Power dissipated across the 8,Omega resistor | vedclass

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(D) The $8\,\Omega$ resistor is in parallel with the series combination of $1\,\Omega$ and $3\,\Omega$ resistors.Let $V$ be the potential difference a

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$3$

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(D) The $8\,\Omega$ resistor is in parallel with the series combination of $1\,\Omega$ and $3\,\Omega$ resistors.Let $V$ be the potential difference across the parallel branches.Given,power dissipated in $8\,\Omega$ resistor is $P_8 = 2\,W$.Using $P = \frac{V^2}{R}$,we have $2 = \frac{V^2}{8}$,which gives $V^2 = 16$,so $V = 4\,V$.Since the $1\,\Omega$ and $3\,\Omega$ resistors are in series,the current $I$ flowing through them is $I = \frac{V}{R_{eq}} = \frac{4}{1+3} = \frac{4}{4} = 1\,A$.The power dissipated in the $3\,\Omega$ resistor is $P_3 = I^2 R = (1)^2 	imes 3 = 3\,W$.

Power dissipated across the $8\,\Omega$ resistor in the circuit shown here is $2\,W$. The power dissipated in watt units across the $3\,\Omega$ resistor is

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