NCERT Solutions for Class 10 Science Chapter 11 Exercise

Answer: A continuous and closed path of an electric current is called an electric circuit.

Answer: The unit of electric current is ampere. If 1 coulomb of charge flows through a conductor every second then the current through the conductor is said to be 1 ampere.

Answer: Charge of 1 electron = n × e = 1.6 × 10⁻¹⁹ C

If n electrons make one coulomb then n = 1/e
So, n = 1/(1.6 × 10⁻¹⁹)
Or, n = 6.25 × 10¹⁸

Therefore, 1 coulomb of charge constitutes 6.25 × 10¹⁸ electrons.

Answer: An electric battery.

Answer: The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Answer: Potential Difference = 6 V
Charge = 1 C

Now, Work done in a circuit = Potential Difference × Charge
So, energy given = 6 × 1 = 6

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

Answer: The resistance of a conductor depends upon these factors –

1. Length of the conductor
2. Cross-sectional area of the conductor
3. Material of the conductor
4. Temperature of the conductor

Answer: Resistance (R) of a conductor is inversely proportional to its cross-section area (A). This shows that larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. So, the current will flow more easily through a thick wire than a thin wire of the same material.

Answer: According to Ohm’s law –
V = IR
Or, I = V/R

Now, the potential difference is decreased to half.
New potential difference V’ = V/2
So, the new current I’ = V’/R
Or, I’ = (V/2)/R
Or, I’ = (1/2)(V/R)
This gives us I’ = I/2

Therefore, the amount of current flowing through the electrical component is reduced by half.

Answer: The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of a pure metal. Also, an alloy does not melt easily even at high temperature.

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

Answer: Iron is a better conductor because it has lower resistivity than mercury.

(b) Which material is the best conductor?

Answer: Silver is the best conductor because it has the lowest resistivity.

Answer: Schematic diagram of the given circuit is shown below —

Answer: Schematic diagram of the given circuit is shown below —

Total voltage, V = 3 × 2 = 6 V
Total resistance, R = 5 Ω + 8 Ω +12 Ω = 25 Ω
Reading of ammeter, I = V/R = 6/25 = 0.24 A
Reading of voltmeter, V = IR = 0.24 × 12 = 2.88 V

(a) 1 Ω and 10⁶ Ω

Answer: The net resistance in parallel is given by –

1/R = 1/R₁ + 1/R₂
Here, R₁ = 1 Ω, and R₂ = 10⁶ Ω
So, 1/R = 1/1 + 1/10⁶ = (10⁶ + 1)/10⁶
⇒ R = 10⁶/(10⁶ + 1) ≈ 1 Ω

Therefore, the equivalent resistance is 1 Ω.

(b) 1 Ω and 10³ Ω, and 10⁶ Ω

Answer: The net resistance in parallel is given by –

1/R = 1/R₁ + 1/R₂ + 1/R₃
Here, R₁ = 1 Ω, R₂ = 10³ Ω, R₃ = 10⁶ Ω
So, 1/R = 1/1 + 1/10³ + 1/10⁶ = (10⁶ + 10³ + 1)/10⁶
⇒ R = 10⁶/(10⁶ + 10³ + 1) = 1000000/1001001
⇒ R = 0.999 Ω

Therefore, the equivalent resistance is 0.999 Ω.

Answer: The equivalent resistance (R) of the resistors connected in parallel can be calculated by —

1/R = 1/100 + 1/50 + 1/500 = 16/500
Or, 1/R = 16/500
⇒ R = 500/16 Ω = 31.25 Ω
So the resistance of the electric iron = 31.25 Ω

Now, using Ohm’s law, the current flowing across the circuit —
I = V/R = 220/31.25
⇒ I = 7.04 A

Answer: When the electrical devices are connected in parallel there is no division of voltage among the appliances. The potential difference across the devices is equal to supply voltage. The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

(a) 4 Ω

Answer: We can get a total resistance of 4 Ω by connecting the 2 Ω resistance in series with the parallel combination of 3 Ω and 6 Ω. So, the total resistance R can be achieved by –

R = R₁ + (R₂ × R₃)/(R₂ + R₃)
Or, R = 2 + (3 × 6)/(3 + 6)
Or, R = 4 Ω

(b) 1 Ω

Answer: We can obtain a total resistance of 1 Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel –

