A transformer having efficiency of 80% is working on 200 V and 2 kW power supply. If the current in the secondary coil is 8A, the voltage across the secondary coil and the current in the primary coil respectively are

Options

A. 200 V, 10 A.
B. 450 V, 15 A
C. 450 V, 13.5 V
D. 600 V, 15 A

Explanation

To solve the problem, we need to find the voltage across the secondary coil (V2) and the current in the primary coil (I1) of the transformer. We know the following information:

1. Efficiency of the transformer (η) = 80% = 0.8
2. Input voltage (V1) = 200 V
3. Input power (P_input) = 2 kW = 2000 W
4. Current in the secondary coil (I2) = 8 A

### Step-by-Step Solution:

**Step 1: Calculate the current in the primary coil (I1)**

The input power (P_input) can be expressed as:
\[ P_{\text{input}} = V_1 \times I_1 \]

Given:
\[ P_{\text{input}} = 2000 \, \text{W} \]
\[ V_1 = 200 \, \text{V} \]

We can rearrange the formula to find I1:
\[ I_1 = \frac{P_{\text{input}}}{V_1} \]
\[ I_1 = \frac{2000}{200} = 10 \, \text{A} \]

**Step 2: Calculate the output power (P_output)**

Using the efficiency formula:
\[ \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \]

We can rearrange this to find the output power:
\[ P_{\text{output}} = \eta \times P_{\text{input}} \]
\[ P_{\text{output}} = 0.8 \times 2000 = 1600 \, \text{W} \]

**Step 3: Calculate the voltage across the secondary coil (V2)**

The output power can also be expressed as:
\[ P_{\text{output}} = V_2 \times I_2 \]

We know I2 = 8 A, so we can rearrange this to find V2:
\[ V_2 = \frac{P_{\text{output}}}{I_2} \]
\[ V_2 = \frac{1600}{8} = 200 \, \text{V} \]

### Final Answers:
- Voltage across the secondary coil (V2) = 200 V
- Current in the primary coil (I1) = 10 A

### Summary:
The voltage across the secondary coil is 200 V and the current in the primary coil is 10 A.