Question

An electric iron consumes energy at a rate of 840 W when heating is at the maximum. The voltage is 220 V. What are the current and the resistance in each case?​

Answer 1

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

hope it will help u

Answer 2

Answer:

Given,

Voltage of the mains supply, V = 220 V

Power is given by, P = VI

Case 1: When heating is at maximum rate,

Power, P = 840 W

Potential difference, V = 220 V

Current, I =PV=849W220V=3.82APV=849W220V=3.82A

∴ Resistance of the electric iron, R = VI=220V3.82A=57.60Ω

VI=220V3.82A=57.60Ω

Case 2: When heating is at the minimum rate,

Power, P = 360 W

Potential difference, V = 220 V

Current, I = PV=360W220V=1.64 APV=360W220V=1.64A

∴ Resistance of the electric iron,

R=VI=220V1.64A=134.15Ω

R=VI=220V1.64A=134.15Ω

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