The output voltage of an ideal transformer connected to 240 V a.c. mains is 24 V. When this transformer is used to light a bulb with ratings 24 V, 24 W, the primary current would be-

0.1 A

The correct answer is Option (2) → 0.1 A

The power in the secondary circuit is -

$P=V_s.I_s$

$⇒I_s=\frac{p}{V_s}=\frac{24W}{24V}=1A$

Using transformer voltage ratio,

$\frac{V_p}{V_s}=\frac{N_p}{N_s}=\frac{I_s}{I_p}$

$⇒I_p=I_s×\left(\frac{V_s}{V_p}\right)$

$=\frac{1×24}{240}=0.1A$