Question

An electric iron is rated 220V, 1KW. Under normal working conditions, find

(1) the resistance of the heating element in the iron.

(2) the amount of current that will flow.

(3) if the line voltage falls to 180V, find the power consumed.

Answer 1

Dear Student ,
Here in this case from the given data we can write that ,
P = 1 KW = 1000 W and V =220 volt .
$$\begin{gathered} \left(i\right)Resis\tan ce,R=\frac{V^2}{P}=\frac{{\left(220\right)}^2}{1000}=48·4\Omega \\ \left(ii\right)Currentthroughtheelementis, \\ I=\frac{P}{V}=\frac{1000}{220}=4·54A \\ \left(iii\right)Ifthelinevoltagefallsto180Vthenthepowerconsumedisgivenby, \\ P=\frac{V^2}{R}=\frac{{\left(180\right)}^2}{48·4}=675W \end{gathered}$$

Regards