Question

An electric circuit operating at 250 V has following appliances connected in parallel and the energy cost ₹ 5.40 per kVAh.

1. 4 bulbs of 100 W each, working 8 hours a day.
2. Two room heaters of resistance 50 Ω each, working 2 hours a day.
3. Two air conditioners drawing a current of 8A, working 6 hours a day.
4. Three coolers of `1/3` H.P each, working 10 hours a day.
   1. Calculate the monthly bill
   2. Calculate the minimum fuse rating.

Answer 1

(a) Power of 4 bulbs = 4 × 100 = 400 W

∴ Energy consumed by the bulbs in a day = 400 W × 8 h = 3200 Wh

\[\text{(b) Power of one heater }\text{‘}\mathrm{P}\text{’} = \frac{\mathrm{V}^2}{\mathrm{R}}\]

\[= \frac{250 \times 250}{50}\]

= 1250 W

∴ Power of 2 heaters = 1250 W × 2 = 2500 W

∴ Energy consumed by 2 heaters in a day = 2500W × 2h = 5000 Wh

(c) Power of one air conditioner, P = I × V = 8 × 250 = 2000 W

∴ Power of 2 air conditioners = 2000 W × 2 = 4000 W

∴ Energy consumed by two air conditioners in a day = 4000 W × 6h = 24000 Wh.

\[\text{(d) Power of 3 coolers} = 3 \times \frac{1}{3}\text{HP} = 1\text{HP} = 750\text{W}\]

∴ Energy consumed by 3 coolers in a day = 750 W × 10 h = 7500 Wh.

(i) Total energy consumed by all appliances in a day = (3200 + 5000 + 24000 + 7500) Wh

= 39700 Wh

= 39.7 kWh

∴ Total energy consumed in one month = 39.7 kWh × 30 = 1191 kWh

\[\therefore \text{Monthly bill} = 1191\text{kWh} \times \frac{\text{₹}5.40}{\text{kWh}} = \text{₹}6431.40\]

(ii) Total power of appliances in circuit = 400 W + 2500 W + 4000 W + 750 W = 7650 W
\[ \text {∴ Minimum fuse rating }=\frac{\text{P}}{\text{~V}}\]

\[=\frac{7650}{250}\]

\[=30.6\]

\[\simeq {31 A}\]