It sounds like a specific part of the assignment is for YOU to be able to select and justify the method you use. In doing so, you are being asked to think about the various options and their pros and cons. If we just tell you what method to use, you will be robbed of the opportunity to learn from that part of the exercise.

So, instead, why not propose as many alternatives as you can think of and try to list any pros and cons you can. We can then comment on those and perhaps lead you to consider some others, too.

 

I was thinking Thevenin’s Theorem but the capacitor makes things more confusing. But I assume the capacitor will act as a resistor in an AC circuit their must be a formula to convert the capacitor into a resistor.

 

I was thinking Thevenin’s Theorem but the capacitor makes things more confusing. But I assume the capacitor will act as a resistor in an AC circuit their must be a formula to convert the capacitor into a resistor.

The problem statement implies that you are already expected to be conversant with the notion of impedance. What is the impedance of a capacitor at 1000 Hz?

 

Hint, black box with two terminals, how would you establish the Z between
the terminals ? Force a current and measure the V, then Z = V / I.

An approach is to remove load from circuit, short the V source to ground,
place a current source at load position, and solve for V at current source.
Then Z = V / I

"Normally" one would use LaPlace transform to do this, because there is a reactive
component in the network. Can also do it with complex algebra. Eg. Xc = 1/(jwC)
w = radian frequency, Pi = Pi, C = capacitance..

Regards, Dana.

 

 

How do you multiply 159.16 Ω by j and get (-j)(159.16)Ω

You need to be more careful about setting things up correctly instead of having them magically fix themselves based on how you know (or want) the answer should behave.

 

You have done the calculation using a frequency of 1 Khz instead of 100 hz.
Edit. I now see you were answering the question in post #4. I had assumed you were doing the calculation for the original circuit in post #1
Les.

 

You have done the calculation using a frequency of 1 Khz instead of 100 hz.

Les.

Good catch -- that's probably my fault since I asked for it at 1000 Hz because I looked at the image too quickly as I typed.

 

How do you multiply 159.16 Ω by j and get (-j)(159.16)Ω

You need to be more careful about setting things up correctly instead of having them magically fix themselves based on how you know (or want) the answer should behave.

I was presenting the answer in rectangular form as it has an imaginary element 90degrees phase shift = -j

I think for this question the phase shift is not that important

 

I was presenting the answer in rectangular form as it has an imaginary element 90degrees phase shift = -j

I think for this question the phase shift is not that important

My point is that your equation is not an equation (meaning that the left side is not equal to the right side).

 

I was presenting the answer in rectangular form as it has an imaginary element 90degrees phase shift = -j

I think for this question the phase shift is not that important

Hi,

Because there is a cap in there you will most likely end up with a complex impedance not just a resistance. That means there will be a phase shift, but if you stick to using complex numbers you wont have to deal with the phase shift just the real and imaginary parts.

 

Hi everyone. Thanks for all your input I greatly appreciate it. Here is the work I have done so far.

Please see attached. I have used a combination or Norton's, Thevenin's & Nodal Analysis this is by far the most complicated workings out I have done on a seemingly simple circuit.

I am not sure if it's 100% correct tomorrow I will simulate it using SIMetrix.

Any more input and or advice would be greatly appreciated.

Thanks again

 

Hi everyone. Thanks for all your input I greatly appreciate it. Here is the work I have done so far.

Please see attached. I have used a combination or Norton's, Thevenin's & Nodal Analysis this is by far the most complicated workings out I have done on a seemingly simple circuit.

I am not sure if it's 100% correct tomorrow I will simulate it using SIMetrix.

Any more input and or advice would be greatly appreciated.

Thanks again

You shouldn't need to simulate it to determine if it is correct. One of the beauties of most engineering problems is that the correctness of the solution can be determined from the solution itself.

Use your solution to determine the voltage and current across the 500 Ω load. Then analyze the original circuit using Mesh or Nodal analysis and see if you get the same result.

Remember what I said about the need to be more careful and avoid magical behavior? Well, not heeding that advice bit you here.

At the very start of your work you have

Xc = 1591.55 Ω

You then have

Z = -jX and get Z = 1591.55 Ω

How can this be? How can you multiply a real number by -j and end up with a real number?

But because you have gotten in the habit of being sloppy, this didn't raise any alarm bells and so you slugged on and replaced the capacitor with a 1.6 kΩ resistor, when it should have been replaced with a -j·1.6 kΩ impedance.

BTW: Your work is very well organized and well presented. It makes it very easy to follow -- that will pay off when it is graded.

 

You might consider finding the short-circuit current instead of the open-circuit voltage. I think you will find this problem lends itself much better to that approach.

 

Hi everyone. Thanks for all your input I greatly appreciate it. Here is the work I have done so far.

Please see attached. I have used a combination or Norton's, Thevenin's & Nodal Analysis this is by far the most complicated workings out I have done on a seemingly simple circuit. This means you dont get a result like 1000 Ohms, as that would only be right if the cap was a resistor with the calculated value.
So it's not about replacing the cap with a resistor, it's about replacing the cap with an impedance that happens to only have an imaginary part and no real part. After you combine it with some other resistive elements, you come out with a real part in addition to the imaginary part but dont loose the imaginary part.

I am not sure if it's 100% correct tomorrow I will simulate it using SIMetrix.

Any more input and or advice would be greatly appreciated.

Thanks again

Hi,

I just looked at your first pdf file, and i can say that it looks good except when you put a cap in parallel with a resistance you dont get a pure resistance.
For example, for a capacitor C and resistor R in parallel we get:
Z=(R-j*w*C*R^2)/(w^2*C^2*R^2+1)

and for example if w was equal to 1 then we would get:
(R-j*C*R^2)/(C^2*R^2+1)

and you see the 'j' never goes away because we always have an imaginary part as well as a real part:
real: R/(C^2*R^2+1)
imag: -j*C*R^2/(C^2*R^2+1)

From there you deal with the network using complex arithmetic.
Try again and you'll probably get it this time assuming you've done complex math before. If not, you should learn.

 

Hi,

I just looked at your first pdf file, and i can say that it looks good except when you put a cap in parallel with a resistance you dont get a pure resistance.
For example, for a capacitor C and resistor R in parallel we get:
Z=(R-j*w*C*R^2)/(w^2*C^2*R^2+1)

and for example if w was equal to 1 then we would get:
(R-j*C*R^2)/(C^2*R^2+1)

and you see the 'j' never goes away because we always have an imaginary part as well as a real part:
real: R/(C^2*R^2+1)
imag: -j*C*R^2/(C^2*R^2+1)

From there you deal with the network using complex arithmetic.
Try again and you'll probably get it this time assuming you've done complex math before. If not, you should learn.

So is w lower case omega? i.e. (2)(pi)(f) ???

 

So is w lower case omega? i.e. (2)(pi)(f) ???

Hi,

Yes, but you should think about how to do that as it is not hard. It's the same as connecting two resistors in parallel except you have to use complex math:
Z=Z1*Z2/(Z1+Z2)

and the Z's can be resistances, pure imaginary, or complex with real and imaginary parts.