Question

A heating element is designed to dissipate \(2400\,\text{W}\) when connected to \(240\,\text{V}\). The power it dissipates when it is connected to \(120\,\text{V}\) is \((\text{Assume that resistance of the filament is constant})\)

• A. \(600\,\text{W}\)
• B. \(1200\,\text{W}\)
• C. \(1800\,\text{W}\)
• D. \(400\,\text{W}\)

Hint:

For a resistor with constant resistance, \[ P=\frac{V^2}{R}. \] Hence, power varies as the square of the applied voltage: \[ P\propto V^2. \] If the voltage is halved, the power becomes one-fourth.

Answer 1

The Correct Option is A

Step 1: Determine heater resistance from rated values.
R = V₁²/P₁ = 240²/2400 = 24 Ω.

Step 2: Calculate power at reduced voltage.
P₂ = V₂²/R = 120²/24 = 600 W. Alternatively, P ∝ V² → P₂ = P₁(V₂/V₁)² = 2400 × (1/2)² = 600 W.

Step 3: Final Answer:
600 W.