A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW.If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are :

Options

A. 220 V and 20 A
B. 440 V and 20 A
C. 440 V and 5 A
D. 220 V and 10 A

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Output Voltage
Given the output power (P_out) and the current in the secondary coil (I_s), we can calculate the output voltage (V_out) using the formula:
\[ P_{\text{out}} = V_{\text{out}} \times I_s \]

Substituting the known values:
\[ 2200 \, \text{W} = V_{\text{out}} \times 10 \, \text{A} \]

Now, solving for \( V_{\text{out}} \):
\[ V_{\text{out}} = \frac{2200 \, \text{W}}{10 \, \text{A}} = 220 \, \text{V} \]

### Step 2: Use the Turns Ratio to Find Input Voltage
The transformer turns ratio relates the primary and secondary voltages to the number of turns in the coils:
\[ \frac{V_{\text{p}}}{V_{\text{s}}} = \frac{N_{\text{p}}}{N_{\text{s}}} \]

Where:
- \( N_{\text{p}} = 300 \) (turns in primary)
- \( N_{\text{s}} = 150 \) (turns in secondary)
- \( V_{\text{p}} \) is the input voltage
- \( V_{\text{s}} = V_{\text{out}} = 220 \, \text{V} \)

Rearranging the formula to find \( V_{\text{p}} \):
\[ V_{\text{p}} = V_{\text{s}} \times \frac{N_{\text{p}}}{N_{\text{s}}} \]

Substituting the known values:
\[ V_{\text{p}} = 220 \, \text{V} \times \frac{300}{150} \]
\[ V_{\text{p}} = 220 \, \text{V} \times 2 = 440 \, \text{V} \]

### Step 3: Calculate the Input Current
Using the power conservation in the transformer, we can find the input current (I_p) using the formula:
\[ P_{\text{out}} = V_{\text{p}} \times I_p \]

Rearranging to find \( I_p \):
\[ I_p = \frac{P_{\text{out}}}{V_{\text{p}}} \]

Substituting the known values:
\[ I_p = \frac{2200 \, \text{W}}{440 \, \text{V}} = 5 \, \text{A} \]

### Final Results
- Input Voltage \( V_{\text{p}} = 440 \, \text{V} \)
- Input Current \( I_p = 5 \, \text{A} \)

### Summary
The input voltage is 440 V and the input current is 5 A.