The Node Voltage Method is an organized methods of analyzing a circuit. The Node Voltage Method is based on Kirchhoff's Current Law. This technique is embedded inside the popular circuit simulator, $\small \text{SPICE}$SPICEstart text, S, P, I, C, E, end text.

What is the circuit analysis challenge? Solving any circuit means creating and solving $2E$2E2, E independent equations, where $E$EE is the number of elements (components and sources). Half of the equations come from the individual element laws (like Ohm's Law), and the other half come from the connections between elements.

No matter what procedure we use to solve the circuit, there is no getting around the requirement to solve $2E$2E2, E equations. Even for simple circuits, managing $2E$2E2, E equations can be a lot of work. But, there are ways to organize the effort to make it very efficient. The Node Voltage Method is one of two very efficient procedures we have for solving circuits. (The other one is the Mesh Current Method.)

The Node Voltage Method is not new science. It processes the same amount of information contained in $2E$2E2, E equations, but it is quite clever and efficient in how it organizes that information.

We will demonstrate the Node Voltage Method with the same circuit we solved using the fundamental laws:

## Definition: node voltage

We need to define a new term: *node voltage*. Up to now, we've talked about the *element voltage* that appears across the terminals of a single element (also called a *branch voltage*). When we use the term *node voltage*, we are referring to the potential difference between two nodes of a circuit.

We select one of the nodes in our circuit to be the *reference node*. All the other node voltages are measured with respect to this one reference node. If node $\greenD c$cstart color #1fab54, c, end color #1fab54 is assigned as the reference node, we establish two node voltages at nodes $\greenD a$astart color #1fab54, a, end color #1fab54 and $\greenD b$bstart color #1fab54, b, end color #1fab54.

The reference node is almost always called the *ground node*, and it gets a ground symbol in the schematic, as shown above. The potential of the ground node is defined to be $0\,\text V$0V0, start text, V, end text. The potentials of all the other nodes are measured relative to ground.

## Node Voltage Method

The Node Voltage Method breaks down circuit analysis into this sequence of steps,

- Assign a reference node (ground).
- Assign node voltage names to the remaining nodes.
- Solve the easy nodes first, the ones with a voltage source connected to the reference node.
- Write Kirchhoff's Current Law for each node. Do Ohm's Law in your head.
- Solve the resulting system of equations for all node voltages.
- Solve for any currents you want to know using Ohm's Law.

### Assign a reference node and node voltages

We already did this above, but let's do it again. Our example circuit has three nodes, $\greenD a$astart color #1fab54, a, end color #1fab54, $\greenD b$bstart color #1fab54, b, end color #1fab54, and $\greenD c$cstart color #1fab54, c, end color #1fab54, so $N=3$N=3N, equals, 3. Node $\greenD c$cstart color #1fab54, c, end color #1fab54 has a lot of connections, $4$44, and it connects directly to both sources. This makes it a good candidate to play the role of reference node. Node $\greenD c$cstart color #1fab54, c, end color #1fab54 has been marked with the ground symbol to let everyone know our choice for reference node.

We also call out $N-1 = 2$N−1=2N, minus, 1, equals, 2 node voltages on the schematic, labeled in orange as $v_a$vav, start subscript, a, end subscript and $v_b$vbv, start subscript, b, end subscript.

(There is an obvious opportunity here to simplify the two parallel resistors, $6\,\Omega$6Ω6, \Omega with $5\,\Omega$5Ω5, \Omega. We will not do that, because we want to study the Node Voltage Method procedure.)

### Node voltages control the current arrow

Notice something missing from the schematic. There is no orange label on the voltage across the $20\,\Omega$20Ω20, \Omega resistor. When we need to know that voltage, we express it in terms of the node voltages.

$v_{\text{R}} = v_a - v_b\qquad$vR=va−vbv, start subscript, start text, R, end text, end subscript, equals, v, start subscript, a, end subscript, minus, v, start subscript, b, end subscript or $\qquad v_{\text{R}} = v_b - v_a$vR=vb−vav, start subscript, start text, R, end text, end subscript, equals, v, start subscript, b, end subscript, minus, v, start subscript, a, end subscript

#### First important Node Voltage skill - control the current arrow

The node voltages control the direction of the current arrow!

