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

Step 1: Identify the given data.
The electrical appliance has: \[ P = 2.2 \, \text{kW} = 2200 \, \text{W}, \quad V = 220 \, \text{V} \]
Step 2: Calculate the resistance of the appliance.
Using the power formula: \[ P = \frac{V^2}{R} \implies R = \frac{V^2}{P} \] \[ R = \frac{220^2}{2200} = \frac{48400}{2200} = 22 \, \Omega \]
Step 3: Determine power when operated at 110 V.
New voltage: \(V' = 110 \, \text{V}\) \[ P' = \frac{V'^2}{R} = \frac{110^2}{22} = \frac{12100}{22} = 550 \, \text{W} \]
Step 4: Calculate heat produced in 10 minutes.
Time in seconds: \(t = 10 \, \text{min} = 600 \, \text{s}\) \[ H = P' \cdot t = 550 \times 600 = 330000 \, \text{J} \]
Final Answers:
(i) Resistance of the appliance: \(R = 22 \, \Omega\)
(ii) Heat produced in 10 minutes: \(H = 3.3 \times 10^5 \, \text{J}\)