Question

An electric power line having total resistance of \(2\,\Omega\), delivers \(1\,\text{kW}\) of power at \(250\ \text{V}\). The percentage efficiency of the transmission line is _________.

• A. 92.5
• B. 96.9
• C. 86.5
• D. 100

Hint:

Transmission efficiency improves when current is low, which reduces \(I^2R\) losses.

Answer 1

The Correct Option is B

To determine the percentage efficiency of the transmission line, we need to calculate both the input power and the output power, and then use the efficiency formula. 

• First, let's calculate the output power, which is given as 1 kW (i.e., 1000 W).

This power is delivered at 250 V. From the formula of power, we know:

\(P = V \times I\)

where \(P\) is the power, \(V\) is the voltage, and \(I\) is the current. Rearranging the formula to solve for current, we have:

\(I = \frac{P}{V} = \frac{1000}{250} = 4 \, \text{A}\)

• Now, let's calculate the input power to the transmission line. The power loss in the line due to resistance can be calculated using the formula for power loss:

\(P_{\text{loss}} = I^2 \times R = 4^2 \times 2 = 16 \times 2 = 32 \, \text{W}\)

• This loss occurs in the line itself due to its resistance of \(2 \, \Omega\).

The total input power is the sum of the output power and the power loss:

\(P_{\text{input}} = P_{\text{output}} + P_{\text{loss}} = 1000 + 32 = 1032 \, \text{W}\)

Finally, the efficiency of the transmission line can be calculated using the efficiency formula:

\(\text{Efficiency} = \left( \frac{P_{\text{output}}}{P_{\text{input}}} \right) \times 100 = \left( \frac{1000}{1032} \right) \times 100 \approx 96.9\%\)

• Therefore, the percentage efficiency of the transmission line is approximately \(96.9\%\).