1/R = 1/R₁ + 1/R₂ + 1/R₃
Or, 1/R = 1/2 + 1/3 + 1/6
Or, 1/R = 6/6
⇒ R = 1 Ω

Answer (a) : To get the highest resistance, we will connect the four resistances in series. Their equivalent resistance R can be achieved by —
R = 4 + 8 + 12 + 24 = 48 Ω

Answer (a) : To get the lowest resistance, we will connect the four resistances in parallel. Their equivalent resistance R can be achieved by —

1/R = 1/4 + 1/8 + 1/12 + 1/24
Or 1/R = 12/24
⇒ R = 24/12 = 2 Ω

Answer: The heating element of an electric heater is made up of alloy. Alloys have very high resistance and this characteristic allows them to become hot and glow red when current flows through the heating element. But the cord of the same electric heater does not glow because it is made up of copper which has very low resistance.

Answer: Voltage V = 50 V
Time, t = 1 h = 1 × 60 × 60 s.

Now, the amount of heat (H) produced is given by the Joule’s law of heating as —
H = VIt
Or, H = V.q/t.t
Or, H = V.q
Or, H = 50 × 96000
Or, H = 4.8 × 10⁶ J

Therefore, the heat produced while transferring the charge is H = 4.8 × 10⁶ J.

Answer: Resistance, R = 20 Ω
Current I = 5 A
Time, t = 30 s

Now, the amount of heat (H) produced is given by the joule’s law of heating as —
H = Vlt
Also, V = I × R
So, H = (I × R)It
H = (5 × 20) 5 × 30
H = 1.5 × 10⁴ J

Therefore, the amount of heat developed in the electric iron is 1.5 × 10⁴ J.

Answer: The rate of consumption of electric energy in an electric appliance is known as electric power. Hence, the power of the appliance determines the rate at which energy is delivered by a current.

Answer: Power of the electric motor, P = VI
Substituting the values, P = 220 × 5 = 1100 W

The energy consumed by the motor, E = Power × time
So, E = 1100 × 7200 = 7920000 J

Therefore, the power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 10⁶ J.

Answer: (d) 25

Answer: (b) IR²

Answer: (d) 25 W

Answer: (c) 1:4

Answer: A voltmeter should be connected in parallel in the circuit to measure the potential difference between two points.

Answer: Resistance, R = ρ l/A
Or, l = RA/ρ

Now, the cross-section area of the wire —
A = π (Diameter/2)²
So, l = R π (Diameter/2)² / ρ
Substituting the values,
l = 10 × 3.14 × (0.0005/2)²/(1.6 × 10⁻⁸)
Or, l = 122.72 m

Doubling the diameter of the wire, we get new wire of diameter 1 mm (= 0.001 m)
Hence, new resistance, R’ = ρ l/A = 1.6 × 10⁻⁸ × 122.72/π (0.001/2)² = 2.5 Ω
Therefore, the length of the wire is 122.72 m and the new resistance is 2.5 Ω.

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

Answer: The plot between voltage and current is called VI characterstic.
The voltage is plotted on x-axis and current plotted on y-axis.

The slope of the line gives the value of resistance (R) —
Slope = 1/R = BC/AC = 2/6.8
⇒ R = 6.8/2 = 3.4 Ω

Answer: According to Ohm’s law —
V = IR
Or, R = V/I

Substituting the values —
R = 12/0.0025 = 4.8 × 10³ Ω

Answer: Total resistance, R = 0.2 + 0.3 + 0.4 + 0.5 + 12
Or, Total R = 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = V/R
Or, I = 12 / 13.4 = 0.67 A
There is no current division occurring in a series circuit. So, the current through the 12 Ω resistor will be same as 0.67 A.

Answer: Let’s consider the number of resistors required as ‘x’.
Then, the equivalent resistance of the parallel combination of resistor R is given by –
1/R = x × 1/176
R = 176/x

Now, using Ohm’s law. the number of resistors can be calculated as follows –
R = V/I
Substituting the values –
176/x = V/I
Or, x = (176 × 5)/220 = 4
Therefore, the number of resistors required is 4.

(i) 9 Ω

Answer: Consider the image below —

Two 6 Ω resistors are connected in parallel. Their equivalent resistance, R is given by
1/R = 1/6 + 1/6 = 2/6
Or, R = 6/2 = 3 Ω

The third 6 Ω resistor is connected in series with this equivalent 3 Ω. Hence, the equivalent resistance of the circuit is –
R = 6 Ω + 3 Ω = 9 Ω.

