Q128 / 160 · TS EAMCET 2014 · Physics
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
- Correct answerA$5 \mathrm{~J} / \mathrm{s}$
- B$4 \mathrm{~J} / \mathrm{s}$
- C$9 \mathrm{~J} / \mathrm{s}$
- D$10 \mathrm{~J} / \mathrm{s}$
Solution
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 $2 \Omega$ $$ \begin{aligned} P=V i_1 & \\ & =i_1^2 \times R \\ & =\left(\frac{\sqrt{250}}{10}\right)^2 \times 2 \\ & =\frac{250}{100} \times 2=\frac{25}{10} \times 2=\frac{25}{5}=5 \mathrm{~J} / \mathrm{s} \end{aligned} $$
$$ \begin{aligned} \Rightarrow i_2^2 & =\frac{P}{R}=\frac{50}{5}=10 \mathrm{Amp}^2 \\ V & \left.=i_2 R_{(5} \Omega\right) \\ & =\sqrt{10} \times 5=\sqrt{250} \mathrm{~V} \end{aligned} $$ 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 $2 \Omega$ $$ \begin{aligned} P=V i_1 & \\ & =i_1^2 \times R \\ & =\left(\frac{\sqrt{250}}{10}\right)^2 \times 2 \\ & =\frac{250}{100} \times 2=\frac{25}{10} \times 2=\frac{25}{5}=5 \mathrm{~J} / \mathrm{s} \end{aligned} $$
Hence, the correct answer is option: 'A'.