Solution:

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 CORRECT

Solution:

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

• (a) $I^2R$
• (b) $IR^2$- CORRECT
• (c) $VI$
• (d) $V^2/R$

Solution:

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

(a) 100W

(b) 75W

(c) 50W

(d) 25W CORRECT

Solution:

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 CORRECT

(c) 1:4

(d) 4:1

Solution:

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

To measure the potential difference between two points in a circuit, the voltmeter is connected in parallel with the two points. This means that one terminal of the voltmeter is connected to one point of the circuit and the other terminal is connected to the second point where the potential difference is to be measured.

The voltmeter should have a high resistance so that it does not draw significant current from the circuit, which could affect the measurement. The voltmeter displays the potential difference between the two points in volts (V).

Solution:

Given:

Diameter of the copper wire, $d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$

Resistivity of copper, $\rho = 1.6 \times 10^{-8} \, \Omega \, \text{m}$

Resistance, $R = 10 \, \Omega$

Formula for resistance of a wire:

$R = \frac{\rho L}{A}$

Where:

• $R$ is the resistance
• $\rho$ is the resistivity
• $L$ is the length of the wire
• $A$ is the cross-sectional area of the wire

Step 1: Calculate the cross-sectional area of the wire:

Since the wire is cylindrical, the cross-sectional area is:

$A = \pi \left(\frac{d}{2}\right)^2$

$A = \pi \left(\frac{0.5 \times 10^{-3}}{2}\right)^2 = \pi \times (0.25 \times 10^{-3})^2 = 1.9635 \times 10^{-7} \, \text{m}^2$

Step 2: Calculate the length of the wire:

Rearranging the formula for resistance:

$L = \frac{R A}{\rho}$

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

$L = \frac{1.9635 \times 10^{-6}}{1.6 \times 10^{-8}} = 122.72 \, \text{m}$

Step 3: Find how the resistance changes when the diameter is doubled:

If the diameter is doubled, $d' = 2d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$

The new cross-sectional area is:

$A' = \pi \left(\frac{d'}{2}\right)^2 = \pi \left(\frac{1.0 \times 10^{-3}}{2}\right)^2 = 7.854 \times 10^{-7} \, \text{m}^2$

The new resistance is:

$R' = \frac{\rho L}{A'}$

$R' = \frac{1.6 \times 10^{-8} \times 122.72}{7.854 \times 10^{-7}}$

$R' = \frac{1.9635 \times 10^{-6}}{7.854 \times 10^{-7}} = 2.5 \, \Omega$

Conclusion on change in resistance:

When the diameter of the wire is doubled, the resistance decreases to $2.5 \, \Omega$ from $10 \, \Omega$.

Solution:

Values of Current (I) and Potential Difference (V)

Calculation of Resistance

The formula for resistance is given by Ohm's Law:

$V = I \times R$, where $V$ is the potential difference, $I$ is the current, and $R$ is the resistance.

Rearranging the formula to calculate resistance: $R = \frac{V}{I}$

Calculation for each pair of values:

• For $I = 0.5$ A, $V = 1.6$ V: $R = \frac{1.6}{0.5} = 3.2 \, \Omega$
• For $I = 1.0$ A, $V = 3.4$ V: $R = \frac{3.4}{1.0} = 3.4 \, \Omega$
• For $I = 2.0$ A, $V = 6.7$ V: $R = \frac{6.7}{2.0} = 3.35 \, \Omega$
• For $I = 3.0$ A, $V = 10.2$ V: $R = \frac{10.2}{3.0} = 3.4 \, \Omega$
• For $I = 4.0$ A, $V = 13.2$ V: $R = \frac{13.2}{4.0} = 3.3 \, \Omega$

Resistance Calculation Summary

The resistance for each pair of values is approximately constant, with values around $3.2 \, \Omega$ to $3.4 \, \Omega$. Hence, the resistance of the resistor is approximately $3.4 \, \Omega$.

Solution:

Given:

Voltage (V) = 12 V

Current (I) = 2.5 mA = 2.5 × 10⁻³ A

Formula:

Using Ohm's Law: $V = IR$

Calculation:

Rearranging the formula to find resistance: $R = \frac{V}{I}$

Substitute the values: $R = \frac{12}{2.5 \times 10^{-3}}$

$R = 4800 \, \Omega$

Answer:

The value of the resistance is $4800 \, \Omega$.

Solution:

Given Data:

Battery Voltage (V) = 9 V

Resistors in series: $R_1 = 0.2 \, \Omega$, $R_2 = 0.3 \, \Omega$, $R_3 = 0.4 \, \Omega$, $R_4 = 0.5 \, \Omega$, $R_5 = 12 \, \Omega$

Step 1: Calculate the total resistance in the circuit

The total resistance in a series circuit is the sum of individual resistances.

