(D) Reasons for parallel connection:
$(i)$ Each appliance receives the same potential difference (voltage) as the supply source.
$(ii)$ If one appliance fails or is switched off,the others continue to operate independently.
$(b)$ Calculation:
Given: Fuse rating $= 5 \, A$,Power $(P) = 1.5 \, kW = 1500 \, W$,Voltage $(V) = 220 \, V$.
Using the formula $I = \frac{P}{V}$:
$I = \frac{1500}{220} \approx 6.82 \, A$.
Since the current required by the heater $(6.82 \, A)$ is greater than the fuse rating $(5 \, A)$,the fuse will melt (blow off) and the circuit will break.
$(c)$ Required change:
$A$ fuse with a higher rating,typically $10 \, A$,should be installed to accommodate the current drawn by the heater.