Question

A step-down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit

• A. 4 A
• B. 0.2 A
• C. 0.4 A
• D. 2 A

Answer 1

The Correct Option is B

To solve this problem, we need to apply the concept of a transformer and the relationship between power, voltage, and current. A transformer consists of two coils, known as the primary and secondary coils, and operates on the principle of electromagnetic induction.

Given:

• Primary voltage, \(V_p = 220 \, \text{V}\) 
• Secondary voltage, \(V_s = 11 \, \text{V}\)
• Power consumed by the lamp, \(P = 44 \, \text{W}\)
• Power losses in the transformer are ignored.

We need to find the current in the primary circuit, \(I_p\).

Solution Steps:

1. First, calculate the current in the secondary circuit, \(I_s\), using the formula for power:

\(P = V_s \times I_s\)

1. Rearranging the formula to solve for \(I_s\) gives:

\(I_s = \frac{P}{V_s} = \frac{44 \, \text{W}}{11 \, \text{V}} = 4 \, \text{A}\)

1. Using the formula for an ideal transformer where power in the primary is equal to power in the secondary (\(P_p = P_s\)), we have:

\(V_p \times I_p = V_s \times I_s\)

1. Substitute the known values to solve for \(I_p\):

\(220 \, \text{V} \times I_p = 11 \, \text{V} \times 4 \, \text{A}\)

\(220 \, \text{V} \times I_p = 44 \, \text{W}\)

1. Rearrange to find \(I_p\):

\(I_p = \frac{44 \, \text{W}}{220 \, \text{V}} = 0.2 \, \text{A}\)

Thus, the current in the primary circuit is 0.2 A. Therefore, the correct answer is 0.2 A.