Question

Find the current carrying capacity of wire from meter to main distribution board having three light/fan circuits of 800 W each and two 15 A power circuits of 1.5 kW each. Take the permissible power factor as 0.8 and safety factor as 1.5?
1. 50 A
2. 30 A
3. 45 A
4. 65 A

Answer 1

Correct Answer - Option 3 : 45 A

Given,

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,

\(I=\frac{P}{Vcosϕ}= \frac{5400}{240×0.8}=28.125\ A\)

At safety factor (S_f) = 1.5, permissible current (I_p) flow through the circuit will be,

I_p = I × S_f = 28.125 × 1.5 = 42.1875 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