We can express the voltage across the $20\,\Omega$20Ω20, \Omega resistor as the difference between the two node voltages. This can be done two ways, with either $v_a$vav, start subscript, a, end subscript or $v_b$vbv, start subscript, b, end subscript in the first position in the voltage difference equation. The first term in the equation is the one we consider the more positive of the two. Since we use the sign convention for passive components, the choice we make for voltage polarity determines the direction of the current arrow. The current arrow points *into* the positive sign on the resistor voltage.

Above left, $v_a$vav, start subscript, a, end subscript is the more positive voltage compared to $v_b$vbv, start subscript, b, end subscript. The orange arrow representing $v_{\text{R}}$vRv, start subscript, start text, R, end text, end subscript points in the direction of node $\greenD a$astart color #1fab54, a, end color #1fab54, and the current arrow points into the resistor from left to right.

Above right, $v_b$vbv, start subscript, b, end subscript is now defined as the more positive voltage compared to $v_a$vav, start subscript, a, end subscript. The orange arrow representing $v_{\text{R}}$vRv, start subscript, start text, R, end text, end subscript points towards node $\greenD b$bstart color #1fab54, b, end color #1fab54, and the current arrow points into the positive end of the resistor.

We are going to use our new skill immediately, to control the direction of the current arrow in the first term of the KCL equation coming up next.

### Solve the easy nodes

The voltage $v_a$vav, start subscript, a, end subscript is easy to figure out. Node $\greenD a$astart color #1fab54, a, end color #1fab54 connects to a voltage source that connects to reference node $\greenD c$cstart color #1fab54, c, end color #1fab54. That makes it an easy node. The voltage at node $\greenD a$astart color #1fab54, a, end color #1fab54 is $v_a = 140\,\text V$va=140Vv, start subscript, a, end subscript, equals, 140, start text, V, end text.

### Kirchhoff's Current Law at the remaining node

#### Second important Node Voltage skill - scribble on the schematic

The challenging part of circuit analysis is getting the signs right. Scribble on the schematic all you want. Drawing voltage signs and current arrows helps you get the signs in the KCL equation correct.

##### Third important Node Voltage skill - do Ohm's Law in your head as you write KCL

As you write each term in the KCL equation, do Ohm's Law in your head and immediately write the current in terms of node voltages divided by resistance.

We now write a KCL equation for the remaining unsolved node, $\greenE b$bstart color #0d923f, b, end color #0d923f. Node voltage $v_b$vbv, start subscript, b, end subscript is the independent variable.

The current (blue arrow) flowing *into* node $\greenE b$bstart color #0d923f, b, end color #0d923f from the $20\,\Omega$20Ω20, \Omega resistor can be written as $+\dfrac{(140 - v_b)}{20}$+20(140−vb)plus, start fraction, left parenthesis, 140, minus, v, start subscript, b, end subscript, right parenthesis, divided by, 20, end fraction.

The current in the $6\,\Omega$6Ω6, \Omega and $5\,\Omega$5Ω5, \Omega resistors instantly goes into the equation as $-\dfrac{v_b}{6}$−6vbminus, start fraction, v, start subscript, b, end subscript, divided by, 6, end fraction and $-\dfrac{v_b}{5}$−5vbminus, start fraction, v, start subscript, b, end subscript, divided by, 5, end fraction.

We have just one node to deal with, node $\greenE b$bstart color #0d923f, b, end color #0d923f.KCL says the sum of the currents flowing *into* node $\greenE b = 0$b=0start color #0d923f, b, end color #0d923f, equals, 0.

$+\dfrac{(140 - v_b)}{20} - \dfrac{v_b}{6} - \dfrac{v_b}{5} + 18 = 0$+20(140−vb)−6vb−5vb+18=0plus, start fraction, left parenthesis, 140, minus, v, start subscript, b, end subscript, right parenthesis, divided by, 20, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 6, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 5, end fraction, plus, 18, equals, 0

This is pretty cool. Without too much effort, we have one equation with one unknown. When we did this in a previous article using just the fundamental laws, we had to manage $10$1010 equations with $10$1010 unknowns.

### Find the node voltages

Our system of equations happens to be just one equation. Let's solve it to find the node voltage.

