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In the circuit shown, the heat produced in $5 \Omega$ resistance due to current through is $50 \mathrm{~J} / \mathrm{s}$. Then, the heat generated per second in $2 \Omega$ resistance is
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Verified Answer
The correct answer is:
$5 \mathrm{~J} / \mathrm{s}$
The given $P=50 \mathrm{~J} / \mathrm{s}$
$$
P=V i \Rightarrow P=i^2 R
$$
$$
\begin{aligned}
\Rightarrow i_2^2 & =\frac{P}{R}=\frac{50}{5}=10 \mathrm{Amp}^2 \\
V & \left.=i_2 R_{(5} \Omega\right) \\
& =\sqrt{10} \times 5=\sqrt{250} \mathrm{~V}
\end{aligned}
$$
and $2 \Omega$ and $8 \Omega$ are in series.
So, the required resistance
$$
\begin{aligned}
& =2 \Omega+8 \Omega=10 \Omega \\
i_1 & =\frac{V}{R_{(10 \Omega)}} \\
i_1 & =\frac{\sqrt{250}}{10} \mathrm{~A}
\end{aligned}
$$
The heat generated/second in $2 \Omega$
$$
\begin{aligned}
P=V i_1 & \\
& =i_1^2 \times R \\
& =\left(\frac{\sqrt{250}}{10}\right)^2 \times 2 \\
& =\frac{250}{100} \times 2=\frac{25}{10} \times 2=\frac{25}{5}=5 \mathrm{~J} / \mathrm{s}
\end{aligned}
$$
$$
P=V i \Rightarrow P=i^2 R
$$
$$
\begin{aligned}
\Rightarrow i_2^2 & =\frac{P}{R}=\frac{50}{5}=10 \mathrm{Amp}^2 \\
V & \left.=i_2 R_{(5} \Omega\right) \\
& =\sqrt{10} \times 5=\sqrt{250} \mathrm{~V}
\end{aligned}
$$
and $2 \Omega$ and $8 \Omega$ are in series.
So, the required resistance
$$
\begin{aligned}
& =2 \Omega+8 \Omega=10 \Omega \\
i_1 & =\frac{V}{R_{(10 \Omega)}} \\
i_1 & =\frac{\sqrt{250}}{10} \mathrm{~A}
\end{aligned}
$$
The heat generated/second in $2 \Omega$
$$
\begin{aligned}
P=V i_1 & \\
& =i_1^2 \times R \\
& =\left(\frac{\sqrt{250}}{10}\right)^2 \times 2 \\
& =\frac{250}{100} \times 2=\frac{25}{10} \times 2=\frac{25}{5}=5 \mathrm{~J} / \mathrm{s}
\end{aligned}
$$
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