Question

A power transmission line feeds input power at $2300\, V$ to a step down transformer with its primary windings having $4000$ turns, giving the output power at $230\, V$. If the current in the primary of the transformer is $5\, A$, and its efficiency is $90\%$, the output current would be :

• A. $50\, A$
• B. $45\, A$
• C. $25\, A$
• D. $20\, A$

Answer 1

The Correct Option is B

To solve this problem, we will use the principles of transformer operation along with the efficiency formula.

Given data:

• Primary voltage, \( V_p = 2300 \, \text{V} \)
• Primary turns, \( N_p = 4000 \)
• Secondary voltage, \( V_s = 230 \, \text{V} \)
• Primary current, \( I_p = 5 \, \text{A} \)
• Efficiency, \(\eta = 90\% = 0.9\)

First, we calculate the power input to the transformer using:

\(P_{\text{in}} = V_p \times I_p = 2300 \times 5 = 11500 \, \text{W}\)

The efficiency of the transformer is given by the formula:

\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\)

Rearranging to find the output power \( P_{\text{out}} \):

\(P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 11500 = 10350 \, \text{W}\)

Next, we find the output current \( I_s \) using the formula for power in terms of voltage and current:

\(P_{\text{out}} = V_s \times I_s\)

Substitute the known values to find \( I_s \):

\(I_s = \frac{P_{\text{out}}}{V_s} = \frac{10350}{230} = 45 \, \text{A}\)

Therefore, the output current is \(45\, A\). Hence, the correct answer is:

• \(45\, A\)