Problem-Solving Strategy for solving the Nodal Analysis
STEP 1. Determine the number of nodes in the circuit. Select one node as the reference node. Assign a node voltage between each nonreference node and the reference node. All node voltages are assumed positive with respect to the reference node. For an $N$-node circuit, there are $N−1$ node voltages. As a result, $N−1$ linearly independent equations must be written to solve for the node
voltages.
STEP 2. Write a constraint equation for each voltage source—independent or dependent—
in the circuit in terms of the assigned node voltages using KVL. Each constraint equation represents one of the necessary linearly independent equations, and $N_v$ voltage sources yield $N_v$ linearly independent equations. For each dependent voltage source, express the controlling variable for that source in terms of the node voltages. A voltage source—independent or dependent—may be connected between a nonreference node and the reference node or between two nonreference nodes. A supernode is formed by a voltage source and its two connecting nonreference nodes.
STEP 3. Use KCL to formulate the remaining $N−1−N_v$ linearly independent equations.
First, apply KCL at each nonreference node not connected to a voltage source. Second, apply KCL at each supernode. Treat dependent current sources like independent current sources when formulating the KCL equations. For each dependent current source, express the controlling variable in terms of the node voltages.
PROBLEM:
Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.
If there are any errors in solution, please let me know in the comments
SOLUTION:
Applying KCL to each node is as follows:
$\frac{V_1+V_x+12}{1k}+\frac{V_1-V_x+12}{1k}+\frac{V_1-V_x}{1k}+\frac{V_1-V_o}{1k}=0$
$\frac{V_o-V_x}{2k}+\frac{V_o-V_1}{1k}+\frac{V_o}{1k}=0$
$-4mA+\frac{V_x-12}{2k}+\frac{V_x}{2k}+\frac{V_o}{1k}=0$
Combining the equations yields the three equations
$4V_1-V_x-V_o=-24$
$5V_o-2V_1-V_x=0$
$V_x+V_o=10$
Solving these equations, we obtain
$V_o=0.5V$
Related Post: Loop Analysis Technique
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