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)

5.0 ampere

Answer 2

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, $\eta=\frac{\text{Output power}}{\text{Input power}}$
$\eta=\frac{V_sI_s}{V_pI_p}$
or $I_p=\frac{V_sI_s}{\eta V_p}=\frac{(440 \, V)(2 \, A)}{\big(\frac{80}{100}\big)(220 \, V)}$
$=\frac{(440 \, V)(2 \, A)(100)}{(80)(220 \, V)}$
$=5 A$