Question

A transformer has an efficiency of 80% andworks at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is :

• A. 1.59 A
• B. 13.33 A
• C. 1.33 A
• D. 15.1 A

Answer 1

The Correct Option is B

Given: - Efficiency of the transformer: \( \eta = 80\% = 0.8 \) - Input power: \( P_{\text{input}} = 4 \, \text{kW} = 4000 \, \text{W} \) - Secondary voltage: \( V_{\text{secondary}} = 240 \, \text{V} \)

Step 1: Calculating the Output Power

The output power (\( P_{\text{output}} \)) is given by:

\[ P_{\text{output}} = \eta \times P_{\text{input}} \]

Substituting the given values:

\[ P_{\text{output}} = 0.8 \times 4000 \, \text{W} \] \[ P_{\text{output}} = 3200 \, \text{W} \]

Step 2: Calculating the Secondary Current

The power in the secondary coil is related to the secondary voltage and secondary current (\( I_{\text{secondary}} \)) by:

\[ P_{\text{output}} = V_{\text{secondary}} \times I_{\text{secondary}} \]

Rearranging to find \( I_{\text{secondary}} \):

\[ I_{\text{secondary}} = \frac{P_{\text{output}}}{V_{\text{secondary}}} \]

Substituting the values:

\[ I_{\text{secondary}} = \frac{3200 \, \text{W}}{240 \, \text{V}} \] \[ I_{\text{secondary}} = 13.33 \, \text{A} \]

Conclusion: The current in the secondary coil is \( 13.33 \, \text{A} \).