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

To solve the problem regarding the transformer's efficiency and current in the secondary coil, we need to apply the basic principles of transformers and efficiency calculations. 

Given:

• Efficiency (\(\eta\)) = 80% = 0.8
• Primary voltage (\(V_p\)) = 10 V
• Power (\(P_p\)) = 4 kW = 4000 W (since 1 kW = 1000 W)
• Secondary voltage (\(V_s\)) = 240 V

The efficiency formula related to transformers is given by:

\(\eta = \frac{P_s}{P_p}\),

where \(P_s\) is the power in the secondary coil.

First, calculate the power in the secondary coil:

\(P_s = \eta \times P_p = 0.8 \times 4000 \, \text{W} = 3200 \, \text{W}\)

Next, we find the current in the secondary coil using the formula:

\(P_s = V_s \times I_s\)

where \(I_s\) is the current in the secondary coil.

Rearrange the formula to solve for \(I_s\):

\(I_s = \frac{P_s}{V_s} = \frac{3200 \, \text{W}}{240 \, \text{V}} = 13.33 \, \text{A}\)

Thus, the current in the secondary coil is 13.33 A.

Therefore, the correct answer is:

13.33 A

Justification for other options:

• 1.59 A, 1.33 A, 15.1 A: These values do not satisfy the power and efficiency equations used in the transformer calculations under the given conditions.

Remember, the power and efficiency calculations should be consistent with the transformer's behavior and the given conditions of primary and secondary voltages.

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 \( 13.33 \, \text{A} \).