Power dissipated across the 8,Omega res | vedclass

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Question

As valtage dropacioss $8\, \Omega=\sqrt{2 	imes 8}=4 \mathrm{\,V}\left(\because P=\frac{\mathrm{V}^{2}}{\mathrm{R}}ight)$

Therefore voltage drop

Vedclass · Accepted Answer

$3$

Vedclass · Answer

As valtage dropacioss $8\, \Omega=\sqrt{2 	imes 8}=4 \mathrm{\,V}\left(\because P=\frac{\mathrm{V}^{2}}{\mathrm{R}}ight)$

Therefore voltage drop across $3\, \Omega=3 \mathrm{\,V}$

$[\because 4 \mathrm{\,V}  $is divided in ratio of resistances between $1\, \Omega $ and $ 3 \,\Omega$]

Hence power dissipated in $3 \Omega=\frac{(3)^{2}}{3}=3$ $\mathrm{watt}$

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

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