$+\dfrac{140}{20} - \dfrac{v_b}{20} - \dfrac{v_b}{6} - \dfrac{v_b}{5} = -18$+20140−20vb−6vb−5vb=−18plus, start fraction, 140, divided by, 20, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 20, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 6, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 5, end fraction, equals, minus, 18

$- \dfrac{v_b}{20} - \dfrac{v_b}{6} - \dfrac{v_b}{5} = -18 - 7$−20vb−6vb−5vb=−18−7minus, start fraction, v, start subscript, b, end subscript, divided by, 20, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 6, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, 5, end fraction, equals, minus, 18, minus, 7

$\left (- \dfrac{3}{60} - \dfrac{10}{60} - \dfrac{12}{60} \right ) \cdot v_b = -25$(−603−6010−6012)⋅vb=−25left parenthesis, minus, start fraction, 3, divided by, 60, end fraction, minus, start fraction, 10, divided by, 60, end fraction, minus, start fraction, 12, divided by, 60, end fraction, right parenthesis, dot, v, start subscript, b, end subscript, equals, minus, 25

$v_b = -25 \cdot \left (- \dfrac{60}{25} \right )$vb=−25⋅(−2560)v, start subscript, b, end subscript, equals, minus, 25, dot, left parenthesis, minus, start fraction, 60, divided by, 25, end fraction, right parenthesis

$v_b = 60 \,\text V$vb=60Vv, start subscript, b, end subscript, equals, 60, start text, V, end text

### Solve for unknown currents using Ohm's Law

Now we have both node voltages, and we can solve for all the unknown currents using Ohm's Law.

$i_{20\,\Omega} = \dfrac{(v_a-v_b)}{20} = \dfrac{(140-60)}{20} = 4\,\text A$i20Ω=20(va−vb)=20(140−60)=4Ai, start subscript, 20, \Omega, end subscript, equals, start fraction, left parenthesis, v, start subscript, a, end subscript, minus, v, start subscript, b, end subscript, right parenthesis, divided by, 20, end fraction, equals, start fraction, left parenthesis, 140, minus, 60, right parenthesis, divided by, 20, end fraction, equals, 4, start text, A, end text

$i_{6\,\Omega} = \dfrac{v_b}{6} = \dfrac{60}{6} = 10\,\text A$i6Ω=6vb=660=10Ai, start subscript, 6, \Omega, end subscript, equals, start fraction, v, start subscript, b, end subscript, divided by, 6, end fraction, equals, start fraction, 60, divided by, 6, end fraction, equals, 10, start text, A, end text

$i_{5\,\Omega} = \dfrac{v_b}{5} = \dfrac{60}{5} = 12\,\text A$i5Ω=5vb=560=12Ai, start subscript, 5, \Omega, end subscript, equals, start fraction, v, start subscript, b, end subscript, divided by, 5, end fraction, equals, start fraction, 60, divided by, 5, end fraction, equals, 12, start text, A, end text

Ta daaa! Done. The circuit is analyzed.

## Steps in the Node Voltage Method

- Assign a reference node (ground).
- Assign node voltage names to the remaining nodes.
- Solve the easy nodes first, the ones with a voltage source connected to the reference node.
- Write Kirchhoff's Current Law for each node. Do Ohm's Law in your head.
- Solve the resulting system of equations for all node voltages.
- Solve for any currents you want to know using Ohm's Law.

### Reflection: Is the Node Voltage Method magic?

The Node Voltage Method seems like a lot less work than creating, managing, and solving a system of $2E$2E2, E independent equations with $2E$2E2, E unknown voltages and currents. Is the Node Voltage Method magic?

No, there is no magic. The Node Voltage Method is simply a cleverly organized way to approach the same $2E$2E2, E equations. The main innovations are,

- We convinced ourselves we can do Ohm's Law in our heads. We did this while writing the KCL equations. And as we finished up, we used Ohm's Law again to find element currents, which didn't seem like much of a chore. Telling ourselves Ohm's Law is simple makes half of the independent equations seem like not such a big deal.
- Using the concept of
*node voltage*instead of element voltage is a brilliant move that basically annotates the KVL equations right on the schematic, so we don't have to write KVL equations. - We recognize that a few node voltages have trivial solutions, the ones connected to a voltage source whose other terminal is ground. This knocks out one or two equations.
- What remains is the few KCL equations at the non-trivial nodes.

#### How did the Node Voltage Method get the KVL equations to "go away"?

With the Node Voltage Method, we don't even bother to write down the KVL equations. Let's write them anyway and see why.

Our circuit has three meshes, in the left, middle, and right 'windows' of the schematic.

