# Complex Impedance and Phasors

## Homework Statement

An emf Vosinwt is applied to a circuit which consists of a self inductance L of negligible resistance in series with a variable capacitor C. The capacitor is shunted by a variable resistance R. Find the value of C for which the amplitude of the current independent of R.

## The Attempt at a Solution

We can use Phasor Algebra to make the problem easy. But I don't have any idea how to start.
Shall I take the equation V = jVL + VR ? , which comes out by taking VR along x-axis and VL along Y axis. If we use this equation then there is no capacitance anywhere.

Help me out!

Are you familiar with complex algebra? Anyway, there are normally 2 ways:
1/ Deal with complex algebra. I think it would be quite long and tedious.
2/ Draw phasor diagram. First draw the V(RC) vector (denoting voltage phasor across C and R). The I(C) vector will be perpendicular to it and leading it by 90 deg. The I(R) vector is in the same direction. Just draw out the vectors with arbitrary length. Then draw I(L) = I(C) + I(R) (vector summation), followed by V(L) and then V = V(L) + V(RC) (again, vector summation!). Playing around with simple geometry, you will get an expression for I(L).
The third way is to deal with systems of differential equations. Wanna try?

There is actually one shortcut for this specific problem. If I(L) is independent of R, I(L) when R = 0 must be equal to I(L) when R = infinity. When R = 0, the branch is short-circuited, and thus, there is only L left in the circuit; you can easily compute I(L) in this case. When R = infinity, there is no current going through R and thus the circuit basically has only L and C. You have I(L) unchanged in any case; then from here, you can compute C. This method has its own name: the extreme cases (or perhaps I devise the name but it's not novel and already considered as a scientific method). It's very useful when you need to do intuitive analysis or quick check on something

2/ Draw phasor diagram. First draw the V(RC) vector (denoting voltage phasor across C and R). The I(C) vector will be perpendicular to it and leading it by 90 deg. The I(R) vector is in the same direction. Just draw out the vectors with arbitrary length. Then draw I(L) = I(C) + I(R) (vector summation), followed by V(L) and then V = V(L) + V(RC) (again, vector summation!). Playing around with simple geometry, you will get an expression for I(L).

I ended up in a figure. You did not tell how to do the math

There is actually one shortcut for this specific problem. If I(L) is independent of R, I(L) when R = 0 must be equal to I(L) when R = infinity. When R = 0, the branch is short-circuited, and thus, there is only L left in the circuit; you can easily compute I(L) in this case. When R = infinity, there is no current going through R and thus the circuit basically has only L and C. You have I(L) unchanged in any case; then from here, you can compute C. This method has its own name: the extreme cases (or perhaps I devise the name but it's not novel and already considered as a scientific method). It's very useful when you need to do intuitive analysis or quick check on something

Yeah its a very smart method. It comes handy in maths usually. Never thought of applying it here

I don't get one thing here. You have to assume that XC>XL other wise you don't get a meaningful expression. Why is it so?

1/ See figure. Try to prove:
_ $$V^2 = V_{RC}^2 + V_L^2 - 2V_{RC}V_Lcos\alpha$$ (hint: This is purely math. Quite trivial. I think you know this identity )
_ $$cos\alpha = sin\beta$$ (hint: again, this is purely math. And again, it's trivial )
The next big step is to deal with big big big expression

2/ You don't have to assume so. We have: VL = IL|XL-XC|, so it can be either way: XL>XC or XL<XC.

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1/ See figure. Try to prove:
_ $$V^2 = V_{RC}^2 + V_L^2 - 2V_{RC}V_Lcos\alpha$$ (hint: This is purely math. Quite trivial. I think you know this identity )
_ $$cos\alpha = sin\beta$$ (hint: again, this is purely math. And again, it's trivial )
The next big step is to deal with big big big expression

I think you marked alpha wrong. It should be the angle between VRC and VL.

Alpha and Beta are related by Alpha = 90 + Beta

Next equation is IL2 = IC2 + IR2

Where is the big equation ?

