If power dissipated in the $9 \,\Omega$ resistor in the circuit shown is $36\,W$, the potential difference across the $2 \,\Omega$ resistor is .......... $V$
AIPMT 2011, Medium
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Current flows through the $9\, \Omega$ resistor is
${I_{1}^{2}=\frac{36}{9}=4} $ $\left( {As\,\,P = {I^2}R} \right)$
${I_{1}=2 \,\mathrm{A}}$
As the resistors $9\, \Omega$ and $6\, \Omega$ are connected in parallel, therefore potential difference across them is
same.
$\therefore \quad 9 I_{1}=6 I_{2} ; I_{2}=\frac{9 \times 2}{6}=3\, \mathrm{A}$
Current drawn from the battery is
$I=I_{1}+I_{2}=(2+3) \mathrm{A}=5\, \mathrm{A}$
The potential difference across the $2\, \Omega$ resistor is
$=(5 \,A)(2 \,\Omega)=10\, \mathrm{V}$
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