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 basic principles of transformers and the conservation of power. The given question involves a step-down transformer, which is used to reduce the voltage from 220 V (primary) to 11 V (secondary) for a lamp with a power rating of 44 W.

The relationship of power in ideal transformers (ignoring losses) is such that the input power (Pin) is equal to the output power (Pout). Mathematically, this is represented as:

\(P_{\text{in}} = P_{\text{out}}\)

The power on the secondary side (Pout) is given by:

\(P_{\text{out}} = V_s \cdot I_s\)

where \(V_s = 11 \, \text{V}\) is the secondary voltage and \(I_s\) is the secondary current.

We know the lamp uses 44 W, hence:

\(44 \, \text{W} = 11 \, \text{V} \cdot I_s\)

Solving for \(I_s\), we find:

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

Since the power on both sides must be equal for an ideal transformer, it follows:

\(P_{\text{in}} = V_p \cdot I_p = V_s \cdot I_s\)

Given \(V_p = 220 \, \text{V}\), we need to find \(I_p\):

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

Solving for \(I_p\), we have:

\(I_p = \frac{11 \, \text{V} \cdot 4 \, \text{A}}{220 \, \text{V}} = \frac{44}{220} = 0.2 \, \text{A}\)

Hence, the current in the primary circuit is 0.2 A, which matches the given correct answer. This supports choice b (0.2 A).