A 220 V input is supplied to a transformer. The output circuit draws a current of 2 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

Options

A. 5.0 A
B. 3.6 A
C. 2.8 A
D. 2.5 A

Answer

To find the current drawn by the primary windings of the transformer, we can use the formula for the efficiency of a transformer, which is given by:

\[ \text{Efficiency} (\eta) = \frac{V_s \cdot I_s}{V_p \cdot I_p} \]

Where:
- \( V_s \) = Secondary voltage
- \( I_s \) = Secondary current
- \( V_p \) = Primary voltage
- \( I_p \) = Primary current

Given:
- \( V_s = 440 \, V \)
- \( I_s = 2 \, A \)
- \( V_p = 220 \, V \)
- Efficiency \( \eta = 80\% = 0.8 \)

We can rearrange the efficiency formula to solve for \( I_p \):

\[ I_p = \frac{V_s \cdot I_s}{V_p \cdot \eta} \]

Now, substituting the known values into the equation:

1. Calculate \( V_s \cdot I_s \):
\[ V_s \cdot I_s = 440 \, V \cdot 2 \, A = 880 \, VA \]

2. Calculate \( V_p \cdot \eta \):
\[ V_p \cdot \eta = 220 \, V \cdot 0.8 = 176 \, V \]

3. Now substitute these values into the equation for \( I_p \):
\[ I_p = \frac{880 \, VA}{176 \, V} \]

4. Perform the division:
\[ I_p = 5 \, A \]

Thus, the current drawn by the primary windings of the transformer is \( 5 \, A \).

### Final Answer:
The current drawn by the primary windings of the transformer is **5 A**.

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