KVL for the left mesh:

$+140 - (140 - v_b) - v_b = 0$+140−(140−vb)−vb=0plus, 140, minus, left parenthesis, 140, minus, v, start subscript, b, end subscript, right parenthesis, minus, v, start subscript, b, end subscript, equals, 0

This left mesh equation really illustrates the point of node voltages. We express the voltage across the $20\,\Omega$20Ω20, \Omega resistor in terms of node voltages instead of its own element voltage. With this notation, the equation collapses to $0 = 0$0=00, equals, 0.

KVL for the middle mesh:

$+v_b - v_b = 0$+vb−vb=0plus, v, start subscript, b, end subscript, minus, v, start subscript, b, end subscript, equals, 0

KVL for the right mesh:

$+v_b - v_b = 0$+vb−vb=0plus, v, start subscript, b, end subscript, minus, v, start subscript, b, end subscript, equals, 0

All three mesh equations reduce to $0 = 0$0=00, equals, 0 and basically drop out of the procedure. This is what your textbook means if it says something like, "With the Node Voltage Method, the KVL equations are written implicitly on the schematic."

## Guided example

**Solve this circuit using the Node Voltage Method.**

If you want to work this problem on your own, go for it!. Copy this schematic and work through the steps of the Node Voltage Method listed above. Even if you don't intend to carry the computation all the way through, I encourage you to do the steps up to writing KCL. That will really help you understand the Node Voltage Method.

**Assign a reference node.**

[Assign the reference node.]

**Assign node voltage names to the remaining nodes.**

[Name the nodes and assign node voltages.]

**Solve the easy nodes first.**

[Solve the easy nodes.]

** Write Kirchhoff's Current Law for each node. Do Ohm's Law in your head.**

[Write Kirchhoff's Current Law for the remaining nodes, with scribbling.]

** Solve the resulting system of equations for all node voltages.**

[Solve the system of equations.]

** Solve for any currents you want to know using Ohm's Law.**

[Solve for currents using Ohm's Law.]

## A twist - floating voltage source

Sometimes you come across a circuit where a voltage source does not have either of its terminals connected to the ground node. We say the voltage source is *floating*. A floating source is a problem for the Node Voltage Method, but it is not too big of a challenge.

In this circuit, battery $\text V_2$V2start text, V, end text, start subscript, 2, end subscript is floating. Let's use the Node Voltage Method and see what happens.

- The reference node has been selected and marked with the ground symbol.
- The other three nodes have been named and assigned node voltages, $v_a$vav, start subscript, a, end subscript, $v_b$vbv, start subscript, b, end subscript and $v_c$vcv, start subscript, c, end subscript.
- The first analysis step is to solve the easy node, $v_a$vav, start subscript, a, end subscript. Since this node is connected to a voltage source that goes to ground, we immediately know $v_a = \text V_1$va=V1v, start subscript, a, end subscript, equals, start text, V, end text, start subscript, 1, end subscript. One node down, two to go.

Next, let's write a KCL equation at node $b$bb,

$i_{\text R2} + i_{\text R_3} + i_{\text V_2} = 0$iR2+iR3+iV2=0i, start subscript, start text, R, end text, 2, end subscript, plus, i, start subscript, start text, R, end text, start subscript, 3, end subscript, end subscript, plus, i, start subscript, start text, V, end text, start subscript, 2, end subscript, end subscript, equals, 0

$\dfrac{(v_a - v_b)}{\text R2} - \dfrac{v_b}{\text R3} + i_{\text V_2}? = 0$R2(va−vb)−R3vb+iV2?=0start fraction, left parenthesis, v, start subscript, a, end subscript, minus, v, start subscript, b, end subscript, right parenthesis, divided by, start text, R, end text, 2, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, start text, R, end text, 3, end fraction, plus, i, start subscript, start text, V, end text, start subscript, 2, end subscript, end subscript, question mark, equals, 0

Oops, what should we put in for the current in the floating battery, $i_{\text V2}$iV2i, start subscript, start text, V, end text, 2, end subscript? The defining equation for the battery does not talk about current. Its defining equation is $v = \text V_2$v=V2v, equals, start text, V, end text, start subscript, 2, end subscript, and there is no $i$ii term involved. Batteries don't tell us what their current is. That's up to the rest of the circuit. So what do we write for this term in the KVL equation if we don't know $i$ii in the battery?

