An electric iron consumes energy at a rate of 840 W when heating is at

1.	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 rate. The applied voltage is 220 V. What is the value of current and the resistance in each case?
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, \(R = \frac{V}{I} = \frac{220V}{1.64 A} = 134.15 \Omega\)

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 rate. The applied voltage is 220 V. What is the value of current and the resistance in each case?

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,

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

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