Given. The network of Figure 2. The ideal current source $I_s$ sits in series between the $V_{s1}/V_{s2}/R_1$-$R_4$ sub-network and the $R_5$-$R_6$-$R_7$-$R_L$ sub-network; because an ideal current source forces exactly $I_s$ into node N2 regardless of what is connected to its other terminal, everything to the left of $I_s$ is electrically irrelevant to the Thevenin equivalent seen by $R_L$.
Given data
Quantity
Value
$R_1,R_2,R_3,R_4$
$50,\ 100,\ 50,\ 100\ \Omega$
$R_5,R_6,R_7$
$100,\ 20,\ 80\ \Omega$
$V_{s1},\ V_{s2}$
$20\text{ V},\ 5\text{ V}$
$I_s$
$20\text{ A}$
Find. $V_{TH}$, $R_{TH}$, the load resistance and power at maximum power transfer, and $P_{R_L}$ for $R_L=100\,\Omega$.
Figure 2: Network for Question 2 (only I_s, R5, R6, R7 set the Thevenin seen by R_L -- the ideal current source isolates the V_s1/V_s2/R1-R4 sub-network)
Approach. Remove $R_L$; find the open-circuit voltage at its terminals ($V_{TH}$) and the resistance seen looking back into those terminals with all sources deactivated ($R_{TH}$); then apply the maximum-power-transfer theorem and the voltage-divider form of the loaded circuit.
a) $V_{TH}$ -- open-circuit voltage. With $R_L$ removed, node N2 (where $I_s$, $R_5$, and $R_6$ meet) sees $R_5$ to ground in parallel with the series path $R_6+R_7$ to the open terminal, so all of $I_s$ splits between them: $$V_{N2} = \dfrac{I_s}{\tfrac{1}{R_5}+\tfrac{1}{R_6+R_7}} = \dfrac{20}{\tfrac{1}{100}+\tfrac{1}{100}} = 1000\text{ V}$$ The open-circuit load voltage is then a voltage division of $V_{N2}$ across $R_6,R_7$ (no current flows into the open terminal, so $R_7$ carries the same current as $R_6$): $$V_{TH} = V_{N2}\cdot\dfrac{R_7}{R_6+R_7} = 1000\cdot\dfrac{80}{100} = \boxed{800\text{ V}}$$
b) $R_{TH}$ -- deactivate sources. Opening $I_s$ (ideal current source → open circuit) disconnects the entire left sub-network from N2, leaving only $R_5$ (N2-to-ground) and $R_6$ in series with $R_7$ (N2-to-load-terminal) visible from the load: $$R_{TH} = R_7\parallel(R_6+R_5) = \dfrac{80(20+100)}{80+20+100} = \boxed{48\,\Omega}$$
c) Maximum power transfer. By the maximum-power-transfer theorem, $R_L$ draws maximum power when $R_L=R_{TH}$: $$R_{L,opt} = R_{TH} = \boxed{48\,\Omega}, \qquad P_{max} = \dfrac{V_{TH}^2}{4R_{TH}} = \dfrac{800^2}{4(48)} = \boxed{3333.3\text{ W}}$$
d) Power at $R_L=100\,\Omega$. Using the loaded Thevenin circuit as a voltage divider, $$P_{R_L} = \left(\dfrac{V_{TH}}{R_{TH}+R_L}\right)^2 R_L = \left(\dfrac{800}{148}\right)^2(100) = \boxed{2921.8\text{ W}}$$ (less than $P_{max}$, as expected since $100\,\Omega \ne R_{TH}$).
