# Problem 1-10: Solving by Nodal Analysis - Circuit with Four Nodes

Solve the circuit by nodal analysis and find $V_a$ .

Solution
a) Choose a reference node, label node voltages:

b) Apply KCL to each node:

Node 1:
$-Is_2 +\frac{V_1 - V_2}{R_3}+\frac{V_1 - V_3}{R_1}=0 \to 6V_1 - V_2 - 5V_3=5$ (1)

Node 2:
$Is_3 +\frac{V_2 - V_1}{ R_3} + \frac{V_2}{R_2}=0 \rightarrow - 4V_1+9V_2=-40$ (2)

Node 3:
$-Is_1 - Is_3 +\frac{V_3 - V_1}{ R_1}=0 \rightarrow V_3 - V_1=4$ (3)
(1), (2) and (3) imply that $V_1=37v, \, V_2= 12v$ and $V_3=41$ .

c) Find the required quantities:

If we apply KVL in the loop shown above:

$-V_1 - V_a +V_2=0 \rightarrow V_a= -25 v$ .

## 3 thoughts on “Problem 1-10: Solving by Nodal Analysis - Circuit with Four Nodes”

1. Joseph kyei baffour says:

Wow.

2. aray says:

Would it not be easier if we had taken v2 in the circuit as reference node?

3. hagos says:

your solution beast for eletrical engineering.