Problem 4-88

Use superposition to find the voltage v_o.

Solve the circuit for v_o one independent source at a time and add the results. Note that the dependent source must always remain in the circuit.

First "kill" the 5 mA source. I will use mesh analysis.

Define dependent source parameter.

i_f = i₁ - i₂

Express current source value in terms of mesh current. Removing the current source then eliminates the top mesh.

2.2 i_f = i₁
2.2 i₁ - 2.2 i₂ = i₁
1.2 i₁ - 2.2 i₂ = 0
i₁ = 1.8333 i₂ (Equation 1)

KVL around mesh 2 gives

4K (i₂ - i₁) + 20K i₂ = 25 (Equation 2)

Inserting equation 1 into equation 2

4K (i₂ - 1.8333 i₂) + 20K i₂ = 25
16.667K i₂ = 25
i₂ = 1.5 mA

Finally calculate v_o from Ohm's Law.

v_o = 20K i₂ = 20K (1.5m)
v_o = 30 V (Voltage due to 25 volt source alone.)

Next kill the 25 volt source, replacing the 5 mA source.

This time I will use nodal analysis.

There are only two nodes, and if the lower node is the reference, the other is simply v_o.

Define dependent source parameter.

i_f = v_o / 4K

Write KCL at node o. I will sum currents out of the node.

v_o / 20K + 5 mA + v_o / 4K - 2.2 (v_o / 4K) = 0
v_o + 100 + 5 v_o - 11 v_o = 0
5 v_o = 100
v_o = 20 V (Voltage due to 5 mA source alone.)

Finally, the total voltage v_o is just the sum of the voltages due to the individual sources.

v_o = 30 + 20
v_o = 50 V