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

Answer

To solve the problem step by step, we will use the given information about the transformer, including its efficiency, power rating, and voltages.

### Step 1: Understand the given data
- Efficiency (η) = 80% = 0.8
- Power output (P_out) = 4 kW = 4000 W
- Primary voltage (V_p) = 100 V
- Secondary voltage (V_s) = 200 V

### Step 2: Calculate the power input (P_in)
Using the efficiency formula:
\[ \text{Efficiency} = \frac{P_{out}}{P_{in}} \]
Rearranging gives:
\[ P_{in} = \frac{P_{out}}{\text{Efficiency}} = \frac{4000 \, \text{W}}{0.8} = 5000 \, \text{W} \]

### Step 3: Calculate the primary current (I_p)
Using the formula for power:
\[ P_{in} = V_p \times I_p \]
Rearranging gives:
\[ I_p = \frac{P_{in}}{V_p} = \frac{5000 \, \text{W}}{100 \, \text{V}} = 50 \, \text{A} \]

### Step 4: Calculate the secondary current (I_s)
Using the power output formula:
\[ P_{out} = V_s \times I_s \]
Rearranging gives:
\[ I_s = \frac{P_{out}}{V_s} = \frac{4000 \, \text{W}}{200 \, \text{V}} = 20 \, \text{A} \]

### Final Answer
The primary current (I_p) is 50 A and the secondary current (I_s) is 20 A.

### Summary of the solution:
- Primary current (I_p) = 50 A
- Secondary current (I_s) = 20 A

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