A heater rated 1000 W at 100 V is placed in parallel with an unknown resistor R. This parallel pair is in series with a 17 Ω resistor across 100 V, as shown. If heater power is 22.5 W, find R (Ω), rounded to the nearest integer.
Physics · Mixed Chapter
JEE Mains

Question

A heater rated 1000 W at 100 V is placed in parallel with an unknown resistor R. This parallel pair is in series with a 17 Ω resistor across 100 V, as shown. If heater power is 22.5 W, find R (Ω), rounded to the nearest integer.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Heater resistance at rating: Rₕ = V²/P = 100²/1000 = 10 Ω. Step 2 — Actual voltage across the heater and the parallel resistor: Ph=Vₚ²/Rₕ ⇒ Vₚ=√(22.5×10)=15 V. Step 3 — Current through series 17 Ω resistor: Voltage across series resistor is 100−15=85 V. Thus Iₛ=85/17=5 A. Step 4 — Heater and bypass currents: Iₕ = Vₚ/Rₕ=15/10=1.5 A. Iᵣ=Iₛ−Iₕ=5−1.5=3.5 A. Step 5 — Calculate R: R=Vₚ/Iᵣ=15/3.5=30/7≈4.29 Ω. Rounded to nearest integer R=4 Ω. Answer: 4.