Question

A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is_____A.

Answer 1

To find the output current of the transformer, we can use the principles of conservation of energy and the efficiency of the transformer.

Given Values: Input power $P_{\text{in}} = V_{\text{in}} \times I_{\text{in}}$.

$V_{\text{in}} = 2300 \, V$ (input voltage).

$I_{\text{in}} = 5 \, A$ (input current).

Efficiency $\eta = 90\% = 0.9$.

Output voltage $V_{\text{out}} = 230 \, V$.

Calculating Input Power:

$P_{\text{in}} = V_{\text{in}} \times I_{\text{in}} = 2300 \, V \times 5 \, A = 11500 \, W$.

Calculating Output Power: Since the transformer is 90% efficient:

$P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 11500 \, W = 10350 \, W$.

Calculating Output Current: Using the output power and output voltage, the output current $I_{\text{out}}$ can be calculated as:

$P_{\text{out}} = V_{\text{out}} \times I_{\text{out}} \Rightarrow I_{\text{out}} = \frac{P_{\text{out}}}{V_{\text{out}}} = \frac{10350 \, W}{230 \, V}$.

Final Calculation:

$I_{\text{out}} = 45 \, A$.