(ii) 4 Ω

Answer: Consider the image below —

Two 6 Ω resistors are connected in series. Their equivalent resistance, R is given by
R = 6 + 6 = 12 Ω

The third 6 Ω resistor is connected in parallel with this equivalent 12 Ω. Hence, the equivalent resistance of the circuit is –
1/R = 1/12 + 1/6 = 3/12
Or, R = 12/3 = 4 Ω.

Answer: Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of electric bulb, P = 10 W
Now, the power P is given by P = V²/R
Or, R = V²/P
Hence, the resistance R of each bulb –
R = 220²/10 = 4840 Ω

The total resistance R‘ of x number of electric bulbs connected in parallel cab be achieved by –
1/R‘ = 1/R + 1/R + …….. x times.
Or, 1/R‘ = 1/R × x
Or, x = R/R‘ = 4840/R‘
Applying the relation, R = V/I
Total resistance R‘ = 220/5 = 44 Ω
So, x = 4840/44 = 110

Therefore, the total number of electric bulbs connected in parallel are 110.

Answer: Case 1: When the two coils A and B are used separately –
Resistance R = 24 Ω
So, the current I = V/R = 220/24
I = 9.17 A

Case 2: When the two coils are connected in series –
Equivalent resistance R = 24 Ω + 24 Ω = 48 Ω
So, the current I = V/R = 220/48
I = 4.58 A

Case 3: When the two coils are connected in parallel –
Equivalent resistance 1/R = 1/24 + 1/24 = 2/24
So, the current I = V/R = 220 × 2/24
I = 18.33 A

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors

Answer: 1 Ω and 2 Ω resistors are connected in series. So, the equivalent resistance of the circuit, R = 1 + 2 = 3 ohm.
Now, applying Ohm’s law, V = IR
Or, I = V/R = 6/3 = 2 A

In series combination, the current in the circuit remains constant. Therefore, the power is given by –
P = I²R
Or, P = 2² × 2 = 8 W

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors

Answer: 1 Ω and 2 Ω resistors are connected in parallel.
Again, I = V/R = 6/3 = 2 A

In parallel combination, the voltage in the circuit remains constant. Therefore, the power is given by –
P = V²/R = 4²/2 = 8 W

Answer: Since both the bulbs are connected in parallel, the voltage across each of them will be the same.
Now, the Current drawn by the bulb of rating 100 W is given by –
Power, P = V I
Or, I = P/V = 100/220 A
Also, current drawn by the bulb of rating 60 W is given by –
I = P/V = 60/220

So, the total current drawn from the line = 100/220 + 60/220 = 0.727 A

Answer: The energy consumed by an electrical appliances is given by the equation —
H = P × t, where P is the power of the appliance and t is the time.

Now, energy used by 250 W TV set in 1 hour = 250 W × 1 h = 250 Wh
And, energy used by 1200 W toaster in 10 minutes = 1200 × 10/60 = 200 Wh

Comparing the results, we can easily tell that the TV set uses more energy than the toaster.

Answer: R = 8 Ω,
I = 15 A,
t = 2 h

Now, the rate at which heat is developed in the heater is equal to the power –
P = I² × R = (15)² × 8 = 1800 J/s

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Answer: Tungsten has a very high melting point and resistivity. Due to these two properties tungsten can produce and maintain high temperatures.

(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?

Answer: Alloys have higher melting points than a pure metal. They also are preferred in heating devices because of the heat they produce. This is because of their reduced conductivity which also prevents an electric shock since metals are good conductors while alloys are not.

(c) Why is the series arrangement not used for domestic circuits?

Answer: In series arrangement, failure of one component results in failure of the complete circuit. Also, each device will get different voltages in series arrangement. Therefore, parallel connection is preferred over series connection.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer: Resistance (R) of a wire is inversely proportional to its area of cross-section (A). It can be written as R ∝ 1/A. Thus, increasing the area of cross-section of wire (thickness of wire) decreases its resistance. Or, when the area of cross-section of wire decreases, then its resistance increases.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: Copper and aluminium wires are good conductors of electricity because of their low resistivity. Thus, they are usually employed for transmission of electricity.