 10/60 h = 200 Wh.

So the energy used by TV is more as compared to toaster.

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

Solution:
The rate at which the heat develops in the heater can be calculated by;
P = I² R

By putting the values in the equation, we will get;
P = (15)² × 8 Ω = 1800 Watt

Hence, the electric heater produces heat at the rate of 1800 Watt

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 copper and aluminum wires are usually employed for electricity transmission?

Solution:
(a) Electric lamp filaments need a strong metal with a high melting point. Tungsten is used because it can withstand very high temperatures without melting.

(b) Heating device conductors are made from alloys, not pure metals, because alloys have higher resistivity and higher melting points, which prevents them from oxidizing at high temperatures.

(c) Series circuits are avoided in domestic settings because they keep the same current through all appliances, regardless of their resistance. This setup means appliances cannot be turned on or off individually.

(d) The resistance of a wire decreases as its cross-sectional area increases.

(e) Copper and aluminum wires are commonly used for electricity transmission because they are good conductors with low resistivity and can be drawn into thin wires easily.