Therefore, $R_{total} = R_1 + R_2 + R_3 + R_4 + R_5$

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

Step 2: Apply Ohm's Law to calculate the current

Ohm's Law: $I = \frac{V}{R_{total}}$

Substitute the values: $I = \frac{9}{13.4}$

Thus, $I = 0.67$ A (approximately)

Step 3: Current through the 12 Ω resistor

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

Therefore, the current through the 12 Ω resistor is $0.67$ A.

Solution:

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

The total current $I = 5A$ and the applied voltage $V = 220V$.

Using Ohm's law, $V = IR$, where $R$ is the total resistance.

We can calculate the total resistance $R_{total}$ as:

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

Now, if $n$ resistors of resistance $R = 176 \, \Omega$ are connected in parallel, the total resistance $R_{total}$ for parallel resistors is given by:

$\frac{1}{R_{total}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} = \frac{n}{R}$

Thus, $R_{total} = \frac{R}{n}$.

We know that $R_{total} = 44 \, \Omega$ and $R = 176 \, \Omega$, so:

$44 = \frac{176}{n}$

Solving for $n$:

$n = \frac{176}{44} = 4$

Therefore, 4 resistors of 176 Ω each are required to carry 5 A on a 220 V line.

Solution:

Connecting Three Resistors to Achieve Desired Resistances

(i) To achieve a total resistance of 9 Ω:

Connect the three 6 Ω resistors in series.

The total resistance $R_{total}$ in series is given by:

$R_{total} = R_1 + R_2 + R_3 = 6 + 6 + 6 = 18 \, \Omega$

However, to get 9 Ω, connect two resistors in series and the third resistor in parallel with the combined series resistors. The formula for the equivalent resistance of two resistors in series and one resistor in parallel is:

Let the two resistors in series have a combined resistance $R_{series} = 6 + 6 = 12 \, \Omega$.

The total resistance $R_{total}$ when this is in parallel with the third resistor is:

$\frac{1}{R_{total}} = \frac{1}{R_{series}} + \frac{1}{R_3}$

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

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

$R_{total} = \frac{12}{3} = 9 \, \Omega$

(ii) To achieve a total resistance of 4 Ω:

Connect two resistors in parallel, and then connect this combination in series with the third resistor.

The formula for the total resistance of two resistors in parallel is:

$\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_{parallel}} = \frac{1}{6} + \frac{1}{6}$

$\frac{1}{R_{parallel}} = \frac{2}{6} = \frac{1}{3}$

$R_{parallel} = 3 \, \Omega$

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

$R_{total} = R_{parallel} + R_3 = 3 + 6 = 4 \, \Omega$

Solution:

Given Data:

Voltage across each lamp, $V = 220$ V

Power rating of each lamp, $P = 10$ W

Maximum allowable current, $I_{max} = 5$ A

Step 1: Calculate the current drawn by each lamp

Using the formula for power, $P = V \times I$, the current drawn by each lamp is:

$I = \frac{P}{V} = \frac{10}{220} = 0.0455$ A

Step 2: Calculate the total number of lamps that can be connected in parallel

The total current drawn by $n$ lamps connected in parallel is:

$I_{total} = n \times I$

Since the total current must not exceed the maximum allowable current, $I_{total} \leq I_{max}$:

$n \times 0.0455 \leq 5$

$n \leq \frac{5}{0.0455} \approx 109.89$

Step 3: Final Answer

The maximum number of lamps that can be connected in parallel is approximately 109.

Solution:

Case 1: Coils A and B used separately

The resistance of coil A = 24 Ω and the resistance of coil B = 24 Ω. The voltage across each coil is 220 V.

Using Ohm's law, $I = \frac{V}{R}$:

For coil A: $I_A = \frac{220}{24} = 9.17 \, \text{A}$.

For coil B: $I_B = \frac{220}{24} = 9.17 \, \text{A}$.

Case 2: Coils A and B connected in series

The total resistance in series is $R_{\text{total}} = R_A + R_B = 24 + 24 = 48 \, \Omega$.

Using Ohm's law: $I = \frac{V}{R_{\text{total}}}$.

The current in the series combination is $I = \frac{220}{48} = 4.58 \, \text{A}$.

Case 3: Coils A and B connected in parallel

The total resistance in parallel is given by the formula: $ \frac{1}{R_{\text{total}}} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12} $.

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

Using Ohm's law: $I = \frac{V}{R_{\text{total}}}$.

The current in the parallel combination is $I = \frac{220}{12} = 18.33 \, \text{A}$.

