If the power dissipated in the 3,Omega resist | vedclass

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(A) First,calculate the equivalent resistance of the parallel combination of $2\,\Omega$ and $4\,\Omega$ resistors:$R_p = \frac{2 	imes 4}{2 + 4} = \

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$8$

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(A) First,calculate the equivalent resistance of the parallel combination of $2\,\Omega$ and $4\,\Omega$ resistors:$R_p = \frac{2 	imes 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}\,\Omega$.In the circuit,the $3\,\Omega$ resistor is in series with the parallel combination $(4/3\,\Omega)$.Since the current $I$ is the same for components in series,the power $P = I^2 R$ is proportional to resistance $(P \propto R)$.Given power in $3\,\Omega$ resistor is $P_3 = 27\,W$,let $P_p$ be the power in the parallel combination:$\frac{P_3}{P_p} = \frac{R_3}{R_p} \Rightarrow \frac{27}{P_p} = \frac{3}{4/3} = \frac{9}{4}$.$P_p = 27 	imes \frac{4}{9} = 12\,W$.Now,for the parallel combination,the voltage $V_p$ across both resistors is the same. Power $P = V^2/R$,so $P \propto 1/R$.$\frac{P_2}{P_4} = \frac{R_4}{R_2} = \frac{4}{2} = 2 \Rightarrow P_2 = 2 P_4$.Since $P_2 + P_4 = P_p = 12\,W$,we have $2 P_4 + P_4 = 12\,W \Rightarrow 3 P_4 = 12\,W \Rightarrow P_4 = 4\,W$.Therefore,$P_2 = 2 	imes 4 = 8\,W$.

If the power dissipated in the $3\,\Omega$ resistor is $27\,W$,then what is the power dissipated in the $2\,\Omega$ resistor (in $,W$)?

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