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(B) To find the maximum power delivered to the variable resistance $R$, we use the Maximum Power Transfer Theorem. According to this theorem, the powe

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$0.4$

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(B) To find the maximum power delivered to the variable resistance $R$, we use the Maximum Power Transfer Theorem. According to this theorem, the power is maximum when $R = R_{th}$, where $R_{th}$ is the Thevenin equivalent resistance across the terminals of $R$.$1$. Find the Thevenin voltage ($V_{th}$ or $V_{ab}$): Remove $R$ and calculate the open-circuit voltage across the terminals $a$ and $b$. The $6 \, \Omega$ and $3 \, \Omega$ resistors form a voltage divider across the $12 \, V$ source. The voltage across the $3 \, \Omega$ resistor is $V_{ab} = 12 	imes \frac{3}{6+3} = 12 	imes \frac{3}{9} = 4 \, V$.$2$. Find the Thevenin resistance ($R_{th}$ or $R_{ab}$): Short the voltage source. The $6 \, \Omega$ and $3 \, \Omega$ resistors are now in parallel, and this combination is in series with the $8 \, \Omega$ resistor. Thus, $R_{ab} = 8 + \frac{6 	imes 3}{6+3} = 8 + \frac{18}{9} = 8 + 2 = 10 \, \Omega$.$3$. Calculate maximum power: $P_{\max} = \frac{V_{ab}^2}{4 R_{ab}} = \frac{4^2}{4 	imes 10} = \frac{16}{40} = 0.4 \, W$.

The maximum power delivered to resistance $R$ is ............... $W$.

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