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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 8 April · Evening Shift · Physics · Current Electricity

A heater is designed to operate with a power of 1000 ~W in a 100 ~V...

A heater is designed to operate with a power of 1000 W1000 \mathrm{~W} in a 100 V100 \mathrm{~V} line. It is connected in combination with a resistance of 10Ω10 \Omega and a resistance RR, to a 100 V100 \mathrm{~V} mains as shown in figure. For the heater to operate at 62.5 W62.5 \mathrm{~W}, the value of R\mathrm{R} should be _____Ω\Omega. A heater is designed to operate with a power of 1000 ~W in a 100 ~V line. It is connected in combination with a

Correct answer

0

Step-by-step solution

Rheater =V2P=(100)21000=10Ω\mathrm{R}_{\text{heater }}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(100)^2}{1000}=10 \Omega For heater P=V2R⇒V=PRP=\frac{V^2}{R} \Rightarrow V=\sqrt{P R}V=62.5×10V=25v\begin{aligned} & \mathrm{V}=\sqrt{62.5 \times 10} \\ & \mathrm{ V}=25 \mathrm{v}\end{aligned}i1=7510=7.5A,iH=2510=2.5A.iR=i1−iH=5V=IRR=255=5Ω\begin{aligned} & i_1=\frac{75}{10}=7.5 \mathrm{ A}, \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{ A} . \\ & \mathrm{i}_{\mathrm{R}}=\mathrm{i}_1-\mathrm{i}_{\mathrm{H}}=5 \\ & \mathrm{ V}=\mathrm{IR} \\ & \mathrm{R}=\frac{25}{5}=5 \Omega\end{aligned}$$
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