Question

22. An electric iron consumes energy at a rate of 840 W when the heating is at the
minimum rate and 360 W when the heating is at the minimum rate. The applied
voltage is 220 V. What is the value of current and the resistance in each case?​

Answer 1

Answer:

Given,

Voltage of the mains supply, V = 220 V

Power is given by, P = VI

Case 1: When heating is at maximum rate,

Power, P = 840 W

Potential difference, V = 220 V

Current, I = \(\frac{P}{V} = \frac{849 W}{220 V} = 3.82 A\)

\(\therefore\) Resistance of the electric iron, R = \(\frac{V}{I} = \frac{220V}{3.82 A} = 57.60 \Omega\)

Case 2: When heating is at the minimum rate,

Power, P = 360 W

Potential difference, V = 220 V

Current, I = \(\frac{P}{V} = \frac{360 W}{220V} = 1.64 A\)

\(\therefore \) Resistance of the electric iron,

\(R = \frac{V}{I} = \frac{220V}{1.64 A} = 134.15 \Omega\)