Question

8. What current flows through a 1.15 kW electric fire at a potential difference of

230 V? (Remember that 1.15 kW is 1,150 W)
9. Would it be best to use a 3A, 5A or a 13A fuse for the fire above? Explain
why.

Answer 1

I = 5[amp]

Step-by-step explanation:

Electrical power is defined as the product of voltage by current.

`P=V*I`

where:

P = power = 1150 [W]

V = voltage = 230 [V]

I = current [amp]

Now replacing:

`1150=230*I\\I=1150/230\\I=5[amp]`

A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]