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 b $e ______$ Ω.

Question

G1 Physics · Accepted Answer

Image Heater resistance: R(H) = V²/P = 100²/1000 = 10 Ω For heater power 62.5 W: P = V(H)²/R(H) Therefore: V(H) = √[P R(H)] = √(62.5 × 10) = 25 V Thus the voltage across the series 10 Ω resistor is: 100 - 25 = 75 V Current through series resistor: I(total) = 75/10 = 7.5 A Current through heater: I(H) = 25/10 = 2.5 A Current through R: I(R) = 7.5 - 2.5 = 5 A Therefore: R = 25/5 = 5 Ω

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 b e ______ Ω.