A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Ω and a resistance R, to a 100 V mains as shown in figure. For the heater to operate at 62.5 W, the value of R should be _______ Ω.

Solution

Physics sir · 5 steps

Get the heater's resistance from its rating, then find the 25 V it must get at 62.5 W.

1. Heater resistance. The rating fixes the resistance: Rₕ = V²/P = 100²/1000 = 10 Ω. This value stays the same in the new circuit.
2. Voltage across heater. For 62.5 W, V_(BC)² = P Rₕ = 62.5 × 10 = 625, so V_(BC) = 25 V. The heater and R are in parallel between B and C, so R also gets 25 V.
3. Current in the line. The 10 Ω resistor takes the rest of the supply, 100 - 25 = 75 V. So the total current is 75/10 = 7.5 A.
4. Split at B. The heater draws 25/10 = 2.5 A, so R carries 7.5 - 2.5 = 5 A.

   R = V_(BC)/I_R = (25 V)/(5 A) = 5 Ω

5. So. R = 5 Ω. Check: 10 Ω in parallel with 5 Ω is 10/3 Ω. That is one quarter of the total 40/3 Ω, so it gets one quarter of 100 V, which is 25 V.

The common slip: 3.33. Using the full 7.5 A line current through R gives 25/7.5 ≈ 3.33 Ω. Subtract the heater's 2.5 A first.

The idea to keep. A heater's rating gives you its resistance, nothing more. Find the voltage it needs for the new power, then use the series drop and the parallel current split.

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