A uniform heating wire of resistance 36Ω is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is ____________

Solution

Physics sir · 3 steps

Keep the voltage fixed and use P = V²/R: halving the resistance doubles the power in each piece.

1. First case. The full wire has R = 36 Ω across 240 V. With the voltage fixed, use P = V²/R, so P₁ = 240²/36 = 57600/36 = 1600 W.
2. Halve the wire. Resistance is proportional to length, so each half has R = 18 Ω. Each half gets the full 240 V on its own.

   P₂ = 2 × 240²/18 = 2 × 3200 W = 6400 W

3. So. P₁ : P₂ = 1600 : 6400 = 1 : 4, so x = 4.

The common slip: 2. Comparing the full wire with just one half gives 1600 : 3200, that is 2. Both halves dissipate power, so add them.

The idea to keep. When the voltage stays the same, think P = V²/R. Smaller resistance means more power, and pieces on separate supplies add their powers.

Answer from the NTA final answer key. Solution by PrepWiser.