At this point we break away from the standard script for Node Voltage Method, and resort to our own cleverness. It is okay to do this. Remember the Node Voltage script nothing more than an efficient way to create and solve simultaneous equations. The floating battery is giving us a little bit of trouble, but we have not forgotten the point is to create a set of independent equations.

Looking at the circuit, we can make two observations,

The voltage at node $c$cc has a rigid relationship to the voltage at node $b$bb. Namely, $v_c = v_b + \text V_2$vc=vb+V2v, start subscript, c, end subscript, equals, v, start subscript, b, end subscript, plus, start text, V, end text, start subscript, 2, end subscript. We can add this to our system of equations, and it makes up for not knowing the current in battery $\text V_2$V2start text, V, end text, start subscript, 2, end subscript.

We also see that the current in battery $\text V_2$V2start text, V, end text, start subscript, 2, end subscript is the same as the current in resistor $\text R1$R1start text, R, end text, 1.

We can express the battery current in the Node Voltage style as $\dfrac{\text V_1 - v_c}{\text R1}$R1V1−vcstart fraction, start text, V, end text, start subscript, 1, end subscript, minus, v, start subscript, c, end subscript, divided by, start text, R, end text, 1, end fraction.

Even better, we can write the battery current in terms of $v_b$vbv, start subscript, b, end subscript as $\dfrac{\text V_1 - (v_b + \text V_2)}{\text R1}$R1V1−(vb+V2)start fraction, start text, V, end text, start subscript, 1, end subscript, minus, left parenthesis, v, start subscript, b, end subscript, plus, start text, V, end text, start subscript, 2, end subscript, right parenthesis, divided by, start text, R, end text, 1, end fraction.

Now we can complete the KCL equation at node $b$bb.

$\dfrac{(\text V_1 - v_b)}{\text R2} - \dfrac{v_b}{\text R3} + \dfrac{\text V_1 - (v_b + \text V_2)}{\text R1} =0$R2(V1−vb)−R3vb+R1V1−(vb+V2)=0start fraction, left parenthesis, start text, V, end text, start subscript, 1, end subscript, minus, v, start subscript, b, end subscript, right parenthesis, divided by, start text, R, end text, 2, end fraction, minus, start fraction, v, start subscript, b, end subscript, divided by, start text, R, end text, 3, end fraction, plus, start fraction, start text, V, end text, start subscript, 1, end subscript, minus, left parenthesis, v, start subscript, b, end subscript, plus, start text, V, end text, start subscript, 2, end subscript, right parenthesis, divided by, start text, R, end text, 1, end fraction, equals, 0

This equation is a little more complicated than usual, but it is still a solvable one equation in one unknown, $v_b$vbv, start subscript, b, end subscript.

Once we solve for $v_b$vbv, start subscript, b, end subscript, we use our extra equation to immediately get $v_c$vcv, start subscript, c, end subscript.

$v_c = v_b + \text V_2$vc=vb+V2v, start subscript, c, end subscript, equals, v, start subscript, b, end subscript, plus, start text, V, end text, start subscript, 2, end subscript

Done! We have all three node voltages. If you want to find currents, proceed with Ohm's Law as we did earlier.

The floating voltage source is a favorite for teachers to put on tests, to see how you respond to an unexpected circuit configuration. We got through the difficulty by being observant and remembering that it is alright to add an extra equation to the system if needed.

### Supernode

We used the rigid relationship between the two nodes of the floating battery to generate a term to put in the KVL equation, plus an extra equation. Some textbooks call this a *supernode*. In the preceding discussion, we could have used that word, but we just resorted to our creativity to work through the puzzle.

## Node Voltage Method summary

The Node Voltage Method is one of two well-ordered methods of solving a circuit. This technique is embedded inside the popular circuit simulator, $\small \text{SPICE}$SPICEstart text, S, P, I, C, E, end text. The sequence of steps can be summarized as,

- Assign a reference node (ground).
- Assign node voltage names to the remaining nodes.
- Solve the easy nodes first, the ones with a voltage source connected to the reference node.
- Write Kirchhoff's Current Law for each node. Do Ohm's Law in your head.
- Solve the resulting system of equations for all node voltages.
- Solve for any currents you want to know using Ohm's Law.

If the circuit includes a floating source, add extra equations to account for otherwise missing current or voltage variables.