Source Transformation Circuits, Formula, and Examples

Source transformation – From the previous posts, we have learned about series and parallel connection along with wye-delta transformation. Those methods will help us greatly when simplifying an electrical circuit. Another method we can use is Source Transformation.

An electrical circuit is built from both active and passive elements. As you already know, the active elements are able to produce energy while the passive elements only absorb energy or convert into another form of energy. The most common active elements we use are voltage source and current source. Both can be divided again into independent and dependent sources.

When we talk about an independent source, its stored energy (voltage or current) is a fixed value. On the other hand, the dependent source has value depending on other variables in the circuit (voltage or current in a specific spot).

Source Transformation Circuits

The complexity of electrical circuits can vary for every application. Along with the basic Ohm’s Law and Kirchhoff’s Laws, we also have the electrical circuit analysis theorems:

Along with those four above, we can also use Source Transformation. From the name implies, we transform the voltage source into current source and vice versa. Of course there will be specific procedures we must follow. The full explanation will be found below.

Source transformation works with the principle of equivalence. We remember that the term equivalent of a circuit where v-i characteristics are identical with the original circuit.

Basic to these tools is the concept of equivalence. We recall that an equivalent circuit is one whose v-i characteristics are identical with the original circuit.

Source Transformation Formula

We recall that node voltage or mesh current analysis equations are written with a simple inspection of the circuit when the sources are independent voltage source or independent current source.

It is very convenient to be able to do substitution between a voltage source connected series with a resistor and a current source connected parallel with a resistor. The example of this idea can be seen in Figure.(1). Either substitution is known as a source transformation.

source transformation
Figure 1. Source transformation

A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.

At the first glance it seems two circuits in Figure.(2) are different but actually they are equivalent since they have the same v-i characteristics at terminals a-b.

source transformation
Figure 2. Source transformation

How do we know those circuits are equivalent?

When we turn off their sources, the equivalent resistance at a-b in both circuits is R.

When we make the a-b short-circuited:

  • The top-side circuit has short circuit current flowing through a-b at isc=vs/R.
  • The bottom-side circuit has isc=is.

Hence, we need the v-i characteristics vs/R=i to make two circuits are equivalent. From this we conclude that source transformation formula is written as:

(1)   \begin{align*}v_{s}=i_{s}R \quad \mbox{or} \quad i_{s}=\frac{v_{s}}{R}\end{align*}

Not only for independent sources, we can also implement source transformation for dependent sources.

We can see the example in Figure.(2) where we have a dependent voltage source connected series with a resistor. We transform this into a dependent current source connected parallel with the same resistor or vice versa. We also use the Equation.(1) above to fulfill the equivalence rule.

The same with wye-delta transformation, this source transformation won’t affect the rest part of the circuit. This is a powerful method to manipulate circuits for easier analysis.

However, we should keep the following points in mind when dealing with source transformation.

  1. Note from Figures.(1) or (2) that the arrow of the current source is directed toward the positive terminal of the voltage source.
  2. Note from Equation.(1) that source transformation is not possible when R = 0, which is the case with an ideal voltage source. However, for a practical, non ideal voltage source, R ≠ 0. Similarly, an ideal current source with R = ∞ cannot be replaced by a finite voltage source.

Source Transformation Current to Voltage

Observe the circuit example below to understand how to do source transformation current to voltage. Of course the circuit has a current source and we need to transform it into a voltage source.

The source transformation formula will be:

    \begin{align*}v_{s}&=i_{s}R\\&=2\times 5\\&=10V\end{align*}

source transformation

Source Transformation Voltage to Current

Observe the circuit example below to understand how to do source transformation voltage to current. Of course the circuit has a voltage source and we need to transform it into a current source.

The source transformation formula will be:

    \begin{align*}i_{s}&=\frac{v_{s}}{R}\\&=\frac{10}{5}\\&=2A\end{align*}

source transformation

Source Transformation Examples

For better understanding let us review examples below :

1.) Use source transformation to find vo in the circuit of Figure.(3).

source transformation
Figure 3

Solution :

We first transform the current and voltage sources to obtain the circuit in Figure.(4a). The current source will be replaced with a voltage source by:

    \begin{align*}V&=I\times R\\&=3\times 4\\V&=12V\end{align*}

source transformation
Figure 4.(a)

Combining the 4 Ω and 2 Ω resistors in series results in a 6Ω resistor. Transforming the 12 V voltage source gives us Figure.(4b). The current source value is written by:

    \begin{align*}I&=\frac{V}{R}\\&=\frac{12}{6}\\I&=2A\end{align*}

source transformation
Figure 4.(b)

We now combine the 3 Ω and 6 Ω resistors in parallel to get 2 Ω.

We also combine the 2 A and 4 A current sources to get 2 A source. Thus, by repeatedly applying source transformations, we obtain the circuit in Figure.(4c).

source transformation
Figure 4.(c)

We use current division in Figure.(4c) to get

    \begin{align*}i=\frac{2}{2+8}(2)=0.4A\end{align*}

and

    \begin{align*}v_{o}=8i=8(0.4)=3.2V\end{align*}

Alternatively, since 8 Ω and 2 Ω resistors in Figure.(4c) are in parallel, they have the same voltage vo across them. Hence,

    \begin{align*}v_{o}&=(8||2)(2A)\\&=\frac{8\times 2}{10}(2)\\v_{o}&=3.2V\end{align*}

2.) Find vx in Figure.(5) using source transformation.

source transformation
Figure 5

The circuit in Figure.(5) involves a voltage-controlled dependent current source. We transform this dependent current source as well as the 6 V independent voltage source as shown in Figure.(6a). The 18 V voltage source is not transformed because it is not connected in series with any resistor.

source transformation
Figure 6.(a)

The two 2 Ω in parallel combine to give a 1 Ω resistor, which is in parallel with the 3 A current source.

source transformation
Figure 6.(b)

The current source is transformed to a voltage source as shown in Figure.(6b). Notice that the terminals for vx are intact. Applying KVL around the loop in Figure.(6b) gives

(2.1)   \begin{align*}-3+5i+v_{x}+18=0\end{align*}

Applying KVL to the loop containing only the 3 V voltage source, the 1 Ω resistor, and vx yields

(2.2)   \begin{align*}-3+1i+v_{x}=0 \quad \rightarrow \quad v_{x}=3-i\end{align*}

Substituting this into (2.1), we get

    \begin{align*}15+5i+3-i=0 \quad\rightarrow\quad i=-4.5A\end{align*}

Alternatively, we may apply KVL to the loop containing vx, the 4 Ω resistor, the voltage-controlled dependent voltage source, and the 18 V voltage source in Figure.(6b). We get

    \begin{align*}-v_{x}+4i+v_{x}+18=0 \quad\rightarrow\quad i=-4.5A\end{align*}

Thus,

    \begin{align*}v_{x}=3-i=7.5V\end{align*}

Leave a Comment