Q 12-03-096JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
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 the figure. For the heater to operate at $62.5\ \text{W}$, the value of $R$ should be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
Heater resistance: $R_h = \dfrac{100^2}{1000} = 10\ \Omega$.
At $62.5\ \text{W}$: $V_h = \sqrt{62.5\times10} = 25\ \text{V}$, so the heater current is $2.5\ \text{A}$.
The $10\ \Omega$ resistor has $100 - 25 = 75\ \text{V}$, so it carries $7.5\ \text{A}$. The current through $R$ is $7.5 - 2.5 = 5\ \text{A}$:
$$R = \frac{25}{5} = 5\ \Omega$$
Solution by Sreeraj P, M.Sc Physics