Use source transformation to find the Thevenin equivalent circuit with respect to terminals, a, b.
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
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) ]
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
If I did anything wrong, I hope I am on the right track at least?? Thank you
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