An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the
minimum. The voltage is 220 V. What are the current and the resistance in each case?
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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
Explanation:
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Answer:
Solution is in picture
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