Analyzing Circuits via Source Transformation
An electrical network can consist of sources and passive elements. Sources are circuit components which possess their own energy and are capable of transferring this energy to other circuit elements.
There are two basic types of sources: voltage sources and current sources. These can be further classified as independent or dependent. In the case of independent sources, the voltage or current is fixed. If the source is dependent, the value of the voltage or current depends on the amount of current or voltage elsewhere in the circuit.
Passive components do not have their own energy. As a result of this, they are regarded as sinks. However, they influence the amount of current or voltage in a given portion of the circuit. Resistors, capacitors, and inductors are passive components.
The complexity of electrical networks ranges from very simple—e.g., a voltage divider—to very complicated—e.g., the internal structure of an integrated circuit (IC).
A good electrical designer is expected to have a sound knowledge of the entire system irrespective of its complexity. This is absolutely essential when the issue arises of upgrading or troubleshooting the system.
At this point, it is essential to note that analyzing an electrical circuit is sometimes easy and straightforward, taking only a couple of minutes in its entirety. Sometimes, though, it can involve a lot of hard work (or rather, smart work), and it may even force the analyzer to resort to help from software. Nevertheless, the mode of analysis is based on certain basic rules and theorems.
Here's a list of important theorems along with a brief explanation:
1. Superposition Theorem: Aids in finding the current and voltage in a circuit which has multiple sources; the effects produced by each of the sources individually can be summed.
2. Thevenin's Theorem: Aids in circuit simplification; multiple sources and resistances can be represented by an equivalent circuit with just a single voltage source and a single resistor.
3. Norton's Theorem: Aids in circuit simplification; multiple sources and resistances can be represented by an equivalent circuit with just a single current source and a single resistor.
4. Millman's Theorem: A simplification technique involving circuits with parallel branches.
At this point, it should be noted that all these theorems are based on the basic rules governing the field of electronics, namely, Ohm's Law and Kirchoff's Laws.
In addition, we may sometimes find a circuit which has the resistors connected either in a delta/pi or a star/Y/T configuration. In such cases, we can employ star-to-delta or delta-to-star transformation when analyzing the circuit.
Consider the circuit shown in Figure 1; the goal is to find the current (denoted by i) through the central 5 Ω resistor. Here, mesh analysis (Kirchoff's Voltage Law, KVL) cannot be applied readily because the circuit has a branch which has a current source. Thus, we need to devise a method by which we can eliminate this current source from our circuit. However, while doing so, we need to take care that the current and the voltages in the circuit remain unaltered.
Recall Ohm's Law, which states that .
Looking back at the circuit (Figure 1) again, we can see that the 1 A current source has a 10 Ω resistor in parallel with it. Let us now replace this combination with a voltage source, V = 1 A × 10 Ω = 10 V, and a 10 Ω series resistor. You can see what this looks like in Figure 2. Note that the positive terminal of the voltage source is placed to the left, because the current-source arrow was pointing to the left. These two circuits (Figure 1 and Figure 2) are considered to be equivalent: the 1 A current entering node X from node Y has not changed.
The process carried out here is called source transformation. We have transformed an existing current source with a parallel resistor into an equivalent voltage source with a series resistor.
The circuit in Figure 2 can be further simplified as it has a 10 Ω resistor in series with a 5 Ω resistor. These can be replaced by an equivalent 15 Ω (= 10 Ω + 5 Ω) resistor. The simplified circuit is shown in Figure 3(a).
Now we can easily apply mesh analysis to solve the problem before us. However, there is also a much easier graphical way of achieving this: apply source transformation again!
Previously we transformed a current source with a parallel resistor, but we can also apply source transformation to a voltage source with a series resistor. We have two such arrangements, as indicated in Figure 3(b). This circuit is equivalent to the one shown in Figure 3(a).
So here we will apply voltage-to-current source transformation, which is very similar to current-to-voltage source transformation. The process involves replacing the voltage source V in series with a resistor R with an equivalent network which has a current source in parallel with a resistor R. The current source is oriented such that the arrow points towards the positive terminal of the voltage source being replaced (see Figure 4).
Thus, for the leftmost branch, we have a current source in parallel with a 5 Ω resistor. Similarly, for the rightmost branch, we get in parallel with a 15 Ω resistor. The resulting circuit is shown in Figure 4.
The circuit in Figure 4 has two current sources pointing in the same direction, and hence these can be replaced by a single current source whose value is equal to their sum, i.e., .
There are three resistors: two 5 Ω resistors and one 15 Ω resistor, all in parallel. We could replace all three of them with an equivalent resistance (REQ), but our goal is to find the current through the 5 Ω resistor, so we will combine only the other two.
Having made these changes, we get the circuit shown in Figure 5.
Now, let us apply current-to-voltage source transformation once again for the combination indicated in Figure 5.
Here, the voltage source will be of the value , with the positive terminal towards node X, in series with a resistor.
The resulting circuit is shown in Figure 6.
For Figure 6, we can readily apply KVL to obtain the current through the 5 Ω resistor: