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(D) The power dissipated in the $6\,\Omega$ resistor is given by $P = I^2 R_{6}$.Given $P = 6\,W$ and $R_{6} = 6\,\Omega$,the current $I$ flowing thro

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$24$

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(D) The power dissipated in the $6\,\Omega$ resistor is given by $P = I^2 R_{6}$.Given $P = 6\,W$ and $R_{6} = 6\,\Omega$,the current $I$ flowing through the $6\,\Omega$ resistor is:$I = \sqrt{\frac{P}{R_{6}}} = \sqrt{\frac{6}{6}} = 1\,A$.This current $I$ is the total current flowing from the $12\,V$ battery.The equivalent resistance of the parallel combination of $R$ and $8\,\Omega$ is $R_{p} = \frac{8R}{8+R}$.The total resistance of the circuit is $R_{eq} = 6 + \frac{8R}{8+R}$.Using Ohm's law,$I = \frac{V}{R_{eq}}$,we have:$1 = \frac{12}{6 + \frac{8R}{8+R}}$.$6 + \frac{8R}{8+R} = 12$.$\frac{8R}{8+R} = 6$.$8R = 6(8+R) = 48 + 6R$.$2R = 48$.$R = 24\,\Omega$.

In the circuit shown in the figure,the power dissipated as heat in the $6\,\Omega$ resistor is $6\,W$. What is the value of resistance $R$ in the circuit? ................ $\Omega$

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