A transformer having efficiency of `90%` is working on `200 V` and `3 kW` power supply. If the current in the secondary coil is `6A`, the voltage across the secondary coil and current in the primary coil respectively are

Options

A. 300 V, 15 A
B. 450 V, 15 A
C. 450 V, 13.5 A
D. 600 V, 15 A

Explanation

To solve the problem step by step, we will use the concepts of transformer efficiency and the relationships between primary and secondary voltages and currents.

### Step 1: Understand the given data
- Efficiency of the transformer, \( \eta = 90\% = 0.9 \)
- Input voltage, \( V_p = 200 \, V \)
- Input power, \( P_{in} = 3 \, kW = 3000 \, W \)
- Current in the secondary coil, \( I_s = 6 \, A \)

### Step 2: Calculate the output power using efficiency
The efficiency of a transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \]
Rearranging this gives:
\[ P_{out} = \eta \times P_{in} \]
Substituting the known values:
\[ P_{out} = 0.9 \times 3000 = 2700 \, W \]

### Step 3: Relate output power to secondary voltage and current
The output power can also be expressed in terms of the secondary voltage and current:
\[ P_{out} = V_s \times I_s \]
Substituting \( P_{out} \) and \( I_s \):
\[ 2700 = V_s \times 6 \]
Solving for \( V_s \):
\[ V_s = \frac{2700}{6} = 450 \, V \]

### Step 4: Calculate the current in the primary coil
The input power can be expressed in terms of the primary voltage and current:
\[ P_{in} = V_p \times I_p \]
Rearranging gives:
\[ I_p = \frac{P_{in}}{V_p} \]
Substituting the known values:
\[ I_p = \frac{3000}{200} = 15 \, A \]

### Final Results
- Voltage across the secondary coil, \( V_s = 450 \, V \)
- Current in the primary coil, \( I_p = 15 \, A \)

### Summary
The voltage across the secondary coil and the current in the primary coil are \( 450 \, V \) and \( 15 \, A \) respectively.

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