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A $12\ V$, $60\ W$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220\ V$. Assuming the transformer to be ideal, what is the current in the primary winding?
A
0.37 A
B
0.27 A
C
2.7 A
D
3.7 A
Explanation
$i_s=5\ A$ and $\frac{i_p}{i_s}=\frac{V_s}{V_p}$ give $i_p=0.27\ A$.
Detailed Solution
For a transformer: $\frac{i_p}{i_s}=\frac{V_s}{V_p}=\frac{N_s}{N_p}$
For the bulb, $V_s=12\ V$; $i_s=\frac{60}{12}=5\ A$
$\Rightarrow\frac{i_p}{5}=\frac{12}{220}$
$i_p=\frac{60}{220}=\frac{3}{11}=0.27\ A$