Question

an electric iron consumes energy at a rate of 840W when heating is at the maximum rate and 360W when the heating is at the minimum the voltage is 220V . what are the current and the resistance in each case ​

Answer 1

Answer;

When heating at max rate,

power, P = 840W

V = 220V

We know P = VI

⇒ 840 = 220 × I

⇒ I = 840/220

⇒ I = 4 A

R = V/I = 220/4 = 55 Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,

power, P = 360W

V = 220V

We know P = VI

⇒ 360 = 220 × I

⇒ I = 360/220

⇒ I = 1.636 A

R = V/I = 220/1.636 = 134.45 Ω

So current = 1.636A and resistance = 134.45Ω

Answer 2

Answer:

(a) For maximum heating rate:

P = 840 W;

V = 220 Volt

$Using \: \: P = VI$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: I = \frac{P}{V} = \frac{840}{220} = 3.82A$

$And \: \: resistance R = \frac{V}{I} = \frac{220}{3.82} = 57.60Ω.$

(b) for minimum heating rate :

P = 360 W ; V = 220 Volt

$∴ \: \: \: \: \: \: \: \: I = \frac{P}{V} = \frac{360}{220} = 1.636A \\ \\ or \: I = 1.64A.$

$And \: resistance R = \frac{V}{I} = \frac{220}{1.64} = 134.15Ω.$