Node Analysis in AC Circuits

Just like mesh analysis uses Kirchhoff's voltage law to turn a circuit into a series of simultaneous equations, node analysis uses Kirchhoff's current law to do the same things. Why would we want node analysis in a world with mesh analysis? Well some problems are just simpler using one or the other.
In node analysis we write KCL for each node, that means that for every node in the circuit we write an expression for the currents coming into and out of the node and we claim that they all sum to zero. That is, no current enters a node without also leaving it.
The steps to "Node Analysis" are:
  1. Find the impedances of each resistor, capacitor and inductor in the circuit given the frequency of the source(s)
  2. Convert everything in the circuit to phasor form
  3. Choose a ground reference node (choosing the node with the most branches is usually a good choice but you can choose any node and get the right answer)
  4. Assign a variable for the voltage at each node (like \(V_1\) for the voltage at node 1)
  5. Write KCL for each node in the circuit, setting the sum of all currents entering a node to zero. The current from node 1 to node 2 with a resistor R between them is simply: \(\frac{V_1-V_2}{R}\)
  6. Solve the equations simultaneously using your favourite method
You now have the voltage at each and every node (as you'll see once we do an example). From this you can find the voltage or current you were looking for initially. There's a rule for if the circuit has a voltage source between two nodes where neither is the reference node that we will cover later (it's similar to the rule for current sources in mesh analysis if you're familiar with that).

Find the voltage \(V_1\) in the circuit below using node analysis.

First we'll find the impedances of the capacitor and resistor. Obviously the impedance of the resistor is \(Z_R = 200\Omega\). Our \(\omega = 1\) so the impedance of the capacitor is: \(Z_C = -j\frac{1}{10\cdot 10^{-3}}\) \(Z_C = -j100\) Now converting everything to phasors we get the new circuit diagram below: Now we our reference node has already been chosen for us since ground has been drawn in. Now we're going to write KCL for node 1. Note that the voltage source tells us what the voltage at the upper-left node must be since the bottom node is defined as 0V. \(\frac{V_1 - 5\angle 0^\circ}{100\angle -90^\circ} + \frac{V_1}{200\angle 0^\circ} = 0.02\angle 0^\circ\) That's just saying that the currents leaving node 1 to the left and right must sum to give the current entering from the current source in the center. Solve for \(V_1\) we get: \(V_1 = \frac{4-10j}{1+2j} = 4.82\angle -131.4^\circ\)V
Usually our circuit won't be so nice and we'll have several nodes with unknown voltages to deal with. If our circuit has a voltage source that isn't connected to our reference node (0V) then we need to handle it a little differently. First we'll pretend that the two nodes are actually one large node and write KCL for all the branches coming into or out of either of the two connected nodes. Then we write a second equation linking the two nodes' voltages using the voltage source (i.e. V_3 = V_4 + 5V for a 5V battery).
When a voltage source links two non-reference nodes in nodal analysis we join the two nodes into a "super node" and write KCL for both the nodes at once (all currents entering either node must be balanced by a current leaving one of the nodes). Then we write a second equation linking the two nodes' voltages by the current source.

Find the voltage \(V_2\) in the circuit below.

We can see that \(V_3 = 2\angle 0^\circ\). Now \(V_1\) and \(V_2\) are connected by a voltage source so we combine the two nodes into a single "super node" as shown: Now we write KCL for the supernode as a whole: \(\frac{V_1-2\angle 0^\circ}{50} + \frac{V_2-2\angle 0^\circ}{100} + \frac{V_2}{200} = 0\) And we write the equation linking \(V_1\) and \(V_2\): \(V_1 = V_2 + 5\angle 0^\circ\) If we solve these two simultaneously we find that: \(V_2 = -1.14\angle 0^\circ = -1.14\cos(2\pi t)\) Using the fact that \(-\cos(t) = \sin(t+\pi)\): \(V_2 = 1.14\cos(2\pi t + \pi)\)
practice problems