The current drawn from the primary of a transformer which steps down 220 V to 22 V at 0.1 amp. shall be
Answer & explanation
Correct answer: option 4
Here np = 220 V, ns = 22 V, Is = 0.1 amp. and Ip = ?
$\frac{E_s}{E_p} = \frac{I_p}{I_s} = \frac{n_s}{n_p}$
$\Rightarrow I_p = \frac{E_s}{E_p} \times I_s = \frac{22}{220} \times 0.1A = 0.01A$