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  4. A 60 W load is connected to the secondary of an ideal transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, the current in the primary coil is

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Q. A $60\,W$ load is connected to the secondary of an ideal transformer whose primary draws line voltage. If a current of $0.54\,A$ flows in the load, the current in the primary coil is

Alternating Current

Solution:

Here, $P_s = 60\,W$,
$ I_s = 0.54\, A$,
$V_p = 220\,V$
$\therefore V_{s}=\frac{P_{s}}{I_{s}}$
$\frac{60}{0.54}$
$=111\,V$
As $\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}}$, for an ideal transformer
$\therefore I_{p}=\frac{V_{s}}{V_{p}}\times I_{s}$
$=\frac{111}{220}\times0.54 = 0.27\,A$