Question

The current drawn by the primary of a transformer, which steps down $200\,V$ to $20\,V$ to operate a device of resistance $20\,\Omega $. is (Assume the efficiency of the transformer to be $80\, \%$)

• A. $0.125\,A$
• B. $0.225\,A$
• C. $0.325\,A$
• D. $0.425\,A$

Answer 1

Answer: (A)

Given: $V_p = 200\,V$,
$R = 20\,\Omega$,
$V_s=20\,V$, $\eta = 80\%$
Current through the secondary coil is
$I_{s}=\frac{V_{s}}{R}$
$=\frac{20\,V}{20\,\Omega}$
$=1\,A$
Efficiency of transformer $\eta=\frac{Output\, power}{Input \,power}=\frac{V_{s}I_{s}}{V_{p}I_{p}}$
or $I_{p}=\frac{V_{s}\,I_{s}}{V_{p}\,\eta}$
$=\frac{20\times1\times100}{200\times80}$
$=0.125\,A$