What, When, Where, and Why?

What and why?

You've learned several techniques for analyzing and solving circuits; this amount of information can be a bit overwhelming, so we've included a short summary of when and why to use them, so you can feel better prepared to handle any circuit we throw at you!

We will be walking through a few simple examples.

Nodal Analysis

Nodal analysis is best for circuits with several voltage sources, especially if you have been asked to solve for voltage differences at a particular node.

Review: A node is defined as any point in a circuit where two or more components come together.

The general procedure for nodal analysis is as follows:

1. Label the $n$ node voltages, taking care to define their signage consistently (i.e., be sure that each node is defined as positive/negative with respect to the same reference point).
2. Apply KCL at each node, putting all equations in terms of your unknown node voltages.

   a. Use KCL to determine current flow at $N-1$ of the $N$ nodes. This will create a set of $N-1$ linear equations.

   b. Be sure to take advantage of supernodes. Supernodes are used when two nodes are separated by voltage sources, rather than resistors or current sources. These supernodes have fixed values, and provide constraint equations for your circuit.

   c. Compute currents in terms of voltage differences between nodes.

3. Solve the resulting system of linear equations for your unknown voltages.

Nodal Analysis Example

First let's label all our different potentials/nodes and current directions: Now, to make our calculations easier, let's define a ground; we'll arbitrarily choose $V_4$.

Now let's set up all the current equations using the node potentials: 1. $\frac{V_4 + V_{s_1} - V_1}{R_1} = \frac{V_1 - V_2}{R_2} + I_s$ 2. $\frac{V_1 - V_2}{R_2} + I_s = \frac{V_2 - V_3}{R_5}$ 3. $\frac{V_{s_1} - V_1}{R_1} = \frac{V_2 - V_3}{R_5}$ 4. $\frac{V_2 - V_3}{R_5} = \frac{V_3 - V_{s_2} - V_4}{R_6} + \frac{V_3 - V_4}{R_7}$ 5. $\frac{V_{s_1} - V_1}{R_1} = \frac{V_3 - V_{s_2}}{R_6} + \frac{V_3}{R_7}$

Since we defined $V_4$ as ground, we can replace all instances of that variable with 0: 1. $\frac{V_{s_1} - V_1}{R_1} = \frac{V_1 - V_2}{R_2} + I_s$ 4. $\frac{V_2 - V_3}{R_5} = \frac{V_3 - V_{s_2}}{R_6} + \frac{V_3}{R_7}$

These equations look pretty daunting, and solving them by hand would be a huge pain. However, if we re-arrange these equations by solving for our unknowns ($V_1, V_2,$ and $V_3$), we can create a linear system that makes everything look a lot cleaner:

1. $(\frac{1}{R_1}+\frac{1}{R_2})V_1 + (\frac{-1}{R_2})V_2 = \frac{V_{s1}}{R_1} - I_s$
2. $(\frac{-1}{R_1})V_1 +(\frac{1}{R_2}+\frac{1}{R_5})V_2 +(\frac{-1}{R_5})V_3 = I_s$
3. $(\frac{1}{R_1})V_1+(\frac{1}{R_5})V_2 + (\frac{-1}{R_5})V_3 = \frac{V_{s1}}{R_1}$
4. $(\frac{-1}{R_5})V_2 + (\frac{1}{R_5} + \frac{1}{R_6}+\frac{1}{R_7})V_3 = \frac{V_{s2}}{R_6}$
5. $(\frac{1}{R_1})V_1 + (\frac{1}{R_6}+\frac{1}{R_7})V_3 = \frac{V_{s1}}{R_1} + \frac{V_{s2}}{R_6}$

In matrix form:

$\begin{bmatrix} \frac{1}{R_1} + \frac{1}{R_2} & \frac{-1}{R_2} & 0 \\ \frac{-1}{R_1} & \frac{1}{R_2} + \frac{1}{R_5} & \frac{-1}{R_5} \\ \frac{1}{R_1} & \frac{1}{R_5} & \frac{-1}{R_5} \\ 0 & \frac{-1}{R_5} & \frac{1}{R_5} + \frac{1}{R_6} + \frac{1}{R_7} \\ \frac{1}{R_1} & 0 & \frac{1}{R_6} + \frac{1}{R_7}\end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix} = \begin{bmatrix} \frac{V_{s1}}{R_1} - I_s \\ I_s \\ \frac{V_{s1}}{R_1} \\ \frac{V_{s2}}{R_6} \\ \frac{V_{s1}}{R_1} + \frac{V_{s2}}{R_6}\end{bmatrix}$

Given resistor and independent source values, we can solve the above system using Gaussian elimination for our 3 remaining unknown nodes, $V_1, V_2,$ and $V_3$.

Superposition

Superposition is most useful for linear circuits containing multiple independent sources, as it allows us to solve for the current or voltage at a single point as the sum of the contributions of all independent sources acting alone.

The general procedure for superposition is as follows:

1. Repeat the following for each independent voltage and current source:

   a. Replace independent voltage sources with a short circuit (a plain wire).

   b. Replace independent current sources with an open circuit (a gap in the circuit).

   c. Calculate the contribution of the source to the selected point of interest.

2. Sum the individual contributions of each independent source.

Superposition Example

Here is a walkthrough of a superposition example with both voltage sources and current sources:

Thevenin and Norton Equivalent Circuits

Thevenin and Norton equivalencies are useful for reducing complex circuits to simpler ones; i.e. circuits that only use ideal voltage/current sources and series/parallel resistance, respectively.

The general procedure for using Thevenin and Norton equivalencies is as follows:

1. Choose a breaking point in the circuit.

2. Thevenin: Compute the open circuit voltage, $V_{oc}$. Norton: Compute the short circuit current, $I_{sc}$.
3. Compute the Thevenin equivalent resistance, $R_{Th}$.

   a. If there are only independent sources, short circuit all voltage sources and open circuit the current sources (see superposition).

   b. If there are also dependent sources, compute $R_{Th} = \frac{V_{oc}}{I_{sc}}$.
4. Replace the circuit with the equivalent you have just created!

Thevenin and Norton Example

Here is an example of Thevenin and Norton equivalences with both dependent and independent sources: