Source transformation is the process of modifying a linear circuit by transforming voltage sources into current sources using Norton's theorem, and vice versa using Thévenin's theorem. The process is usually combined with series and parallel simplifications of resistor (or impedance in the case of AC circuits), current sources and voltage sources in order o simplify the original circuit or compute the thought variable.

Source transformation can be applied to both independent and dependent sources.

When transforming a circuit from a Norton equivalent circuit to a Thévenin equivalent circuit the value of the resistance remains the same, while the voltage of the voltage source is $$\begin{equation}V=R I\end{equation}$$

When transforming a circuit from a Thévenin equivalent circuit to a Norton equivalent circuit the value of the resistance remains the same, while the current of the current source is $$\begin{equation}I=\frac{V}{R}\end{equation}$$

Consider the circuit shown in Fig. 3(a), where we need to determine the current $I_x$. If we start by performing a source transformation on the independent voltage source, we would lose the variable of interest, which also serves as the control variable for the dependent source. Therefore, we begin by applying the source transformation to the dependent voltage source instead. This source can be replaced by an equivalent current source with current $\frac{3I_x}{3} = I_x$, as illustrated in Fig. 3(b).

Notice that, consistent with the convention established in Fig. 2, the current of the transformed source is directed away from the source and toward the node where the positive terminal of the original voltage source was connected. It is important to follow this convention every time we perform a source transformation.

Next, we can combine the two current sources (see Fig. 4(c)) and make one more source transformation, in which the dependent current source is transformed into a dependent voltage source with voltage $3\cdot\frac{4+I_x}{3}=12+3I_x$ as in Fig. 4(d).

Finally, we can combine the two voltage sources and the resistors and obtain the circuits shown in Fig. 4(5). Since this is a simple circuit (which contains only one loop), we can compute the sought variable by writting KVL $$\begin{equation}23+3I_x + 5I_x =0\end{equation}$$ which can be solved for $$\begin{equation}I_x=-\frac{23}{8}=2.75\ {\class{mjunit}V}\end{equation}$$ which is the same value as the one obtained in Example 3 in the Superposition section.

• Circuit with 2 loops, 2 sources, 2-3 resistors (numerical simplification)
• Circuit with 4 loops, 3 sources, 4 resistors (numerical simplification)