Use nodal analysis method to solve the circuit and find the power of the  3\Omega – resistor.
Nodal Analysis  Dependent Current Source


I. Identify all nodes in the circuit.
The circuit has 3 nodes as shown below.

Nodal Analysis  Dependent Current Source - Circuit has 4 nodes

II. Select a reference node. Label it with the reference (ground) symbol.

The node in the middle is connected to 5 nodes and is the node with the largest number of elements connected to it. Therefore, we select it as the reference node of the circuit.

III. Assign a variable for each node whose voltage is unknown.
We label the remaining nodes as shown below.  V_1 is connected to the reference node through a voltage source. Therefore, it is equal to the voltage of the dependent voltage source:  V_1=-6I_x .

Nodal Analysis  Dependent Current Source - Reference Node - Node Voltages

IV. If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

The voltage of the dependent voltage source is  -I_x. We should find this value in terms of the node voltages.  I_x is the current of the  5\Omega– resistor. The voltage across the resistor is  V_1-V_2. You may ask why not  V_2-V_1. Well, that is also correct; the voltage across the resistor is either  V_1-V_2 or  V_2-V_1 depend on which terminal we choose to be the positive one. In this circuit, we are going to use this voltage drop to determine  I_x. We prefer to use  V_1-V_2 simply because  V_1 is the voltage of the terminal that  I_x entering from. Therefore, the Ohm’s law can be applied in the simple form of  V=R \times I. By using the voltage drop  V_1-V_2, we have  I_x=\frac{V_1-V_2}{5\Omega}.

V. Write down a KCL equation for each node.

Node of  V_1 :
Because there is a voltage source in this node, there is no advantage in writing a KCL equation for this node. All we need to do is to use the voltage of the dependent voltage source and its relation with other node voltages:
 \left\{ \begin{array}{l} I_x=\frac{V_1-V_2}{5\Omega} \\ V_1=-6I_x \end{array} \right. \to V_1=\frac{6}{11}V_2.

Node of  V_2 :
 \frac{V_2-V_1}{5\Omega}+\frac{V_2}{3\Omega}-2A=0 \to -3V_1+8V_2=30.
Substituting  V_1=\frac{6}{11}V_2 ,
 V_2=\frac{33}{7} V \to V_1=\frac{18}{7} V.

All node voltages are obtained. The power of the  3\Omega -resistor is
 R_{3\Omega}=\frac{V_2^2}{3\Omega}=7.408 W.

The PSpice simulation result is illustrated below. The PSpice schematics can be downloaded from

Nodal Analysis  Dependent Current Source - PSpice simulation result

Published by Yaz

Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too.

Join the Conversation


  1. thank you very much, it was helpful on the dependent source part of the circuit.

Leave a comment

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.