Help with complex numbers in AC analysis

In summary, the individual is seeking advice on how to solve for Vo in a nodal analysis equation and how to calculate the total impedance in a circuit. They are unsure about how to apply the same concepts used in DC to these situations involving complex numbers. The expert recommends keeping the equations symbolic until ready to add, subtract, multiply, or divide, and mentions the importance of having the frequency variable ω in the equations for a better understanding of the circuit's response to different signals.
  • #1
e_brock123
14
0

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.
 
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  • #2
e_brock123 said:
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.
 

Related to Help with complex numbers in AC analysis

What are complex numbers and how are they used in AC analysis?

Complex numbers are numbers that have both a real and an imaginary component. In AC analysis, they are used to represent the magnitude and phase of a sinusoidal signal. This allows for a more comprehensive analysis of circuits that involve AC signals.

What is the difference between a complex number and a phasor?

A complex number is a mathematical representation of a sinusoidal signal, while a phasor is a graphical representation of the same signal. Phasors are often used in AC analysis to simplify calculations and visualize the relationships between different signals in a circuit.

How do I perform calculations with complex numbers in AC analysis?

To add or subtract complex numbers, you simply add or subtract their real and imaginary components separately. To multiply two complex numbers, you use the FOIL method. To divide, you multiply by the complex conjugate of the denominator. It is important to keep track of the real and imaginary components separately and to use proper units for the phase.

What is impedance and how is it related to complex numbers in AC analysis?

Impedance is the measure of the opposition to the flow of AC current in a circuit. In AC analysis, it is represented by a complex number, with the real part representing the resistance and the imaginary part representing the reactance. The magnitude of impedance is the hypotenuse of the complex number, and the phase angle is the angle between the real and imaginary parts.

How can I use complex numbers to analyze circuits with multiple AC sources?

In circuits with multiple AC sources, you can use complex numbers to find the total impedance and current for each source, and then add them together using vector addition. This allows for a more efficient and accurate analysis of complex circuits with multiple AC sources.

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