Question

A $220$ volt input is supplied to a transformer. The output circuit draws a current of $2.0$ ampere at $440$ volts. If the efficiency of the transformer is $80\%$, the current drawn by the primary windings of the transformer is

• A. 3.6 ampere
• B. 2.8 ampere
• C. 2.5 ampere
• D. 5.0 ampere

Answer 1

Answer: (D)

Here,
Input voltage, $V_{p}=220 V$
Output voltage, $V_{s}=440 V$
Input current, $I_{p}=$ ?
Output current, $I_{s}=2 A$
Efficiency of the transformer, $\eta=80 \%$
Efficiency of the transformer, $\eta=\frac{\text { Output power }}{\text { Input power }}$
$\eta=\frac{V_{s} I_{s}}{V_{p} I_{p}}$
or $I_{p}=\frac{V_{s} I_{s}}{\eta V_{p}}=\frac{(440 V )(2 A )}{\left(\frac{80}{100}\right)(220 V )}$
$=\frac{(440 V)(2 A)(100)}{(80)(220 V)}$
$=5 A$