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

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(D) Let the voltage of the battery be $V = 12\,V$. The resistor $R$ and the $8\,\Omega$ resistor are in parallel. Let their equivalent resistance be $

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Both $(A)$ and $(C)$

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(D) Let the voltage of the battery be $V = 12\,V$. The resistor $R$ and the $8\,\Omega$ resistor are in parallel. Let their equivalent resistance be $R_p = \frac{8R}{8+R}$.This combination is in series with the $2\,\Omega$ resistor.The total resistance of the circuit is $R_{eq} = R_p + 2 = \frac{8R}{8+R} + 2 = \frac{8R + 16 + 2R}{8+R} = \frac{10R + 16}{8+R}$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{12(8+R)}{10R+16} = \frac{6(8+R)}{5R+8}$.The power dissipated in the $2\,\Omega$ resistor is $P = I^2 	imes 2 = 2 	imes \left[ \frac{6(8+R)}{5R+8} ight]^2 = 72 	imes \frac{(8+R)^2}{(5R+8)^2}$.To maximize $P$,we analyze the function $f(R) = \frac{8+R}{5R+8}$.As $R 	o 0$,$f(R) 	o \frac{8}{8} = 1$. As $R 	o \infty$,$f(R) 	o \frac{1}{5} = 0.2$.Since $f(R)$ is a decreasing function for $R \ge 0$,the maximum value occurs at $R = 0$.At $R = 0$,$P = 72 	imes \left( \frac{8+0}{0+8} ight)^2 = 72 	imes 1^2 = 72\,W$.Thus,both $(A)$ and $(C)$ are correct.

The value of the resistance $R$ in the figure is adjusted such that the power dissipated in the $2\,\Omega$ resistor is maximum. Under this condition:

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