Node Voltage

As its name implies, the node voltage method is used to find a node’s voltage with respect to ground.  While a voltage divider can be used for the same purpose, the primary purpose of a voltage divider is to find voltage drops across resistances rather than with respect to ground.  One disadvantage of using a voltage divider is that a circuit must be simplified to a two resistor series circuit before the equation can be applied.  The larger the circuit, the more difficult it becomes to simplify the circuit, especially when there are multiple sources.  With the node voltage method, no simplification is necessary.  It can be applied to the circuit as is.


The circuit in Figure 1 shows three resistors meeting at a node.  We’ll use this node to show how node voltage works in the most general sense (i.e. without using any numbers).  As you’ll see, the node voltage method is basically Ohm’s law applied to Kirchoff’s current law.

Note: All of the voltages in Figure 1 (and other figures throughout this article) are node voltages, or voltages at a node, which are taken with respect to ground.

Steps for solving a circuit using node voltage

1. Pick a node – This can be any node.  In the problems you’ll see, you’re told which node to use.  In reality, node voltage can be applied to any node because KCL can be applied to any node.  Let’s pick the Vx node for this example.

2. Apply KCL to the node – There are three paths or branches connected to the Vx node and each path has its own current.  These currents are labeled as I1, I2, and I3.  Rather than trying to guess which way these currents flow, assume that all currents leave the node.  If we assume that all currents leave the node, then we’re also assuming no currents enter the node, simplifying the KCL equation as shown below.

\Sigma i_{leaving} = \Sigma i_{entering} \rightarrow \Sigma i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0

Note that it doesn’t matter if our assumption is wrong.  If one of the currents is actually flowing into the node, it will show up later as a negative in the KCL equation (and we’re not afraid of negative numbers).

3. Write equations for each current – Unless one of the branches is a current source, you’ll have to place each current in terms of voltages and resistances according to Ohm’s law.

I_{1} = \frac{V_{x} - V_{1}}{R_{1}}               I_{2} = \frac{V_{x} - V_{2}}{R_{2}}             I_{3} = \frac{V_{x} - V_{3}}{R_{3}}

4. Derive the node voltage equation – Plug the equations you found in step 3 back into the KCL equation in step 2 to derive the node voltage equation for Vx.

i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{x} - V_{1}}{R_{1}} + \frac{V_{x} - V_{2}}{R_{2}} + \frac{V_{x} - V_{3}}{R_{3}}

5. Solve for any variables – In most problems you’ll encounter, all of the variables will be known except the voltage at the node.  In this instance, solve for Vx and you’re finished.  A node voltage equation can be written for any node in a circuit.  For multiple variables, you’ll need multiple equations (more on this later).


Example 1: Find the voltage V1 for the circuit in Figure 2.

Solution: The first step has already been taken care of in the problem statement.  The next step is to write the KCL equation for the V1 node.  Mark the currents as shown in Figure 3, assuming that they all flow away from the node.  The KCL equation becomes:

\Sigma i_{leaving} = \Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

Next, write equations for each current.  The two voltage sources are connected to ground, making it easy to find the voltages at the nodes on the upper left and right corners of the circuit.  Notice that Va’s positive terminal is connected to ground, making the voltage at the negative terminal -8 V.

I_{1} = \frac{V_{1} - (-V_{a} )}{R_{1}} = \frac{V_{1} - (-8)}{150}            I_{2} = \frac{V_{1} - 0}{R_{2}} = \frac{V_{1}}{30}             I_{3} = \frac{V_{1} - V_{b}}{R_{3}} = \frac{V_{1} - 10}{20}

Using these equations, derive the node voltage equation for V1.  Solving for V1 gives the voltage with respect to ground at that node.

\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} + 8}{150} + \frac{V_{1}}{30} + \frac{V_{1} - 10}{20} = 0 300(\frac{V_{1} + 8}{150} + \frac{V_{1}}{30} + \frac{V_{1} - 10}{20} = 0) = 0(300) \rightarrow 2V_{1} + 16 + 10V_{1} + 15V_{1} - 150 = 0 \rightarrow V_{1} = 4.96V

Working with current sources


Example 2: Find the voltage V1 for the circuit in Figure 4.

Solution: The approach doesn’t change when there are current sources in the circuit.  Current sources can actually simplify the math.  Once again, the node has been chosen by the problem statement.  Now write the KCL equation and solve the problem the same way as before.  Note that the current source points into the node and therefore needs to be added as a negative current.

\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

I_{1} = \frac{V_{1} - 6}{3k}           I_{2} = \frac{V_{1}}{1k}          I_{3} = -0.01

\Sigma i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} - 6}{3k} + \frac{V_{1}}{1k} - 0.01 = 0 \rightarrow V_{1} = 9V

Working with Multiple Variables


Example 3: Find the voltages V1 and V2 for the circuit in Figure 6.

Solution: Approach every node voltage problem in the same manner: choose a node and derive its node voltage equation.  If the equation has more than one variable, then move to the next node and write another node voltage equation.  Continue this process until the number of equations equals the number of variables.  In this example, there will be two variables (V1 and V2).  This means two node voltage equations are required to find the two voltages.

This time there are two nodes to choose from.  Start with V1 first.  As always, assume the currents flow away from the node.

node-voltage-currents2\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

Deriving equations for I1 and I2 is the same as before, but I3 is slightly different.  The voltage V2 is unknown, but it can still be placed into the equation.  It will be treated as a variable just like V1.

I_{1} = \frac{V_{1} - V_{a}}{R_{1}} = \frac{V_{1} - 150}{20}                I_{2} = \frac{V_{1} - 0}{R_{2}} = \frac{V_{1}}{80}               I_{3} = \frac{V_{1} - V_{2}}{R_{3}} = \frac{V_{1} - V_{2}}{40}

The node voltage equation for V1 becomes

I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} - 150}{20} + \frac{V_{1}}{80} + \frac{V_{1} - V_{2}}{40} = 0 \rightarrow 7V_{1} - 2V_{2} = 600

Now follow the same procedure to find the node voltage equation for V2.

node-voltage-currents3I_{1} = \frac{V_{2} - V_{1}}{R_{1}} = \frac{V_{2} - V_{1}}{40}

I_{2} = -11.25 A I_{3} = \frac{V_{2} - 0}{R_{4}} = \frac{V_{2}}{4} I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{2} - V_{1}}{40} -11.25 + \frac{V_{2}}{4} = 0 \rightarrow -V_{1} + 11V_{2} = 450

Finally, solve the system of equations to find V1 and V2.

V_{1} = 100V                            V_{2} = 50V

That should just about cover quite a few node-voltage examples and concepts.  If you have any questions feel free to ask them in the questions section of the website, or simply leave a comment below!  Many thanks to Ryan Eatinger ( for contribution of this post. 🙂

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