E X E R C I S E S CHAPTER 11 Electricity SCIENCE NCERT TEXTBOOK (CBSE) CLASS 10:

 E X E R C I S E S

CHAPTER  11  Electricity  

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:
If the wire is cut into 5 equal parts, the resistance of each part will be $\frac{R}{5}$. When connected in parallel, the equivalent resistance $R'$ is given by the formula:

$$\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$

Since all parts are identical, this simplifies to:

$$\frac{1}{R'} = 5 \times \frac{1}{R/5} = \frac{5}{R/5} = \frac{25}{R}$$

Thus, $R' = \frac{R}{25}$, and the ratio $R/R' = 25$.

Correct answer: (d) 25

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

(a) $I^2R$
(b) $IR^2$
(c) $VI$
(d) $\frac{V^2}{R}$

Answer:
Electrical power is given by the formula $P = I^2R$, $P = VI$, or $P = \frac{V^2}{R}$.
The term $IR^2$ does not represent electrical power.

Correct answer: (b) $IR^2$

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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

Answer:
The power of the bulb when operated at 220 V is 100 W. The relationship between power and voltage is $P = \frac{V^2}{R}$. If the voltage is halved (from 220 V to 110 V), the power consumed will decrease by a factor of $\left( \frac{110}{220} \right)^2 = \frac{1}{4}$. Hence, the power consumed will be $\frac{100}{4} = 25 \, W$.

Correct answer: (d) 25 W

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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:
In both cases, the power dissipated is proportional to the resistance. When connected in series, the total resistance is $2R$, and when connected in parallel, the total resistance is $\frac{R}{2}$. The heat produced (or power dissipated) is $P = \frac{V^2}{R}$, so the power in the series case is $\frac{V^2}{2R}$ and in the parallel case is $\frac{V^2}{R/2} = 2 \times \frac{V^2}{R}$. Hence, the ratio of heat produced in series to parallel is 1:2.

Correct answer: (a) 1:2

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

Answer:
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured. This is because a voltmeter measures the potential difference between two points without affecting the current in the circuit.

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

Answer:
The resistance of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area.
The cross-sectional area $A = \pi \left( \frac{d}{2} \right)^2$, where $d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$.
So, $A = \pi \left( \frac{0.5 \times 10^{-3}}{2} \right)^2 = 1.9635 \times 10^{-7} \, \text{m}^2$.
We can now find the length $L$ for $R=10\mathrm{\Omega :}$

$$10 = \frac{1.6 \times 10^{-8} \times L}{1.9635 \times 10^{-7}}$$

Solving for $L$:

$$L = \frac{10 \times 1.9635 \times 10^{-7}}{1.6 \times 10^{-8}} = 122.0 \, \text{m}$$

If the diameter is doubled, the new area will be four times greater (since $A \propto d^2$), which will reduce the resistance by a factor of 4.

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

$I$ (amperes): 0.5, 1.0, 2.0, 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 the resistor.

Answer:
You can plot $V$ vs $I$ on a graph and determine the slope of the line. The slope gives the resistance $R$, since $R = \frac{V}{I}$. The average resistance calculated from the given data is approximately $R = \frac{V}{I}$, and it should give you the value for the resistance of the resistor.

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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:
Using Ohm’s law $V = IR$, the resistance is:

$$R = \frac{V}{I} = \frac{12 \, \text{V}}{2.5 \times 10^{-3} \, \text{A}} = 4800 \, \Omega$$

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9. A battery of 9 V is connected in series with resistors of $0.2 \, \Omega, 0.3 \, \Omega, 0.4 \, \Omega, 0.5 \, \Omega$ and $12\mathrm{\Omega .\; How\; much\; current\; would\; flow\; through\; the\; 12 }$$\Omega$ resistor?

Answer:
First, find the total resistance:

$$R_{\text{total}} = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \, \Omega$$

Now, using Ohm's law, the total current is:

$$I = \frac{V}{R_{\text{total}}} = \frac{9 \, \text{V}}{13.4 \, \Omega} = 0.67 \, \text{A}$$

Since the resistors are in series, the same current flows through all resistors, including the 12 $\Omega$ resistor.

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

Answer:
The total resistance required is:

$$R_{\text{total}} = \frac{V}{I} = \frac{220 \, \text{V}}{5 \, \text{A}} = 44 \, \Omega$$

For $n$ resistors in parallel:

$$\frac{1}{R_{\text{total}}} = n \times \frac{1}{R}$$ $$\frac{1}{44} = n \times \frac{1}{176}$$ $$n = \frac{176}{44} = 4$$

So, 4 resistors of 176 $\Omega$ are required.

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.

Answer:

(i) To obtain a resistance of 9 Ω:

• You can connect two resistors in parallel to reduce the total resistance, and then connect the third resistor in series with the parallel combination.
• When two resistors of 6 Ω each are connected in parallel, the equivalent resistance $R_{\text{eq}}$ is given by:

$$\frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

So, $R_{\text{eq}} = 3 \, \Omega$.

