Question

Resistance physics?

In this problem, you will looks at the resistances and currents in an electric light bulb and kettle.
1.) An electric light bulb is rated at 150 W , and an electric kettle at 2.5 kW .
Calculate the resistance of the light bulb when operating on a main supply of average voltage 240 V .
Give your answer in Ohms (Ω).

2.) Calculate the resistance of the kettle when operating on a main supply of average voltage 240 V .
Give your answer in Ohms (Ω).

3.) What current is drawn from the supply by the bulb in Part A?
Give your answer in Amps.

4.) What current is drawn from the supply by the kettle in Part A?
Give your answer in Amps.

Answer 1

To solve this problem set,you need to know about a formula,

Power (`P`) dissipated across a resistance of `r` due to application of a voltage difference of `V` is related as, `P=V^2/r`

Or, `r=V^2/P`

Now,for the bulb given, `V=240,P=150`

So,`r _ b = (240×240)/150=384 ohm`

And,for the kettle `V=240,P=2.5×10^3`

So, `r_k =(240×240)/(2.5×10^3)=23.04 ohm`

Now,we know from Ohm's Law, `V=Ir` or, `r=V/I`

So,we can write, `P=V^2/(V/I)=VI`

So,current drawn by the bulb `=150/240=0.625 A`

And by the kettle is `=(2.5×10^3)/240=10.42A`