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Current Electricity question

2014 · Shift 0 · Q57
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Current Electricity question

2014 · Shift 0 · Q57

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In a large building, three are 151515 bulbs of 40W40W40W, 555 bulbs of 100W100W100W, 555 fans of 80W80W80W and 111 heater of 1kW.1kW.1kW. The voltage of electric mains is 220V.220V.220V. The minimum capacity of the main fuse of the building will be:
  1. A
    8A8A8A
  2. B
    10A10A10A
  3. C
    12A12A12A
  4. D
    14A14A14A
View written solutionFree

Correct answer: C

  1. Calculate total power consumed

There are:

  • 151515 bulbs of 40 W40\,\text{W}40W
  • 555 bulbs of 100 W100\,\text{W}100W
  • 555 fans of 80 W80\,\text{W}80W
  • 111 heater of 1 kW=1000 W1\,\text{kW} = 1000\,\text{W}1kW=1000W

So total power is

P=15×40+5×100+5×80+1000P = 15\times 40 + 5\times 100 + 5\times 80 + 1000P=15×40+5×100+5×80+1000

Now compute each term:

15×40=600 W15\times 40 = 600\,\text{W}15×40=600W 5×100=500 W5\times 100 = 500\,\text{W}5×100=500W 5×80=400 W5\times 80 = 400\,\text{W}5×80=400W

Therefore,

P=600+500+400+1000=2500 WP = 600 + 500 + 400 + 1000 = 2500\,\text{W}P=600+500+400+1000=2500W
  1. Use the relation between power, voltage, and current

We know,

P=VIP = VIP=VI

So,

I=PV=2500220I = \frac{P}{V} = \frac{2500}{220}I=VP​=2202500​ I≈11.36 AI \approx 11.36\,\text{A}I≈11.36A
  1. Find the minimum fuse capacity

The fuse should have a rating just above the working current. Since the current is about 11.36 A11.36\,\text{A}11.36A, the minimum suitable fuse capacity from the given options is:

12 A12\,\text{A}12A
  1. Check options
  • 8 A8\,\text{A}8A: too small
  • 10 A10\,\text{A}10A: too small
  • 12 A12\,\text{A}12A: suitable
  • 14 A14\,\text{A}14A: works, but not minimum

Hence, the correct option is C.

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