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. Numerical answer type. Work it out, then open the answer.

Answer: 5. Heater resistance Rh = (100²)/(1000) = 10 Ω. At 62.5 W: VBC = √(PRh) = √(62.5 × 10) = 25 V, heater current = 2.5 A. The 10 Ω resistor drops 100 - 25 = 75 V, so total current = 7.5 A. Current in R = 7.5 - 2.5 = 5 A, so R = 25/5 = 5 Ω.

A heater is designed to operate with a power of 1000 W in a 100 V… | JEE Main 2024 (8 Apr, Shift 2)

Q18JEE Main 2024 · 8 APR · SHIFT 2Difficult

Electrical Power & Heating Effect

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

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Heater resistance $R_h = \frac{100^2}{1000} = 10\ \Omega$ .
At 62.5 W: $V_{BC} = \sqrt{PR_h} = \sqrt{62.5 \times 10} = 25$ V, heater current $= 2.5$ A.
The $10\ \Omega$ resistor drops $100 - 25 = 75$ V, so total current $= 7.5$ A.
Current in $R = 7.5 - 2.5 = 5$ A, so $R = \frac{25}{5} = 5\ \Omega$ .

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