Q79 / 90 · JEE Main 2012 (12 May Online) · Physics
Three resistors of $4 \Omega, 6 \Omega$ and $12 \Omega$ are connected in parallel and the combination is connected in series with a $1.5 \mathrm{~V}$ battery of $1 \Omega$ internal resistance. The rate of Joule heating in the $4 \Omega$ resistor is
- A$0.55 \mathrm{~W}$
- B$0.33 \mathrm{~W}$
- Correct answerC$0.25 \mathrm{~W}$
- D$0.86 \mathrm{~W}$
Solution
Resistors $4 \Omega, 6 \Omega$ and $12 \Omega$ are connected in parallel, its equivalent resistance $(R)$ is given by
$$
\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12} \Rightarrow R=\frac{12}{6}=2 \Omega
$$
Again $R$ is connected to $1.5 \mathrm{~V}$ battery whose internal resistance $r=1 \Omega$.
Equivalent resistance now,
$$
R^{\prime}=2 \Omega+1 \Omega=3 \Omega
$$
Current, $I_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \mathrm{~A}$
$$
\begin{aligned}
& I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x \\
& \Rightarrow x=\frac{1}{12}
\end{aligned}
$$
$\therefore$ Current through $4 \Omega$ resistor $=3 x$
$$
=3 \times \frac{1}{12}=\frac{1}{4} \mathrm{~A}
$$
Therefore, rate of Joule heating in the $4 \Omega$ resistor
$$
=I^2 R=\left(\frac{1}{4}\right)^2 \times 4=\frac{1}{4}=0.25 \mathrm{~W}
$$
Hence, the correct answer is option: 'C'.