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Q. A transformer with efficiency 80% works at $4\, kW$ and $100\, V$. If the secondary voltage is $200\,V$, then the primary and secondary currents are respectively

AIIMSAIIMS 2017Alternating Current

Solution:

Here, $\eta$ = 80 %
$P_{i} $ = 4 kW = 4000 W
$V_{p}$ = 100 V, Vs = 200 V
$I_{p}=\frac{p_{i}}{V_{p}}=\frac{4000}{100}=40\,A$
$\eta=\frac{V_{s}\,I_{s}}{V_{p}\,I_{p}};\, \frac{80}{100}=\frac{200I_{s}}{4000}=\frac{I_{s}}{20}$
or $I_{s}=\frac{20\times80}{100}=16\,A$