> For the complete documentation index, see [llms.txt](https://ee16a.gitbook.io/studee16a/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://ee16a.gitbook.io/studee16a/circuits/nodal-analysis.md).

# Nodal Analysis

### What and Why?

Nodal analysis is one of several ways to solve circuits. We'll go over all the different methods you've learned so far later in this GitBook, but we'd like to spend a little more time on nodal analysis, as it is by far one of the more important (and easiest!) methods.

Nodal analysis uses KCL to determine potential difference at nodes throughout a circuit. It produces a **system of linear equations** that can be solved using simple Gaussian elimination. We will walk through the process in detail now:

## Step 1: Choosing Nodes

Let's say we want to analyze the following circuit: ![](https://1435250727-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LnuoWs729nFx1G2RBZA%2F-LnuoXUHo_ncHr0Bp5hM%2F-LnuoYURofqVQHS9I75Y%2FScreen%20Shot%202017-03-16%20at%206.43.14%20PM.png?generation=1567576178937609\&alt=media)

The only *unknown* voltage difference is at the intersection of the two resistors and the current source. We will call this node $$V\_1$$.

> **Understanding checkpoint**: Why is $$V\_1$$ the only node with an unknown voltage?

![](https://1435250727-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LnuoWs729nFx1G2RBZA%2F-LnuoXUHo_ncHr0Bp5hM%2F-LnuoYUTLYb2cuMiQRWA%2FScreen%20Shot%202017-03-16%20at%206.42.54%20PM.png?generation=1567576180302679\&alt=media)

## Step 2: Analyzing Current

There are 3 branches coming out of $$V\_1$$, and thus 3 current values to consider. We will declare the positive direction to be moving *out* of $$V\_1$$.

![](https://1435250727-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LnuoWs729nFx1G2RBZA%2F-LnuoXUHo_ncHr0Bp5hM%2F-LnuoYUVmOts3HD-2dlR%2FScreen%20Shot%202017-03-16%20at%206.53.11%20PM.png?generation=1567576180603927\&alt=media)

> **Understanding checkpoint**: There is $$-20mA$$ of current running out of $$V\_1$$ towards ground. Can you explain to yourself why this is?

We can use Ohm's law, $$V= IR$$, to obtain the following equations: 1. $$i\_1 = \frac{V\_1-5}{100}$$ 2. $$i\_2 = \frac{V\_1}{200}$$

> **Understanding checkpoint**: Can you explain why the $$V$$ in each of these equations has the value it does? Hint: what is the definition of voltage?

Using KCL, we get a third equation: 3. $$i\_1 + i\_2 - 20mA = 0$$.

## Step 3: Using Current to Solve for Voltage

Using replacement, and solving for $$V\_1$$:

$$V\_1 = \frac{(\frac{5}{100}+20)}{(\frac{1}{100}+\frac{1}{200})} \approx 4.67 V$$