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A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volts are to be used from a household supply. The rating of fuse to be used is:

Correct answer: D. 10 A

  • A. 2.5 A
  • B. 5.0 A
  • C. 7.5 A
  • D. 10 A

Explanation

The total power drawn by the appliances is Ptotal = 1500 W + 500 W = 2000 W. The total current drawn from the 200 V supply is I = Ptotal / V = 2000 W / 200 V = 10 A. A 10A fuse is required.

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About Current Electricity

Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.

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