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

Answer: (B)

13.33 A

Answer 2

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 $13.33 \, \text{A}$.