RC circuit with complex exponential

AI Thread Summary
The discussion focuses on solving a parallel RC circuit problem involving a current source and finding the values of resistance (R) and capacitance (C). The user initially attempted source transformation but struggled to match the expected results. Key insights include using the impedance formula Z = Vs(t)/Is(t) to derive R and C from the real and imaginary parts of the admittance equation. The conversation also touches on converting sine to cosine functions for voltage inputs and the implications for current in a series versus parallel circuit. The final clarification emphasizes the correct formulation of the impedance and the approach to solving a second-order differential equation for circuit analysis.
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I am stuck on this...
Given a circuit: current source (Is(t)), R , C - all parallel; Is(t) = e^jt, Vs(t) = 223.6e^j(t - 63.43), Vs(t) is voltage across the current source, which I assume to be the same across R and C since they are ||.
Find R and C. (ans: 500 Ohm, 4mF)

My attemp was to use source transformation and get R from it, but it did not match the answer.
I.e. I tried to make a circuit all series by source transformation: V = IR, so given Vs(t) and Is(t) , R = 223.6e^(-j63.43) and that does not turn out to be 500 Ohm

Another way I am thinking is that Vs(t) is the response on the C, so I will have to work backwards from the equations for phase and Vm (response magnitude).

Any help on this is GREATLY appreciated...this is as clear as mudd... :cry:
 
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The impedance of the parallel RC circuit is Z = Vs(t)/Is(t).
But Z = R + jC. The real part of Z is R and the imaginary part is C.
 
Thanks for helping, SGT, although we have not talked about impedance.
So, I got Z = 223.6e^-j63.43, and corresponding representation as 223.6cos(-63.43) + j 223.6 sin(-63.43) and I still do not get 500 for the real part :frown:
Could you explain in little more detail, I do not think I am quite getting it.
 
My mistake.
The admittance of the parallel RC circuit is Y = 1/R+jωC (it is the complex equivalent of conductance).
Where ω is the frequency of the voltage/current. In your example ω=1.
But Y = Is(t)/Vs(t) = 1/223.6et+63.43
So, 1/R is the real part and C the imaginary part of Y.
 
Thanks, that worked! I will have to look a little ahead to understand it better.

Umm, I have a related question though, different problem, didn't want to start another thread:
if I have input excitation Vs(t) = 100sin(10t) and I see all examples in the book with cosines, so converting it I get 100cos(10t - 90). Then I want to be sure if this is correct to assume:
1. Vs(t) = 100sin(10t) = 100cos(10t - 90) = Re[Vs e^j(10t - 90)]
2. Vc(t) = Vs cos(10t + phita) or is it (10t + phita - 90)?

And let's say I need IL(t) and not Vc(t) in diff equation for the circuit, is the response form different for current? is Vs just a const that marks amplitude/magnitude?
Appreciate your help very much.
 
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Vs is the voltage of your source, VC the voltage across the capacitor and VL the voltage across the inductor.
In a series circuit all elements have the same current and different voltages. In a parallel circuit all elements have the same voltage and different currents.
If you have an RC series circuit, the impedance Z = R - jXc, where Xc = 1/(ωC).
The current across the elements is I = V/Z and the voltage across the capacitor is Vc = IXc.
 
Well, the circuit is RLC series, so I have 2-order diff equation and we have to use IL as a variable in the equation.
The input is Vs(t) = 100 sin(10t) with LRC given, we need to find IL(t) and then Vc(t), so in this case if IL is the variable, does my diff equation end like this:

... + (R/L)deriv[IL(t)] + IL/(LC) = deriv(Vs)/L?

but then it is in sinus form, so I converted it in cos(10t - 90) and then I do not know how response looks like.

Thanks again.
P.S. we did not start phasors yet.

<edit> I think I got it </edit>
 
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You have a 2nd order non-homogeneous differential equation. It's solution is the sum of the solution to the homogeneous equation and a particular solution.
You should not derive the voltage of the source. If you have a series circuit you should use voltages:
vL + vR + vC = Vs
But vR = R.i and vL = Ldi/dt and i = CdvC/dt.
If you replace the values in your equation you will have a 2nd order non-homogeneous differential equation in vC.
The solution of the homogeneous equation depends on the relative values of R, L and C.
The particular solution is of the form Acos(10t+φ)
If you replace this solution and it's two first derivatives in your equation you can solve for A and φ.
 
SGT said:
The impedance of the parallel RC circuit is Z = Vs(t)/Is(t).
But Z = R + jC. The real part of Z is R and the imaginary part is C.
I think you made a typo that can be misleading...should be Z = R + jX, where X = -1/(w*C) or w*L (w is omega = 2*pi*f).
 
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SGT said:
...the impedance Z = R - jXc, where Xc = 1/(ωC)...
I should have read all the posts...you caught that one.
 
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