Help with complex numbers in AC analysis

  • Thread starter e_brock123
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Homework Statement



Ok so before I begin I've done my share of research and have had no luck as people seem to be moving to calculators for answers these days. I know my rectangular and polar conversions in AC but I haven’t done too much work with complex numbers.

I basically have two questions, one is solving Vo after applying nodal analysis on a AC circuit, and the other is working out the total impedance in a circuit.



The Attempt at a Solution



Nodal analysis equation:
(Vo-0)/-j25 + (Vo-0)/j20 + (Vo-20angle(-15))/60 = 0

Some advice on how to go about solving Vo would be great like I normally find a common factor and cancel the denominators in DC and move Vo to one side and the numbers to the other. But unfortunately for this I have no idea on how to simplify.

Total impedance:

(200+j80)//(-j100) then in series with a (8-j200)

Once again I’m not quite sure on how to apply the same concept of DC to this new situation.
Any help will be greatly appreciated.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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Nodal analysis equation:
(Vo-0)/-j25 + (Vo-0)/j20 + (Vo-20angle(-15))/60 = 0

Some advice on how to go about solving Vo would be great like I normally find a common factor

It's easier and less prone to error to solve it symbolically and wait to plug in actual complex impedances until the end.

For this example, let Z1=-j25, Z2=j20, Z3=20∠(-15)

Then rewrite your equation:

Vo/Z1 + Vo/Z2 + (Vo - Z3)/60 = 0

Some algebraic manipulation:

Vo = (Z3/60)*(60*Z1*Z2)/(60*Z2+60*Z1+Z1*Z2)

Then substitute. When you multiply and divide complex numbers, it is easier to do that in polar form. When you add and subtract, it's easier to do in (a+jb) form.

Total impedance:

(200+j80)//(-j100) then in series with a (8-j200)

Again, it is easier to keep it symbolic until you are ready to add, subtract, multiply, divide.

You have (Z1 // Z2) + Z3 = Z1*Z2 / (Z1 + Z2) + Z3, plug in values.


Is there a reason you are plugging in a value for the frequency variable ω in these equations? (Maybe you are an electrician so ω is always 50/60Hz?)

It's normally preferable to have the resulting impedance in jω form with the frequency explicit. Conceptually it emphasizes that the 'resistance' (magnitude of the impedance) of circuit elements change with frequency and also it will allow an idea of how the circuit responds to signals other than sinusoids.

Any arbitrary signal can be expressed as a summation of sinusoids by the Fourier transform, so how the circuit responds at different frequencies tells something about how those arbitrary input signals will be modified. Normally we talk about filtering action like a lowpass, bandpass or highpass filter, which can be spotted from the complex impedance of the transfer function with frequency explicitly in there.
 

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