We have noticed that series-parallel combination and wye-delta transformation help simplify circuits.

This circuit analysis theorems are classified as:

*Source transformation*is another tool for simplifying circuits. 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.This circuit analysis theorems are classified as:

- Superposition theorem
- Source transformation
- Thevenin theorem
- Norton theorem
- Maximum power transfer

## Source Transformation

In previous posts about node voltage or mesh current equations can be obtained by mere inspection of a circuit when the sources are all independent current or independent voltage sources. It is therefore expedient in the circuit analysis to be able to substitute a voltage source in series with a resistor for a current source in parallel with a resistor, or vise versa as can be seen in Figure.(1). Either substitution is known as a

**source transformation**.Figure 1. Source transformation |

Asource transformationis the process of replacing a voltage sourcevin series with a resistor_{s}Rby a current sourceiin parallel with a resistor_{s}R, or vice versa.

The two circuit in Figure.(1) are equivalent - provided they have the same voltage-current relation at terminals

*a-b*. It is easy to show that they are indeed equivalent. If the sources are turned off, the equivalent resistance at terminals*a-b*in both circuits is R. Also, when terminals*a-b*are short-circuited, the short circuit current flowing from*a*to*b*is*i*in the circuit on the left-hand side and_{sc}= v_{s}/R*i*for the circuit on the right-hand side. Thus, v_{sc}= i_{s}_{s}/R =*i*in order for the two circuits to be equivalent. Hence, source transformation requires that_{s}(1) |

Source transformation also applies to dependent sources, provided we carefully handle the dependent variable. As shown in Figure.(2), a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa where we make sure Equation.(1) is satisfied.

Figure 2 |

Like the wye-delta transformation we studied before, a source transformation does not affect the remaining part of the circuit. When applicable, source transformation is a powerful tool that allows circuit manipulation to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation.

- Note from Figures.(1) or (2) that the arrow of the current source is directed toward the positive terminal of the voltage source.
- 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, nonideal voltage source,*R*≠ 0. Similarly, an ideal current source with*R*= ∞ cannot be replaced by a finite voltage source.

## Source Transformation Examples

For better understanding let us review examples below :**1.Use source transformation to find**

*v*in the circuit of Figure.(3)._{o}Figure 3 |

__Solution :__
We first transform the current and voltage sources to obtain the circuit in Figure.(4a). Combining the 4 Ω and 2 Ω resistors in series and transforming the 12 V voltage source gives us Figure.(4b). 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).

Figure 4 |

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

and

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

*v*across them. Hence,_{o}**2.Find**

*v*in Figure.(5) using source transformation._{x}Figure 5 |

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

Figure 6 |

The current source is transformed to a voltage source as shown in Figure.(6b). Notice that the terminals for v

_{x}are intact. Applying KVL around the loop in Figure.(6b) gives(2.1) |

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

*v*yields_{x}(2.2) |

Substituting this into (2.1), we get

Alternatively, we may apply KVL to the loop containing

*v*, the 4 Ω resistor, the voltage controlled dependent voltage source, and the 18 V voltage source in Figure.(6b). We get_{x}
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