Question

The secondary coil of an ideal step down transformer is delivering $500\, W$ power at $12.5 \,A$ 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

$P=V I$
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}=2.5 A $
$\because \frac{N_{1}}{N_{2}}=\frac{V_{1}}{V_{2}}$
$ \Rightarrow \frac{5}{1}=\frac{V_{1}}{40}$