Example 12.10 - Chapter 12 Class 10 - Electricity (Term 2)

Last updated at March 30, 2023 by Teachoo

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 minimum. The voltage is 220 V. What are the current and the resistance in each case?

Example 12.10 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 minimum. The voltage is 220 V. What are the current and the resistance in each case?
Given,
Potential difference = V = 220
Maximum power = 840 W
Minimum power = 360 W
We know that,
P = V I
∴ 𝑃/𝑉 = I
I = 𝑃/𝑉
And by Ohm’s law,
V = I R
∴ R = 𝑉/I
For P = 840 W
Finding current
I = 𝑃/𝑉
I = 840/220
I = 84/22
I = 42/11
I = 3.82 A
For P = 360 W
Finding current
I = 𝑃/𝑉
I = 360/220
I = 36/22
I = 18/11
I = 1.64 A
Finding resistance
R = 𝑉/I
R = 220/3.82
R = 220/(42/11)
R = (220 × 11)/42
R = (110 × 11)/21
R = 1210/21
R = 57.61 Ω
Finding resistance
R = 𝑉/I
R = 220/1.64
R = 220/(18/11)
R = (220 × 11)/18
R = (110 × 11)/9
R = 1210/9
R = 134.44 Ω
When heating is at maximum rate (P = 840 W)
Current is 3.82 A
Resistance is 57.6 Ω
When heating is at minimum rate (P = 360 W)
Current is 1.64 A
Resistance is 134.44 Ω

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo

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