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 source and connecting a test source in place of R_o. I will determine the voltage developed across the circuit by a 1 amp source and use mesh analysis to accomplish this. The circuit we must solve is thus

Define dependent source parameters in terms of the mesh currents.

i_D = I_B

V_D = 2 I_A

Define current source values in terms of mesh currents, then delete the current sources to form supermeshes.

I_C = - 1

2 V_D = I_B - I_C
2 (2 I_A) = I_B + 1
4 I_A - I_B = 1 (equation 1)

We now see that there is only one mesh remaining. Write KVL around mesh 1.

2 I_A + 5 (I_A - I_B) + 4 i_D = 0
2 I_A + 5 (I_A - I_B) + 4 I_B = 0
7 I_A - I_B = 0 (equation 2)

Solving equations 1 and 2 yields

I_A = - 333.3 mA
I_B = - 2.333 A

KVL around the perimeter gives

2 I_A + 4 I_B + V = 0
V = - 2 I_A - 4 I_B
V = - 2 (- 0.333) - 4 (- 2.333)
V = 10 V

From Ohm's Law

R_Th = V / 1 = 10

So for maximum power transfer to the load

R_o = 10 W

We can either insert a 10 ohm resistor in the circuit and analyze it, or we can find the Thevenin or Norton equivalent circuit and connect the 10 ohm resistor to the equivalent circuit. I will choose the latter course.

Find I_sc.

Define dependent source parameters in terms of the mesh currents.

i_D = I_B

V_D = 2 I_A

Define current source value in terms of mesh currents, then delete the current source to form a supermesh.

2 V_D = I_B - I_C
2 (2 I_A) = I_B - I_C
4 I_A - I_B + I_C = 0 (equation 3)

Now write KVL around mesh A and supermesh B/C.

Mesh A:

2 I_A + 5 (I_A - I_B) + 4 i_D - 60 = 0
2 I_A + 5 (I_A - I_B) + 4 I_B = 60
7 I_A - I_B = 60 (equation 4)

Supermesh B/C:

4 I_B - 4 i_D + 5 (I_B - I_A) = 0
4 I_B - 4 I_B + 5 (I_B - I_A) = 0
- 5 I_A + 5 I_B = 0 (equation 5)

Solving equations 3, 4, and 5 yields

I_A = 10 A
I_B = 10 A
I_C = - 30 A

Thus

I_sc = - 30 A

and the Norton equivalent circuit is

If we connect a load of 10 ohms to this circuit, exactly half of the current will go through the load (current division), thus the maximum power absorbed by the load is

P = (- 15)² (10)
P = 2250 W

Define dependent source parameters in terms of the mesh currents.

i_D = I_B

V_D = 2 I_A

Define current source value in terms of mesh currents, then delete the current source to form a supermesh.

2 V_D = I_B
2 (2 I_A) = I_B
4 I_A - I_B = 0 (equation 6)

Removing the current source open mesh B to the external world, so no KVL needed around mesh B.

KVL around mesh A:

2 I_A + 5 (I_A - I_B) + 4 i_D - 60 = 0
2 I_A + 5 (I_A - I_B) + 4 I_B = 60
7 I_A - I_B = 60 (equation 7)

Solving equations 6 and 7 gives

I_A = 20 A
I_B = 80 A

Finally, KVL around the perimeter of the circuit gives

2 I_A + 4 I_B + V_oc - 60 = 0
V_oc = - 2 I_A - 4 I_B + 60
V_oc = - 2 (20) - 4 (80) + 60
V_oc = - 300 V

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

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IMPORTANT NOTE: The novice's first thought is that this question can be answered by saying 50%. The (faulty) logic is thus:

If the load equals the Thevenin equivalent resistance, then each resistor is absorbing exactly half of the power generated by the Norton (or Thevenin) equivalent source.

Although this sometimes reflects what actually happens in the original circuit in terms of power generation, it is certainly not always true. A simple, easy to understand counter-example is shown below.

The Thevenin equivalent for this circuit is (details left to the reader)

Thus the load resistance should equal 10 ohms for maximum power transfer to the load.

If a 10 ohm resistor is connected to the Thevenin equivalent circuit, the load resistance absorbs 2.5 watts and the Thevenin equivalent source generates 5 watts, thus indeed the load absorbs 50% of the power generated by the Thevenin equivalent source.

However, if a 10 ohm resistor is connected to the original circuit, the load resistance still absorbs 2.5 watts, but the 10 V source in the original circuit generates 15 watts, thus the percentage of power generated in the original circuit that is absorbed by the load is only 16.7%.

The moral: Until your level of understanding of circuits is quite sophisticated, you will have to work out a problem like this using the original circuit, NOT the Thevenin or Norton equivalent.

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Hooking up a 10 ohm resistor (the Thevenin resistance of the original circuit in the book as well as the counter example above) to the original circuit, we get

Using mesh analysis:

Define dependent source parameters in terms of the mesh currents.

i_D = I_B

V_D = 2 I_A

Define current source value in terms of mesh currents, then delete the current source to form a supermesh.

2 V_D = I_B - I_C
2 (2 I_A) = I_B - I_C
4 I_A - I_B + I_C = 0 (equation 8)

Now write KVL around mesh A and supermesh B/C.

Mesh A:

2 I_A + 5 (I_A - I_B) + 4 i_D - 60 = 0
2 I_A + 5 (I_A - I_B) + 4 I_B = 60
7 I_A - I_B = 60 (equation 9)

Supermesh B/C:

4 I_B + 10 I_C - 4 i_D + 5 (I_B - I_A) = 0
4 I_B + 10 I_C - 4 I_B + 5 (I_B - I_A) = 0
- 5 I_A + 5 I_B + 10 I_C = 0 (equation 10)

Solving equations 8, 9, and 10 yields

I_A = 15 A
I_B = 45 A
I_C = - 15 A

We already calculated that the load absorbs 2250 watts (which is easily verified with the value of I_C calculated immediately above). We thus need to determine the total power generated in the circuit.

Power generated by the 60 volt source:

P₆₀ = 60 I_A = 60 (15)
P₆₀ = 900 W

Power generated by the dependent voltage source:

P_DVS = 4 i_D (I_B - I_A)
P_DVS = 4 I_B (I_B - I_A)
P_DVS = 4 (45) (45 - 15)
P_DVS = 5400 W

Power generated by the dependent current source:

First determine the voltage across the dependent current source (+ reference at bottom). Note that this is the same as the voltage across the load resistor.

V_DCS = - 10 I_C (arrow enters negative reference!)
V_DCS = - 10 (- 15)
V_DCS = 150 V

Now calculate the power:

P_DCS = V_DCS (2 V_D)
P_DCS = V_DCS (2 [2 I_A])
P_DCS = 150 (4) (15)
P_DCS = 9000 W

The total power generated in the circuit is thus

P_Total = 900 + 5400 + 9000
P_Total = 15300 W

Therefore the percentage of the power generated by the circuit that is absorbed by the load is:

P_Load / P_Total X 100% = 2250 / 15300 X 100% = 14.7%