Power dissipated across the 8 ,,Omega resisto | vedclass

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Question

Resistance of series combination of $3\,\Omega $ and $1\,\Omega $ is

$R_{1}=3+1=4 \,\Omega, R_{2}=8\, \Omega$

Let $i$ be the total cur

Vedclass · Accepted Answer

$3$

Vedclass · Answer

Resistance of series combination of $3\,\Omega $ and $1\,\Omega $ is

$R_{1}=3+1=4 \,\Omega, R_{2}=8\, \Omega$

Let $i$ be the total current in the circuit.

Current through $R_{1}$ is $i_{1}=\frac{i 	imes R_{2}}{R_{1}+R_{2}}=\frac{i 	imes 8}{12}=\frac{2 i}{3}$

Current through $R_{2}$ is $i_{2}=\frac{i 	imes R_{1}}{R_{1}+R_{2}}=\frac{i 	imes 4}{12}=\frac{i}{3}$

Power dissipated in $3\, \Omega$ resistor is

$P_{1}=i_{1}^{2} 	imes 3$       ....$(i)$

Power dissipated in $8 \Omega$ resistor is

$P_{2}=i_{2}^{2} 	imes 8$       ....$(ii)$

$	herefore \quad \frac{P_{1}}{P_{2}}=\frac{i_{1}^{2} 	imes 3}{i_{2}^{2} 	imes 8}$ or, $\frac{P_{1}}{P_{2}}=\frac{(2 i / 3)^{2} 	imes 3}{(i / 3)^{2} 	imes 8}=\frac{12}{8}=\frac{3}{2}$

$P_{1}=\frac{3}{2} 	imes P_{2}=\frac{3}{2} 	imes 2=3 \,	ext { watt }$

$	herefore$ Power dissipated across $3 \Omega$ resistor is $3\,watt.$

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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