In this worked example we use source transformation to simplify a circuit.

Source transformation is explained here.

Explained by Willy McAllister.

## Review

Source transformation between Thévenin and Norton forms,

The resistor value is the same for the Thévenin and Norton forms, $\text R_\text T = \text R_\text N$.

Convert Thévenin to Norton: set $\text I_\text N = \text V_\text T / \text R_\text T$.

Convert Norton to Thévenin: set $\text V_\text T = \text I_\text N \, \text R_\text N$.

Thévenin and Norton forms are equivalent because they have the same $i$-$v$ behavior from the viewpoint of the output port.

## General strategy

Think about source transformation when the problem asks you about a single voltage or current for one specific component. Everything besides that one component is a candidate for source transformation.

The one rule: Don’t include the component with the requested $i$ or $v$ in a source transformation.

The general strategy,

• Scan the circuit. Look for the familiar pattern of the two forms,
• Thévenin form is a voltage source in series with a resistor.
• Norton form is a current source in parallel with a resistor.
• Pick candidates for source transformation. Remember the one rule.
• Purpose: transform a source to increase the number of resistors in series or in parallel.
• Simplify the circuit: merge those resistors into their series or parallel equivalent.
• Redraw the circuit and look for another chance to transform sources.
• Solve for the asked-for variable in the simpler circuit.

## Example circuit

Find $\blueD i$.

We could go after this circuit with methods we’ve learned before, like Node Voltage or Mesh Current. But this time we will do it with source transformation.