Nodal Analysis:

• 3 variables, thus 3 equations needed.
• 240 V source directly connects V₃ to the reference node, thus we immediately know V₃, therefore we really only have 2 equations to solve.
• As usual with nodal analysis, we have the extra algebraic work of finding least common denominators, etc.
• When defining the dependent source parameter i_d, we will need to use KCL expressed in terms of node voltages since i_d is not directly associated with a resistor and we cannot use Ohm's Law.

Mesh Analysis:

• 4 variables, thus 4 equations required.
• I₁ is the only mesh current through the19 A source, thus we immediately know I₁, therefore we really only have 3 equations to solve.
• As usual with mesh analysis, the algebraic simplification is a bit easier than with nodal analysis.
• The dependent source parameter i_d is easy to define since it equals one of the mesh currents.

Verdict:

Since the equations are not to be solved manually but with a calculator or computer, mesh analysis would probably be my choice despite the extra equation to avoid the additional algebra and the slight difficulty with i_d, but it is a close call. If I had to solve the equations manually, I would probably choose nodal analysis.

I have worked this problem by both methods. Look in the homework index under nodal analysis for the nodal solution or click here.

i_b = I₄ - I₂
i_d = I₃

19 Amp source:

19 = I₁ (equation 1)

Dependent current source:

2 i_b = I₃ - I₂
2 (I₄ - I₂ ) = I₃ - I₂
I₂ + I₃ - 2 I₄ = 0 (equation 2)

Removing the current sources to form supermeshes results in the following circuit

Mesh 4:

- 4 i_d + 10 (I₄ - I₃) + 5 (I₄ - I₂) = 0
- 4 I₃ + 10 (I₄ - I₃) + 5 (I₄ - I₂) = 0
5 I₂ + 14 I₃) - 15 I₄ = 0 (equation 3)

Supermesh 2/3:

5 (I₂ - I₄) + 10 (I₃ - I₄) - 240 + 40 (I₂ - I₁) = 0
Replace I₁ using equation 1.
5 (I₂ - I₄) + 10 (I₃ - I₄) - 240 + 40 (I₂ - 19) = 0
45 I₂ + 10 I₃ - 15 I₄ = 1000 (equation 4)

I₁ = 19 A
I₂ = 26 A
I₃ = 10 A
I₄ = 18 A

ia = I₁ - I₂ = 19 - 26
ia = - 7 A

ib = I₄ - I₂ = 18 - 26
ib = - 8 A

ic = I₄ - I₃ = 18 - 10
ic = 8 A

id = I₃
id = 10 A

ie = I₄
ie = 18 A

These answers match the solutions obtain via nodal analysis, so they are almost certainly correct.

In addition, the total power is calculated below which also checks.

I will assume power is being generated by all sources and calculate power generated (P = VI with arrow leaving positive reference) for all four sources. If one or more is actually absorbing power, the value for that source will be negative.

19 A source

P_gen = 19 (40 ia) = 19 (-280) = - 5320 W

Dependent voltage source

P_gen = 4 id (ie) = 4 (10) (18) = 720 W

Dependent current source (I will get the voltage across the source as the sum of the 10 ohm voltage and the 240 volt source).

P_gen = 2 ib (- ic - 240) = 2 (- 8) (-80 - 240) = 5120 W

240 V source

P_gen = 240 id 240 (10) = 2400 W

Net power generated by the sources is

PgenTotal = - 5320 + 720 + 5120 + 2400 = 2920 W

I will calculate power absorbed for all resistors (P = I²R).

40 ohm resistor

P_abs = ia² (40) = 49 (40) = 1960 W

5 ohm resistor

P_abs = ib² (5) = 64 (5) = 320 W

10 ohm resistor

P_abs = ic² (10) = 64 (10) = 640 W

Total power absorbed by the resistors

PabsTotal = 1960 + 320 + 640 = 2920 W

Since power generated by the sources equals power absorbed by the resistors, our answers are probably correct.