(D) The function of a fuse is to protect electric circuits and appliances by stopping the flow of excessively high electric current. It is connected in series with the live wire in a domestic circuit.
$(b)$ Given: Power $(P) = 1.5 \, kW = 1500 \, W$,Voltage $(V) = 220 \, V$.
The current $(I)$ drawn by the electric iron is calculated as:
$I = \frac{P}{V} = \frac{1500 \, W}{220 \, V} \approx 6.82 \, A$.
Since the current required by the electric iron $(6.82 \, A)$ is significantly higher than the fuse rating $(3 \, A)$,the fuse wire will get heated and melt due to the excessive current,breaking the circuit. Consequently,the electric iron will stop working.