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## Homework Statement

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

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## Homework Equations

Voltage Division: (V in)*(R1/R1+R2)

Thevenin / Norton / source transformation procedures

RTh = RNo

VTh = INo*RNo

Polar Form conversion:

Z = (R

^{2}+ X

^{2})

^{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:

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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:

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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)/(8000Ωj

^{2}j + 6000Ω

^{2}j

^{2})

=[ (40Ωj - 80Ω)/(8000Ω

^{2}j - 6000Ω

^{2}) ] * [ (-8000Ω

^{2}j - 6000Ω

^{2})/-8000Ω

^{2}j - 6000Ω

^{2}) ]

= (-320000Ω

^{3}j

^{2}- 240000Ω

^{3}j + 640000Ω

^{3}j + 480000Ω

^{3}) / (-64000000Ω

^{4}j

^{2}+ 36000000Ω

^{4})

= [(80 x 10

^{4})Ω

^{3}+ (40 x 10

^{4})Ω

^{3}]/(100 x 10

^{6}Ω

^{4}

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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