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?
Examples from NCERT Book
Examples from NCERT Book
Last updated at April 16, 2024 by Teachoo
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