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)

$0.125\,A$

Answer 2

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 $I_{p}=\frac{V_{s}\,I_{s}}{V_{p}\,\eta}$ $=\frac{20\times1\times100}{200\times80}$ $=0.125\,A$