Question

An electric iron rated \( 2.2 \, \text{kW}, 220 \, \text{V} \) is operated at \( 110 \, \text{V} \) supply. Find:
its resistance, and
heat produced by it in 10 minutes.

Hint:

If voltage changes but resistance stays constant: Power changes as \( P \propto V^2 \). Halving voltage reduces power to one-fourth.

Answer 1

Concept: For an electrical appliance:

Power relation: \[ P = \frac{V^2}{R} \]
Heat produced: \[ H = P t \]

Step 1: Given data. \[ P = 2.2 \, \text{kW} = 2200 \, \text{W}, \quad V = 220 \, \text{V} \]
Step 2: Find resistance of iron. Using: \[ R = \frac{V^2}{P} \] \[ R = \frac{220^2}{2200} = \frac{48400}{2200} = 22 \, \Omega \]
Step 3: Operated at 110 V. New power consumed: \[ P' = \frac{V'^2}{R} \] \[ P' = \frac{110^2}{22} = \frac{12100}{22} = 550 \, \text{W} \]
Step 4: Heat produced in 10 minutes. Time: \[ t = 10 \, \text{min} = 600 \, \text{s} \] \[ H = P' t = 550 \times 600 = 330000 \, \text{J} \] Final Answers:

[(i)] Resistance = \( 22 \, \Omega \)
[(ii)] Heat produced = \( 3.3 \times 10^5 \, \text{J} \)

Answer 2

The iron is rated 2.2 kW at 220 V, and we need its resistance plus the heat it produces when run at a lower voltage, 110 V, for 10 minutes. Instead of jumping straight to \( R = V^2/P \), let's start from the current the iron draws at its rated voltage.

Step 1: At the rated condition, power equals voltage times current, so the rated current is \[ I = \frac{P}{V} = \frac{2200 \, \text{W}}{220 \, \text{V}} = 10 \, \text{A}. \]

Step 2: The resistance of the heating element is fixed by its construction and does not change with the applied voltage, so using Ohm's law at the rated condition, \[ R = \frac{V}{I} = \frac{220 \, \text{V}}{10 \, \text{A}} = 22 \, \Omega. \]

Step 3: Since \( R \) stays the same when the iron is connected to 110 V instead of 220 V, the power it draws scales with the square of the voltage ratio: \[ \frac{P'}{P} = \left(\frac{V'}{V}\right)^2 = \left(\frac{110}{220}\right)^2 = \frac{1}{4}. \] So \[ P' = \frac{2200}{4} = 550 \, \text{W}. \]

Step 4: Over 10 minutes, that is \( t = 600 \, \text{s} \), the heat produced is \[ H = P' t = 550 \times 600 = 330000 \, \text{J} = 3.3 \times 10^5 \, \text{J}. \]

So the resistance of the iron is \( 22 \, \Omega \), and the heat it produces at 110 V in 10 minutes is \( 3.3 \times 10^5 \, \text{J} \).