Power dissipated across the 8 ,Omega resistor | vedclass

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(C) The circuit consists of two parallel branches. The upper branch has resistors $1 \,\Omega$ and $3 \,\Omega$ in series,so its total resistance is $

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$3$

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(C) The circuit consists of two parallel branches. The upper branch has resistors $1 \,\Omega$ and $3 \,\Omega$ in series,so its total resistance is $R_1 = 1 + 3 = 4 \,\Omega.$ The lower branch has a resistor $R_2 = 8 \,\Omega.$Since the branches are in parallel,the potential difference $V$ across both branches is the same.The power dissipated in a resistor is given by $P = \frac{V^2}{R}.$For the $8 \,\Omega$ resistor,$P_2 = \frac{V^2}{8} = 2 \,	ext{W},$ which implies $V^2 = 16 \,	ext{V}^2.$The current $i_1$ flowing through the upper branch is $i_1 = \frac{V}{R_1} = \frac{V}{4}.$The power dissipated in the $3 \,\Omega$ resistor is $P_{3\Omega} = i_1^2 	imes 3 = \left(\frac{V}{4}ight)^2 	imes 3 = \frac{V^2}{16} 	imes 3.$Substituting $V^2 = 16,$ we get $P_{3\Omega} = \frac{16}{16} 	imes 3 = 3 \,	ext{W}.$

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

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