Question

A 60 W load is connected to the secondary coil of an ideal transformer whose primary coil draws line voltage. If a current of 0.54 A flows in the load, the current in the primary coil is

• A. 0.27 mA
• B. 2.7 A
• C. 0.27 A
• D. 10 A

Answer 1

Answer: (C)

0.27 A

Here, $P_s=60W,I_s=0.54A,V_p=220V$

$\therefore V_s=\frac{P_s}{I_s}=\frac{60}{0.54}=111V$

As $\frac{V_s}{V_p}=\frac{I_p}{I_s}.$ for an ideal transformer

$\therefore I_p=\frac{V_s}{V_p}\times I_s=\frac{111}{220}\times 0.54=0.27A$