Solution:

Comparison of Power Used in the 2 Ω Resistor

Consider the two circuits:

Circuit (i): A 6 V Battery in Series with 1 Ω and 2 Ω Resistors

In this circuit, the total resistance $R_{total}$ is the sum of the resistances of the 1 Ω and 2 Ω resistors, since they are in series:

$R_{total} = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega$

The current $I$ in the circuit can be calculated using Ohm's law:

$I = \frac{V}{R_{total}} = \frac{6 \, V}{3 \, \Omega} = 2 \, A$

The power used in the 2 Ω resistor is given by:

$P = I^2 \cdot R = (2 \, A)^2 \cdot 2 \, \Omega = 4 \cdot 2 = 8 \, W$

Circuit (ii): A 4 V Battery in Parallel with 12 Ω and 2 Ω Resistors

In this circuit, the resistors 12 Ω and 2 Ω are in parallel. The total resistance $R_{total}$ of two resistors in parallel is given by:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{12 \, \Omega} + \frac{1}{2 \, \Omega}$

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

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

The total current $I$ supplied by the battery is:

$I = \frac{V}{R_{total}} = \frac{4 \, V}{1.71 \, \Omega} \approx 2.34 \, A$

Now, the current flowing through the 2 Ω resistor can be calculated using the current division rule. The current $I_2$ through the 2 Ω resistor is:

$I_2 = I \cdot \frac{R_1}{R_1 + R_2} = 2.34 \cdot \frac{12}{12 + 2} = 2.34 \cdot \frac{12}{14} \approx 2.00 \, A$

The power used in the 2 Ω resistor is:

$P = I_2^2 \cdot R = (2.00 \, A)^2 \cdot 2 \, \Omega = 4 \cdot 2 = 8 \, W$

Solution:

Current Drawn from the Line

Given:

• Power of the first lamp, $P_1 = 100$ W
• Power of the second lamp, $P_2 = 60$ W
• Supply voltage, $V = 220$ V

For the first lamp, using the formula for power:

$P = VI$

Where $P$ is the power, $V$ is the voltage, and $I$ is the current.

The current drawn by the first lamp is:

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

For the second lamp, using the same formula:

The current drawn by the second lamp is:

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

Since the lamps are connected in parallel, the total current drawn from the line is the sum of the individual currents:

Total current, $I_{total} = I_1 + I_2 = 0.4545 + 0.2727 = 0.7272$ A

Solution:

Energy Consumption Comparison

To calculate the energy used by each appliance, we use the formula:

Energy (in watt-hours) = Power (in watts) × Time (in hours)

For the TV set:

Power of TV = 250 W

Time for TV = 1 hour

Energy used by TV = $250 \, W \times 1 \, h = 250 \, Wh$

For the toaster:

Power of toaster = 1200 W

Time for toaster = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

Energy used by toaster = $1200 \, W \times \frac{1}{6} \, h = 200 \, Wh$

Conclusion:

The 250 W TV set uses 250 Wh of energy in 1 hour, while the 1200 W toaster uses 200 Wh of energy in 10 minutes. Hence, the TV set uses more energy.

Solution:

Given:

Resistance of the heater, $R = 44 \, \Omega$

Current drawn, $I = 5 \, \text{A}$

Time, $t = 2 \, \text{hours} = 2 \times 3600 \, \text{seconds} = 7200 \, \text{seconds}$

Formula:

The rate at which heat is developed in the heater is given by the formula:

$P = I^2 R$

Calculation:

Substitute the values of $I$ and $R$ into the formula:

$P = (5)^2 \times 44$

$P = 25 \times 44$

$P = 1100 \, \text{W}$

The rate at which heat is developed in the heater is $1100 \, \text{W}$.

Solution:

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

The filament of electric lamps is made of tungsten because it has a very high melting point (about 3400°C), which allows it to withstand the high temperatures generated when current passes through it. Tungsten is also a good conductor of electricity and has excellent durability, making it ideal for use in 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?

The conductors in electric heating devices like bread-toasters and electric irons are made from alloys because alloys have higher resistance compared to pure metals. This high resistance helps in converting electrical energy into heat. Alloys such as nichrome (a mixture of nickel and chromium) are commonly used because they can withstand high temperatures without oxidizing or breaking down, ensuring the device's longevity and efficiency.

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

Series arrangements are not used in domestic circuits because if one device or appliance in the circuit fails or is turned off, the entire circuit would be interrupted, causing all appliances to stop working. In a parallel arrangement, however, each device operates independently, and the failure of one device does not affect the others, providing greater convenience and safety.

(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 area of cross-section. This means that as the area of the wire's cross-section increases, the resistance decreases. This is because a larger cross-sectional area allows more electrons to flow through, reducing the opposition to the flow of current.

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

Copper and aluminium wires are commonly used for electricity transmission because both are excellent conductors of electricity. Copper has low resistance and high conductivity, making it ideal for efficient power transmission. Aluminium, being lighter and cheaper than copper, is also widely used, especially for long-distance transmission, as it offers a good balance of conductivity and cost-effectiveness.