Question

The secondary coil of an ideal step down transformer is delivering $500W$ power at $12.5A$ current. If the ratio of turns in the primary to the secondary is $5:1$, then the current flowing (in ampere) in the primary coil will be

Answer 1

Answer: 2.50

$P=VI$
For secondary:
$V_2=\frac{P_2}{I_2}=\frac{500}{12.5}=40.V$
For an ideal transformer ( $100\%$ efficient),
$P_{\text{input }}=P_{\text{output }}$
$\Rightarrow V_1{I}_1=V_2{I}_2$
$\Rightarrow I_1=\frac{V_2{I}_2}{V_1}$
$=\frac{40(12.5)}{40\times 5}=2.5A$
$\because \frac{N_1}{N_2}=\frac{V_1}{V_2}$
$\Rightarrow \frac{5}{1}=\frac{V_1}{40}$