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Step-by-Step Solution
Step 1: Identify the known quantities
• Primary (input) voltage, $V_p = 220\,\text{V}$
• Secondary (output) voltage, $V_s = 11\,\text{V}$
• Power of the lamp at the secondary side, $P_{\text{out}} = 44\,\text{W}$
• Assume transformer is ideal (no power loss).
Step 2: Express power conservation in an ideal transformer
For an ideal transformer, the input power equals the output power:
$P_{\text{in}} = P_{\text{out}}\,.$
Thus,
$V_p \times I_p = V_s \times I_s\,.$
Step 3: Relate the output power to the secondary current
The lamp consumes $44\,\text{W}$ on the secondary side. Hence,
$P_{\text{out}} = V_s \times I_s = 44\,\text{W}\,.$
From this, the current in the secondary circuit is
$I_s = \frac{44}{V_s} = \frac{44}{11} = 4\,\text{A}.$
Step 4: Use power equivalence to find the primary current
Since $V_p \times I_p = V_s \times I_s = 44\,\text{W}$ (because $P_{\text{out}}=44\,\text{W}$ and there are no losses),
$220 \times I_p = 44\,.$
Solving for $I_p$ gives
$I_p = \frac{44}{220} = 0.2\,\text{A}\,.$
Step 5: State the final answer
The current in the primary circuit of the transformer is $0.2\,\text{A}$.