Now, connect this 3 Ω in series with the third resistor of 6 Ω:

$$R_{\text{total}} = 6 + 3 = 9 \, \Omega$$

(ii) To obtain a resistance of 4 Ω:

• You can connect two resistors in series to give a total resistance of 12 Ω, and then connect this combination in parallel with the third resistor (6 Ω).
• The equivalent resistance $R_{\text{eq}}$ of the two resistors in series:

$$R_{\text{eq}} = 6 + 6 = 12 \, \Omega$$

Now, connect this 12 Ω in parallel with the third resistor of 6 Ω:

$$\frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12}$$

So, $R_{\text{total}} = 4 \, \Omega$

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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?

Answer:

The power consumed by each bulb is 10 W, and the supply voltage is 220 V. We can find the current drawn by each bulb using the formula:

$$P = VI \quad \Rightarrow \quad I = \frac{P}{V}$$

For each bulb:

$$I_{\text{bulb}} = \frac{10}{220} = 0.0455 \, \text{A}$$

Now, if the maximum allowable current is 5 A, the number of bulbs that can be connected in parallel is:

$$\text{Number of bulbs} = \frac{5}{0.0455} \approx 110 \, \text{bulbs}$$

So, 110 bulbs can be connected in parallel.

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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 resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

(i) Both coils A and B used separately:

• The current in each coil is given by Ohm’s law $I = \frac{V}{R}$
• For each coil:

$$I_{\text{coil}} = \frac{220}{24} = 9.17 \, \text{A}$$

(ii) Both coils A and B connected in series:

• The total resistance in series is $R_{\text{total}} = 24 + 24 = 48 \, \Omega$
  
• The current is:

$$I_{\text{series}} = \frac{220}{48} = 4.58 \, \text{A}$$

(iii) Both coils A and B connected in parallel:

• The total resistance in parallel is:

$$\frac{1}{R_{\text{total}}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12}$$

So, $R_{\text{total}}=12\mathrm{\Omega }$

• The current is:

$$I_{\text{parallel}} = \frac{220}{12} = 18.33 \, \text{A}$$

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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.

Answer:

(i) Power in the 2 Ω resistor (in series with 1 Ω and 2 Ω resistors):

• The total resistance is $R_{\text{total}} = 1 + 2 = 3 \, \Omega$
  
• The current in the circuit is $I=\frac{V}{R}=\frac{6}{3}=2\text{A}$
• The power in the 2 Ω resistor is $P=I^{2}R=2^2\times 2=8\text{W}$

(ii) Power in the 2 Ω resistor (in parallel with 12 Ω resistor):

• The total resistance in parallel is:

$$\frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{2} = \frac{1}{12} + \frac{6}{12} = \frac{7}{12}$$

So, $R_{\text{total}}=\frac{12}{7}\approx 1.71\mathrm{\Omega }$

• The current is $I = \frac{V}{R} = \frac{4}{1.71} \approx 2.34 \, \text{A}$
  
• The power in the 2 Ω resistor is $P=I^{2}R={2.34}^2\times 2\approx 10.96\text{W}$

Thus, the power used in the 2 Ω resistor in the second case is higher.

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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?

Answer:

For the 100 W lamp:

$$I_1 = \frac{P_1}{V} = \frac{100}{220} \approx 0.4545 \, \text{A}$$

For the 60 W lamp:

$$I_2 = \frac{P_2}{V} = \frac{60}{220} \approx 0.2727 \, \text{A}$$

The total current drawn from the line is:

$$I_{\text{total}} = I_1 + I_2 = 0.4545 + 0.2727 \approx 0.7272 \, \text{A}$$

So, the total current drawn from the line is approximately 0.727 A.

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

Answer:

Energy is given by $E = P \times t$, where $P$ is the power and $t$ is the time.

• For the TV set:

$$E_{\text{TV}} = 250 \, \text{W} \times 1 \, \text{hr} = 250 \, \text{Wh}$$

• For the toaster:

$$E_{\text{Toaster}} = 1200 \, \text{W} \times \frac{10}{60} \, \text{hr} = 1200 \times \frac{1}{6} = 200 \, \text{Wh}$$

So, the TV set uses more energy (250 Wh) than the toaster (200 Wh).

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17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

The power dissipated in the heater is $P = I^2 R$, where $I=5\text{A and }$$R = 44 \, \Omega$:

$$P = 5^2 \times 44 = 25 \times 44 = 1100 \, \text{W} = 1.1 \, \text{kW}$$

Thus, the rate at which heat is developed is 1.1 kW.

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18. Explain the following:

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

• Tungsten has a high melting point (around 3422°C) and excellent conductivity, making it ideal for filaments in light bulbs. It can withstand the high temperatures needed for the filament to emit light without melting.

(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?

• Alloys have higher resistances than pure metals, which makes them better for heating purposes. The higher resistance helps generate more heat. Additionally, alloys are more durable and resistant to oxidation at high temperatures.

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

• In a series arrangement, if one device fails, the entire circuit is broken, which would turn off all devices. This is inconvenient for domestic use, where we want individual control over devices.

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

• The resistance of a wire is inversely proportional to its cross-sectional area. As the area increases, the resistance decreases, and vice versa. This relationship is given by $R = \rho \frac{L}{A}$.

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

• Copper and aluminum are good conductors of electricity and are relatively cost-effective. Copper has a higher conductivity, but aluminum is lighter and cheaper, making it a common choice for power transmission over long distances.