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when electric iron consume energy at the rate (a) 840 w when heating is at the maximum rate and (b) 367 heating is at the minimum voltage is 220 V. what are the current and resistance in each case.
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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
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
Pawan079:
Thank you sir
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