A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents are respectively

Options

A. 40 A, 16 A
B. 16 A, 40 A
C. 20 A, 40 A
D. 40 A, 20 A

Explanation

To solve the problem of finding the primary and secondary currents of a transformer with given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Efficiency (η) = 80% = 0.80
- Primary Power (Pp) = 4 kW = 4000 W
- Primary Voltage (Vp) = 100 V
- Secondary Voltage (Vs) = 200 V

2. **Calculate Primary Current (Ip):**
- The formula for power is given by:
\[ P = V \times I \]
- Rearranging this formula to find the primary current (Ip):
\[ I_p = \frac{P_p}{V_p} \]
- Substituting the known values:
\[ I_p = \frac{4000 \, \text{W}}{100 \, \text{V}} = 40 \, \text{A} \]

3. **Use Efficiency to Find Secondary Current (Is):**
- The efficiency of a transformer can be expressed as:
\[ \eta = \frac{V_s \times I_s}{V_p \times I_p} \]
- Rearranging to find the secondary current (Is):
\[ I_s = \frac{\eta \times V_p \times I_p}{V_s} \]
- Substituting the known values:
\[ I_s = \frac{0.80 \times 100 \, \text{V} \times 40 \, \text{A}}{200 \, \text{V}} \]
- Calculating:
\[ I_s = \frac{3200 \, \text{W}}{200 \, \text{V}} = 16 \, \text{A} \]

4. **Final Results:**
- Primary Current (Ip) = 40 A
- Secondary Current (Is) = 16 A

### Conclusion:
The primary and secondary currents are respectively **40 A** and **16 A**.

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