Question

An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at 'minimum heating' it consumes a power of 360 W but at 'maximum heating' it takes a power of 840 W. Calculate the current and resistance in each case.

Answer 1

In the case of minimum heating:

P = 360 W
V = 220 V
where the symbols used have their usual meanings.
We know that P = VI
I = $\frac{\mathrm{P}}{\mathrm{V}}$
$I=\frac{360}{220}=1.64\mathrm{A}$

Resistance, R =$\frac{V}{I}$
R = $\frac{220}{1.64}=134.15$Ω

In the case of maximum heating:

P = 840 W
V = 220 V
where the symbols used have their usual meanings.
We know that P = VI
I = $\frac{\mathrm{P}}{\mathrm{V}}$

$I=\frac{840}{220}=3.82\mathrm{A}$

Further, R =$\frac{\mathrm{V}}{\mathrm{I}}$

R = $\frac{220}{3.82}=57.60$ Ω