Question

an electric heater consumes energy at a rate of 840W when heating is at maximum rate the voltage supply in the circuit is 220V what rating of fuse one should use in this circuit ​

Answer 1

Answer:

When heating at max rate,

power, P = 840W

V = 220V

We know P = VI

⇒ 840 = 220 × I

⇒ I = 840/220

⇒ I = 4 A

R = V/I = 220/4 = 55 Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,

power, P = 360W

V = 220V

We know P = VI

⇒ 360 = 220 × I

⇒ I = 360/220

⇒ I = 1.636 A

R = V/I = 220/1.636 = 134.45 Ω

So current = 1.636A and resistance = 134.45Ω

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Answer 2

Calculation of Fuse Rating

Explanation:

Fuse Rating is defined as the rate of energy consumed to the maximum rate of voltage supply in the circuit multiplied by the 1.25.

Fuse Rating = (Watts / Volts) * 1.25

                    = (840 / 220) * 1.25

                    = 4.772

After calculation, the next highest fuse rating to be used. The calculated fuse rating is 4.772 amps, so use a 5 amp fuse.

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