Three resistors of $4\,\Omega ,6\,\Omega $ and $12\,\Omega $ are connected in parallel and the combination is connected in series with a $1.5\, V$ battery of $1\,\Omega $ internal resistance. The rate of Joule heating in the $4\,\Omega $ resistor is ................ $W$
AIEEE 2012, Diffcult
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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 connectedto $1.5 \,\mathrm{V}$ battery whose internal resistance $r=1\, \Omega$
Equivalent resistance now, $R^{\prime}=2\, \Omega+1 \,\Omega=3\, \Omega$
Current, $I_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \,\mathrm{A}$
$I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x$
$\Rightarrow x=\frac{1}{12}$
$\therefore $ Current through $4\, \Omega$ resistor $=3 x$
$=3 \times \frac{1}{12}=\frac{1}{4}\, A$
Therefore, rate of Joule heating in the $4 \,\Omega$ resistor
$=I^{2} R=\left(\frac{1}{4}\right)^{2} \times 4=\frac{1}{4}=0.25 \,\mathrm{W}$
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