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A boys hostel has following appliances when energy is supplied at 200 V and costs ₹ 5.25 per kWh. (a) 40 bulbs of 100 W each, working 8 hours a day.

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Question

A boys hostel has following appliances when energy is supplied at 200 V and costs ₹ 5.25 per kWh.

  1. 40 bulbs of 100 W each, working 8 hours a day.
  2. 20 fans each drawing a current 0.8 A and working 15 hours a day.
  3. Two T.V. sets, each offering a resistance of 200 Ω and working 4 hours a day.
  4. Two electric motors of 1.5 H.P each and working 4 hours a day
    1. Calculate the monthly bill
    2. Amongst the fuse of 48 A and 50 A which one you will use and why?
Numerical
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Solution

(a) Power of 40 bulbs \[ = 40 \times 100 \] \[ = 4000 \text{ W} \]

∴ Energy consumed by the bulbs in a day \[ = 4000 \text{ W} \times 8 \text{ h} \]

\[ = 32000 \text{ Wh} \]

\[ = 32 \text{ kWh} \]

(b) Power of one fan \[ = V \times I \]

\[ = 200 \times 0.8 \]

\[ = 160 \text{ W} \]

∴ Power of 20 fans \[ = 20 \times 160 \] \[ = 3200 \text{ W} \]

∴ Energy consumed by the fans in a day \[ = 3200 \text{ W} \times 15 \text{ h} \]

\[ = 48000 \text{ Wh} \]

\[ = 48 \text{ kWh} \]

(c) Power of one T.V. set \[ = \frac{V^2}{R} \]

\[ = \frac{200 \times 200}{200} \]

\[ = \frac{40000}{200} \]

\[ = 200 \text{ W} \]

∴ Power of 2 T.V. sets \[ = 2 \times 200 \] \[ = 400 \text{ W} \]

∴ Energy consumed by the T.V. sets in a day \[ = 400 \text{ W} \times 4 \text{ h} \]

\[ = 1600 \text{ Wh} \]

\[ = 1.6 \text{ kWh} \]

(d) Power of one motor \[ = 1.5 \times 750 \] \[ = 1125 \text{ W} \]

∴ Power of 2 motors \[ = 2 \times 1125 \] \[ = 2250 \text{ W} \]

∴ Energy consumed by the motors in a day \[ = 2250 \text{ W} \times 4 \text{ h} \] \[ = 9000 \text{ Wh} \] \[ = 9 \text{ kWh} \]

(i) Total energy consumed in a day \[ = 32 + 48 + 1.6 + 9 \] \[ = 90.6 \text{ kWh} \]

∴ Total energy consumed in one month \[ = 90.6 \times 30 \] \[ = 2718 \text{ kWh} \]

∴ Monthly bill \[ = 2718 \times 5.25 \] \[ = \text{₹ } 14269.50 \]

(ii) Total power of appliances \[ = 4000 + 3200 + 400 + 2250 \] \[ = 9850 \text{ W} \]

∴ Minimum fuse rating \[ = \frac{P}{V} \]

\[ = \frac{9850}{200} \]

\[ = 49.25 \text{ A} \]

Since the required fuse rating is 49.25 A, a 48 A fuse would melt under the normal load. Therefore, use the 50 A fuse.

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Chapter 8: Electric Energy, Power and Household Circuits - NUMERICAL PROBLEMS ON ELECTRIC ENERGY [Page 186]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 8 Electric Energy, Power and Household Circuits
NUMERICAL PROBLEMS ON ELECTRIC ENERGY | Q 1. | Page 186
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