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(A) To find the maximum power consumed by the variable load resistance $R_L$,we use the Thevenin equivalent circuit theorem.First,we find the Thevenin

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$100$

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(A) To find the maximum power consumed by the variable load resistance $R_L$,we use the Thevenin equivalent circuit theorem.First,we find the Thevenin voltage $V_{th}$ across the terminals where $R_L$ is connected. By removing $R_L$,the circuit becomes a voltage divider with $60 \ V$ source,$3 \ \Omega$ resistor,and $6 \ \Omega$ resistor.$V_{th} = 60 	imes \frac{6}{3+6} = 60 	imes \frac{6}{9} = 40 \ V$.Next,we find the Thevenin resistance $R_{th}$ by short-circuiting the $60 \ V$ source. The $3 \ \Omega$ and $6 \ \Omega$ resistors are in parallel,and this combination is in series with the $2 \ \Omega$ resistor.$R_{th} = (3 \parallel 6) + 2 = \frac{3 	imes 6}{3+6} + 2 = 2 + 2 = 4 \ \Omega$.For maximum power transfer,$R_L = R_{th} = 4 \ \Omega$.The maximum power is given by $P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{40^2}{4 	imes 4} = \frac{1600}{16} = 100 \ W$.

In the given circuit,find the maximum power consumed by the variable load resistance $R_L$.

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