Performing source transformations.

 • Replacing series or parallel resistors by an equivalent resistor. 

• Replacing series voltage sources by an equivalent voltage source.

 • Replacing parallel current sources by an equivalent source source

We will first go over voltage source transformation, the transformation of a circuit with a voltage source to the equivalent circuit with a current source.

In order to get a visual example of this, let's take the circuit below which has a voltage source as its power source:

Using source transformation, we can change or transform this above circuit with a voltage power source and a resistor, R, in series, into the equivalent circuit with a current source with a resistor, R, in parallel, as shown below:

We transform a voltage source into a current source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula,I=V/R.

Use source transformation to find the Thevenin equivalent circuit with respect to terminals, a, b. 

2. Relevant equations Voltage Division:

(V in)*(R1/R1+R2) Thevenin / Norton / source transformation procedures

 RTh = RNo VTh = INo*RNo Polar Form 

conversion: 

Z = (R2 + X2)1/2 θ = arctan(j / R)

 For Complex numbers: j*j = -1 1/j = -j 3. 

The attempt at a solution I simplified the circuit to this first:

For source transformation, you find VTh by removing the load and then finding the drop from A back to the source right? And you find RTh by finding the eq resistance from A to B with the voltage source suppressed right? So trying to get VTh for the final answer, I think I do this

because I'll want the drop from A right? so for 

VTh: (200∠0deg)V* [(50Ω+60Ωj)/(80Ω-40Ωj)]

 in polar form is (200∠0deg)V* [ (78.1∠50.1)/(61.4∠26.5) ] = (254∠23.6deg)V 

So I think the drop from A is (200∠0)V - (254∠23.6deg)V -> VTh = (-54 ∠ -23.6 deg)V

 Now for RTh: RTh = [ (1/80Ω + 60Ωj) + (-1/100Ωj) ]-1 so = (100Ωj - 80Ω - 60Ω)/(8000Ωj2j + 6000Ω2j2) =[ (40Ωj - 80Ω)/(8000Ω2j - 6000Ω2) ] * [ (-8000Ω2j - 6000Ω2)/-8000Ω2j - 6000Ω2) ] = (-320000Ω3j2 - 240000Ω3j + 640000Ω3j + 480000Ω3) / (-64000000Ω4j2 + 36000000Ω4) = [(80 x 104)Ω3 + (40 x 104)Ω3]/(100 x 106Ω4 RTh =

 0.008Ω + 0.004Ωj 

With Matildo