a) In the circuit below find v_o using nodal analysis. Note that the items in Blue were not on the original diagram but were added to assist in the solution.

I have chosen the bottom node as the reference node and called the top node A, thus the node voltage at that point is V_A.

Next, define the dependent source parameter iD in terms of the node voltage.

To do this, it is helpful to introduce a new voltage variable across the 25 W resistor, V₂₅, to assist us. Note that its polarity was chosen so the reference arrow for iD enters the positive.

We know that the total voltage from node A to the reference node is simply V_A, and that this MUST equal the sum of individual valtages along any path from node A to the reference node. Following the path on the right through the 25 W resistor and the 45 volt source yields

V_A = - V₂₅ + 45

and solving for V₂₅ gives

V₂₅ = 45 - V_A

Ohm's Law gives us the desired current iD.

iD = V₂₅ / 25
iD = (45 - V_A) / 25

Looking ahead a bit, we see that there will be a slight difficulty determining the current through the 5 W resistor since it is in series with a dependent voltage source. The voltage variable V₅ was introduced specifically to deal with this. Similar to the method for finding V₂₅ earlier, we get

V_A = - V₅ + 6.25 iD

Substituting the expression for iD found earlier gives

V_A = - V₅ + 6.25 (45 - V_A) / 25 = - V₅ + 0.25 (45 - V_A) = - V₅ + 11.25 - 0.25 V_A

Solving for V₅ gives

V₅ = 11.25 - 0.25 V_A - V_A
V₅ = 11.25 - 1.25 V_A

Now we can write the KCL equation

Inserting the values for V₅ and iD determined above gives

Now to solve the equation.

Multiply through by 100

45 - V_A + 20 (11.25 - 1.25 V_A) + 4 (45 - V_A) = 0

Collecting terms gives

30 V_A = 450

thus

V_A = 15 volts

Since V_A = v_o

v_o = 15 volts

(Note that I could have simply labeled the top node v_o, but I didn't notice it until I had almost completed this solution and it doesn't seem worth the effort to go back and change everything.)

b) Find the power absorbed by the dependent source. For convenience, I have copied the circuit again, with the addition of a new current to help us find the dependent source power. Recall that the power absorbed expression is positive (P=VI) if the arrow enters the positive reference, so I chose the arrow down so that it enters the positive reference of the dependent source. (One might also make a case for choosing it up so that it would enter the positive reference of V₅.)

As mentioned above P = VI. In this case, V = 6.25 iD and I = i₅.

We found earlier that

iD = (45 - V_A) / 25

Plugging in the calculated value for V_A

iD = 1.2A

We also found that

V₅ = 11.25 - 1.25 V_A

and inserting V_A gives

V₅ = - 7.5

thus by Ohm's Law (which must have a negative sign inserted since the arrow enters the negative reference for V₅)

i₅ = - V₅ / 5 = - (-7.5) / 5 = 1.5

Plugging all this in to P = VI gives

P = 6.25 (iD) (1.5) = 6.25 (1.2) (1.5)
P = 11.25 W absorbed by the dependent source

c) Find the total power developed by the independent sources.

The equation for power developed is positive if the arrow LEAVES the positive reference (one of the few places where this is true). Since this is the case with the 450 mA arrow and v_o, the power developed by the current source is

P₄₅₀ = 15 (450m) = 6.75 W

When considering the 45 volt source, note that the reference arrow for leaves the positive 45 V reference, so again the power developed equation is positive and we get the power developed by the 45 volt source to be

P₄₅ = 45 iD = 45 ( 1.2) = 54 W

Therefore the total power developed by the two independent sources (P_is) is simply the sum of the two, or

P_is = 60.75 W