## 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?

Last updated at Aug. 6, 2019 by Teachoo

Transcript

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

Class 10

Chapter 12 Class 10 - Electricity

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.