SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Power consumed by 2 power circuits = 1.5 kW × 2 = 3 kW = 3000 W
Power consumed by 3 light/fan sub-circuit = 800 × 3 = 2400 W
Total Power consumed by circuit (P) = 3000 W + 2400 W = 5400 W
cos ϕ = 0.8
We know that the supply voltage (V) for distribution = 240 W
Now current flow through the circuit (I) is given by,
$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}\cos \mathrm{\varphi }}$ $=\frac{5400}{240\times 0.8}$ $=28.125\mathrm{A}$
At safety factor $\left({\mathrm{S}}_{\mathit{f}}\right)=1.5$, permissible current $\left({\mathrm{I}}_{\mathrm{p}}\right)$ flow through the circuit will be,
${\mathrm{I}}_{\mathrm{p}}=\mathrm{I}\times {\mathrm{S}}_{\mathrm{f}}=28.125\times 1.5$ $=42.1875\mathrm{A}$
Since current carrying capacity of the cable should be more than the maximum permissible current.
Hence, the current-carrying capacity of cable will be 45 A