a) Find the value of R_o so that it absorbs the maximum possible power from the circuit.

For maximum power transfer to R_o, it should equal the Thevenin equivalent resistance of the circuit to which it is connected, thus we need to determine that equivalent resistance.

I will do this by "killing" the independent sources and cannecting a test source in place of R_o. If a voltage source is chosen, we immediately know i_D which should simplify the analysis a bit. We will thus need to find the current I delivered to the circuit by the test voltage source. I will use nodal analysis to determine I. The new circuit is thus

First, since the 80 ohm resistor is in parallel with the 1 volt source, the dependent source parameter i_D is easily seen to be

i_D = - 1 / 80 = - 12.5 mA

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Next, both voltage sources directly connect two essential nodes forming supernodes. Expressing the value of each source in terms of the node voltages gives

V_B = 1

124 i_D = V_A - V_C

but since i_D = - 12.5 mA

V_A - V_C = 124 (- 12.5m) = - 1.55 (equation 1)

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Since node B is directly connected to the reference node by the 1 volt source, no KCL equation is written at node B.

Nodes A and C form a supernode, and we must write KCL for that supernode. I will sum the currents out of the supernode, and I will also immediately substitute 1 volt for V_B.

V_A / 16 + (V_A -1) / 4 + V_C / 12 + (V_C - 1) / 8 = 0

Multiply through by 48

3 V_A + 12 V_A -12 + 4 V_C + 6 V_C - 6 = 0
15 V_A + 10 V_C = 18 (equation 2)

Solving the two equations gives

V_A = 0.1
V_C = 1.65

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We can now get the current I using KCL at node B

I = (V_B - V_A) / 4 + (V_B - V_C) / 8 + V_B / 80
I = (1 - 0.1) / 4 + (1 - 1.65) / 8 + 1 / 80
I = 156.25 mA

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Finally, Ohm's Law yields the equivalent resistance

R_Th = 1 / I = 1 / 156.25m = 6.4
R_o = 6.4 W

There are basically two ways to proceed here.
One is to find the Thevenin or Norton Equivalent source, then attach the load resistance to the equivalent circuit to determine the power.
The other is to simply insert the load resistance into the original circuit to determine the power absorbed.

I will go the extra mile and find BOTH V_oc and I_sc which will then give us a check of the Thevenin equivalent resistance found directly in part a).

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V_oc: I will use mesh analysis

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Define dependent source parameter

i_D = I_C - I_B

Write KVL around each mesh

Mesh A:

124 i_D + 8 (I_A - I_C) + 4 (I_A - I_B) = 0
124 (I_C - I_B) + 8 (I_A - I_C) + 4 (I_A - I_B) = 0
12 I_A - 128 I_B + 116 I_C = 0 (equation 3)

Mesh B:

4 (I_B - I_A) + 80 (I_B - I_C) + 16 I_B - 100 = 0
- 4 I_A + 100 I_B - 80 I_C = 100 (equation 4)

Mesh C:

8 (I_C - I_A) + 50 + 12 I_C + 80 (I_C - I_B) = 0
- 8 I_A - 80 I_B + 100 I_C = - 50 (equation 5)

Solving equations 3, 4, and 5 gives

I_A = 10.5 A
I_B = 4.7 A
I_C = 4.1 A

Finally, the open circuit voltage is the same as the voltage across the 80 ohm resistor, thus

V_oc = 80 (I_B - I_C) = 80 (4.7 - 4.1)
V_oc = 48 V

So, if we calculated R_Th and V_oc correctly, the power absorbed by the load resistance (maximum) is

P = V_oc² / R_o = 48² / 6.4
P = 360 W

I_sc = V_oc / R_Th = 48 / 6.4 = 7.5 A

So direct determination of I_sc should agree with this value if our previous results are correct.

With the short circuit in place, the circuit is

Before we dive right in, let's consider this circuit for a moment.

First, note that the 80 ohm resistor is shorted out, thus

i_D = 0

This immediately implies that the voltage developed by the dependent source is ZERO.

Since the 80 ohm resistor is shorted out, it can be removed from the circuit without having any effect.
Since the dependent source voltage is zero, it can be replaced with a wire without affecting the circuit. The modified circuit is thus

Although it might appear that we could use source transforms and circuit reduction techniques here, it would be VERY difficult to keep track of the desired current I_sc. Therefore, I will use mesh analysis.

Writing KVL around each mesh gives:

Mesh A:

8 (I_A - I_C) + 4 (I_A - I_B) = 0
12 I_A - 4 I_B - 8 I_C = 0 (equation 6)

Mesh B:

4 (I_B - I_A) + 16 I_B - 100 = 0
- 4 I_A + 20 I_B = 100 (equation 7)

Mesh C:

8 (I_C - I_A) + 50 + 12 I_C = 0
- 8 I_A + 20 I_C = - 50 (equation 8)

Solving equations 6, 7, and 8 gives

I_A = 0 A
I_B = 5 A
I_C = - 2.5 A

Therefore

I_sc = I_B - I_C = 5 + 2.5
I_sc = 7.5 A

Since this agrees with the value we calculated for I_sc using V_oc and R_Th, we have a high degree of confidence that the earlier work is correct.

NOTE: This problem well illustrates the desirability of checking your work by an independent method when possible. When I first solved for R_Th, I dropped a minus sign, thus the value determined for R_Th was incorrect. When my value for I_sc did not agree with V_oc / R_Th, I knew there was an error somewhere.