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: \(H = 3.3 \times 10^5 \, \text{J}\)

Answer 2

Here is a shortcut way to reach the same two results by combining the relations into a single formula before plugging in numbers.

Step 1: The resistance of the iron's heating coil is a fixed property of the coil and does not depend on the voltage applied to it, so it can be found directly from the rated values using \[ R = \frac{V^2}{P} = \frac{(220)^2}{2200} = 22 \, \Omega. \]

Step 2: Because \( R \) does not change, the power consumed at any other voltage \( V' \) is \( P' = \dfrac{V'^2}{R} \). Substituting \( R = \dfrac{V^2}{P} \) gives a single combined formula relating the new power directly to the rated power and the voltage ratio: \[ P' = P \left(\frac{V'}{V}\right)^2. \]

Step 3: Multiplying both sides by the operating time \( t \) gives the heat produced directly, without treating \( P' \) as a separate numeric step: \[ H = P' t = P \left(\frac{V'}{V}\right)^2 t. \]

Step 4: Substituting the numbers in one pass, with \( P = 2200 \, \text{W} \), \( V = 220 \, \text{V} \), \( V' = 110 \, \text{V} \), \( t = 600 \, \text{s} \): \[ H = 2200 \times \left(\frac{110}{220}\right)^2 \times 600 = 2200 \times \frac{1}{4} \times 600 = 550 \times 600 = 330000 \, \text{J}. \]

So the resistance of the iron works out to \( 22 \, \Omega \), and the heat produced in 10 minutes when operated at 110 V is \( 3.3 \times 10^5 \, \text{J} \).