Question

. (a) Define electric power. Express it in terms of potential difference V and resistance R.
(b) An electrical fuse is rated at 2 A. What is meant by this statement?
(c) An electric iron of 1 kW is operated at 220 V. Find which of the following fuses that respectively rated at 1 A,3 A and 5 A can be used in it.
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Answer 1

(a) Define electric power. Express it in terms of potential difference V and resistance R.

Electric power is defined as the rate at which energy is dissipated in a circuit.

We know,

P = work done/time taken

P = energy dissipated/time

P = I²Rt/t

Also,

I = V/R (From ohm's law)

P = (V/R)² x R x t/t

P = V²/R² x R x t/t

P = V²/R

(b) An electrical fuse is rated at 2 A. What is meant by this statement?

This statement means, the maximum current which can flow through the circuit is 2A. If it exceeds, the temperature of the fuse will rise. This will melt the fuse and break the circuit.

(c)

P = 1kW (1kW = 1000 W)

V = 220V

I = P/V

I = 1000/220

I = 4.5A

In the given circuit, a current of 4.5A will flow. Therefore a 5A fuse must be used.

Answer 2

Answer:

(a) Define electric power. Express it in terms of potential difference V and resistance R.

Electric power is defined as the rate at which energy is dissipated in a circuit.

We know,

P = work done/time taken

P = energy dissipated/time

P = I²Rt/t

Also,

I = V/R (From ohm's law)

P = (V/R)² x R x t/t

P = V²/R² x R x t/t

P = V²/R

(b) An electrical fuse is rated at 2 A. What is meant by this statement?

This statement means, the maximum current which can flow through the circuit is 2A. If it exceeds, the temperature of the fuse will rise. This will melt the fuse and break the circuit.

(c)

P = 1kW (1kW = 1000 W)

V = 220V

I = P/V

I = 1000/220

I = 4.5A

In the given circuit, a current of 4.5A will flow. Therefore a 5A fuse must be used.