The value of the resistance R in the figure i | vedclass

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(A) The circuit consists of a $12\,V$ voltage source in series with a parallel combination of an $8\,\Omega$ resistor and a variable resistor $R$,whic

Vedclass · Accepted Answer

$72$

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(A) The circuit consists of a $12\,V$ voltage source in series with a parallel combination of an $8\,\Omega$ resistor and a variable resistor $R$,which is then in series with a $2\,\Omega$ resistor.Let $R_p$ be the equivalent resistance of the parallel combination of $8\,\Omega$ and $R$,given by $R_p = \frac{8R}{8+R}$.The total resistance of the circuit is $R_{eq} = R_p + 2 = \frac{8R}{8+R} + 2$.The current $I$ flowing through the $2\,\Omega$ resistor is $I = \frac{V}{R_{eq}} = \frac{12}{\frac{8R}{8+R} + 2} = \frac{12(8+R)}{8R + 16 + 2R} = \frac{12(8+R)}{10R + 16}$.The power dissipated in the $2\,\Omega$ resistor is $P = I^2 	imes 2 = 2 	imes \left[ \frac{12(8+R)}{10R + 16} ight]^2$.To maximize $P$,we need to maximize the current $I$. The current $I$ is maximum when the total resistance $R_{eq}$ is minimum.$R_{eq} = \frac{8R}{8+R} + 2$. Since $R$ is a resistance,$R \ge 0$. The function $f(R) = \frac{8R}{8+R}$ is an increasing function of $R$. Thus,$R_{eq}$ is minimum when $R = 0$.When $R = 0$,the parallel combination becomes a short circuit,so $R_p = 0$.The total resistance becomes $R_{eq} = 0 + 2 = 2\,\Omega$.The current $I = \frac{12\,V}{2\,\Omega} = 6\,A$.The power dissipated in the $2\,\Omega$ resistor is $P = I^2 R_{fixed} = (6)^2 	imes 2 = 36 	imes 2 = 72\,W$.

The value of the resistance $R$ in the figure is adjusted such that the power dissipated in the $2\,\Omega$ resistor is maximum. Then the power dissipated in the $2\,\Omega$ resistor will be ................ $W$.

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