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A heater is designed to operate with a power of $1000 \mathrm{~W}$ in a $100 \mathrm{~V}$ line. It is connected in combination with a resistance of $10 \Omega$ and a resistance $R$, to a $100 \mathrm{~V}$ mains as shown in figure. For the heater to operate at $62.5 \mathrm{~W}$, the value of $\mathrm{R}$ should be _____$\Omega$.
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The correct answer is:
5
$\mathrm{R}_{\text {heater }}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(100)^2}{1000}=10 \Omega$
For heater $P=\frac{V^2}{R} \Rightarrow V=\sqrt{P R}$
$\begin{aligned} & \mathrm{V}=\sqrt{62.5 \times 10} \\ & \mathrm{~V}=25 \mathrm{v}\end{aligned}$
$\begin{aligned} & i_1=\frac{75}{10}=7.5 \mathrm{~A}, \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A} . \\ & \mathrm{i}_{\mathrm{R}}=\mathrm{i}_1-\mathrm{i}_{\mathrm{H}}=5 \\ & \mathrm{~V}=\mathrm{IR} \\ & \mathrm{R}=\frac{25}{5}=5 \Omega\end{aligned}$
For heater $P=\frac{V^2}{R} \Rightarrow V=\sqrt{P R}$
$\begin{aligned} & \mathrm{V}=\sqrt{62.5 \times 10} \\ & \mathrm{~V}=25 \mathrm{v}\end{aligned}$
$\begin{aligned} & i_1=\frac{75}{10}=7.5 \mathrm{~A}, \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A} . \\ & \mathrm{i}_{\mathrm{R}}=\mathrm{i}_1-\mathrm{i}_{\mathrm{H}}=5 \\ & \mathrm{~V}=\mathrm{IR} \\ & \mathrm{R}=\frac{25}{5}=5 \Omega\end{aligned}$
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