Source Transformation

Is a method used to simplify a circuit by transforming a series voltage source and impedance to a parallel current source and impedance, and vice versa. This process is commonly used to lessen the number of loops in a circuit for easy analysis.

The direction of the arrow of the current source follows the positive sign of the voltage source.

In this circuit, we should identify the voltage at the 4Ω resistor. We will start transforming at the right side of the circuit using simple ohms law.

V= 2A (2Ω)

V= 4 V

 

The resulting circuit would be:

Combine the series impedance's.

Then, transform it again.

I=V/R

I= 4/ 2-j1

I= 1.788∟26.56 A

Combine the parallel impedances.

Z= j3(2-j1)/ j3+(2-j1)

Z= 2.37 ∟18.43 Ω

 

Then transform again so that we can subtract the voltage source and finally obtain V.

V= (1.788∟26.56) (2.25+j0.75)

V= 2.998 + j2.998

V= V1-V2

V= (2.998 +j2.998)-(10)

V= 7.6168 ∟156.82 V

We can now use voltage division to obtain V.

V=(7.616∟156.82)(4)/4+ (2.25+j0.75)

V= 4.84 ∟149.97 V

Superposition

Is a method of circuit analysis and simplification that involves leaving one source in the circuit and killing all the remaining sources, to obtain the unknown value in the circuit. The voltage source would be shorted and the current source would be opened. Here’s a sample problem for this:

First, let’s kill the 12∟0 V source and leave the 6∟0 A source. The resulting circuit would be:

To obtain I, we will use mesh analysis. This loop would create 3 loops but loop 3 is already equal to 6∟0 A and its not necessary for solving I.

Loop 1 would be equal to:

(-2+j1) I1 – (j1)I2=0

Loop 2 would be equal to:

(-j1)I1 + (-1 – j2)I2 = -6

From the loop equations and using matrix, we would obtain:

I= 2.042∟149.78 A

After obtaining I by killing the voltage source, we will now kill the current source and leave the voltage source to solve I. The resulting circuit would be like this:

We will use nodal analysis to solve I. The circuit has a super node, so the resulting equation would be:

(0.5)V1 + (0.1 + j0.7)V2= 0

And for KVL:

V1 – V2 = -12

From the equation and matrix, we would obtain I.

I= 6.51∟40.60 A

To obtain the final I, we will just simply add the 2 I’s we solved.

I= I1 + I2

I= (2.042∟149.78)+(6.51∟40.60)

I= 6.149 ∟58.878 A