MEL TECH
CIRCTRONICS
Source Transformation
AC Source transformation A voltage source with impedance Z in series is the same as a current source with an impedance Z in parallel.
Example:
Consider the circuit shown in Figure 1. Find the value of the voltage measured by the voltmeter.
Figure 1
Solution: The voltmeter measures the voltage across the current source. (The color-coded probes of the voltmeter indicate the reference direction of the voltage measured by the voltmeter.) Figure 2 shows the circuit after the replacing the voltmeter by the equivalent open circuit and adding a label to show the voltage measured by the meter.
Figures 3 through 17 illustrate the use of source transformations and equivalent resistances to simplify the circuit.
Figure 2 The circuit from Figure 1 after the replacing the voltmeter by an open circuit.
Figure 4 The circuit from Figure 3 after doing a source transformation
Figure 3 Separating the circuit from Figure 2 into two parts.
Figure 5 The circuit from Figure 4 after changing the order of parallel elements.
Figure 6 Separating the circuit from Figure 5 into two parts.
Figure 7 The circuit from Figure 6 after replacing parallel resistors with an equivalent resistor.
Figure 8 Separating the circuit from Figure 7 into two parts.
Figure 9 The circuit from Figure 8 after doing a source transformation.
Figure 10 The circuit from Figure 9 after changing the order of series elements.
Figure 11 Separating the circuit from Figure 10 into two parts.
Figure 12 The circuit from Figure 11 after replacing series voltage sources with an equivalent voltage source
Figure 13 Separating the circuit from Figure 12 into two parts.
Figure 14 The circuit from Figure 13 after doing a source transformation.
Figure 15 The circuit from Figure 14 after changing the order of parallel elements.
Figure 16 Separating the circuit from Figure 15 into two parts.
Figure 17 The circuit from Figure 16 after replacing parallel resistors with an equivalent resistor.
Figure 18 The reduced circuit.
Figure 18 shows the simplified circuit after labeling the current i, of the resistor. Applying KCL at the top node of the circuit gives
i + 8 = 2 ⇒ i = − 2 A
3 3
The voltage measured by the meter, vm, is also the voltage across the resistor. Ohm’s law gives
vm = 3 − 2/3 = −2 V