2/ You don't have to assume so. We have: VL = IL|XL-XC|, so it can be either way: XL>XC or XL<XC.

I saw in quoted text that you typed [su b]....[/sub], no need of doing that - There is an X2 button in the advanced editor

This is what I did-
When R=0
I(L)=V/XL
When R=inf
I(L)=V/|XL-XC|

Equating both will give a meaningful expression only if XC>XL

ehild
Homework Helper
You can calculate with complex numbers, can not you? It is easier than the phasor method. The amplitude of the current is equal to the amplitude of the voltage over the magnitude of the impedance (short: impedance). Derive the complex impedance, and determine its magnitude. Find C so as this impedance does not depend on R. You can use Hikaru's method of extremes, this impedance has to be the same that you would get with R=0.

ehild

You can calculate with complex numbers, can not you?

Yeah I know about that. But I am stuck up somewhere-

The impedance of the resistor and capacitor is
1/Z1 = 1/R +1/jXC

My book uses operator 'j' to denote a vector obtained by rotating through 90 degrees anticlockwise i.e. if A is a vector along X-axis, then jA is a vector along Y axis.

I am confused in using +/- j.
Is the above equation correct?

Hikaru's method of extremes

He will get high after reading that

ehild
Homework Helper
The complex impedance of the parallel resistor and capacitor is
$$Z1=\frac{1}{1/R+j\omega C}$$

ehild

Can you explain how you got that?

ehild
Homework Helper
Zc=-j/(ωC) The reciprocals of the impedances add up.

1/Zc=(ωC)/(-j). 1/(-j)=j so 1/Zc=jωC.

For the resultant of resistor and capacitor in parallel is

1/Z=1/R+jωC so

$$Z1=\frac{1}{1/R+j\omega C}$$

ehild

Zc=-j/(ωC)

This is where I am confused. Why did you take -j? Why not +j?
Are the currents in the two branches (capacitor and resistor) in phase?

I just know that the voltage across the capacitor lags behind the current through it by 90 degrees. How do we use this fact in writing the above equation?

ehild
Homework Helper
The voltage is behind the current across the capacitor. See the phasor diagram.

Or: In complex notation we consider U and I in the Euler form

U=Uo exp(jωt), I=Io exp(jωt).

Across a capacitor, U=Q/C. The current is the time derivative of Q.

I=dQ/dt=C dU/dt = C (jω)Uo exp(jωt) --> I=C (jω) U.

The complex impedance is Z=U/I. For a capacitor, it is Zc=1/(jωC)=-j/(ωC).

ehild

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I think you marked alpha wrong. It should be the angle between VRC and VL.

Isn't it the angle between them? Notice that V(RC) can "move up" to the upper side of the parallelogram that forms V.

Next equation is IL2 = IC2 + IR2

Where is the big equation ?

After the equation relating V, V(RC), V(L), express V(RC), V(L) and sin(beta) in term of I(L) and impedances.

This is what I did-
When R=0
I(L)=V/XL
When R=inf
I(L)=V/|XL-XC|
Equating both will give a meaningful expression only if XC>XL

I mean, you don't have to assume so. You would derive that XC > XL after all. And you did.

I saw in quoted text that you typed [su b]....[/sub], no need of doing that - There is an X2 button in the advanced editor

The easier, the lazier

I'll leave the discussion on complex method to ehild and you I'm not very fond of this method (how pathetic, engineers use this method and I'm engineering student ).

ehild
Homework Helper
I'll leave the discussion on complex method to ehild and you I'm not very fond of this method (how pathetic, engineers use this method and I'm engineering student ).

It is such a nice method! Once you understand it, you will adore it. :) What will you do if you have a really complicated circuit with several loops? Using complex impedances, you can use the same methods as in case of DC circuits. This is the big advantage of it.

In this problem, the resultant complex impedance is simply

Z=jωL + 1/(1/R+jωC) or Z=jωL+R/(1+jωCR)

According to your method of extremes, the magnitude of Z should be the same as in case of R=0.

ehild

Last edited:
The voltage is behind the current across the capacitor. See the phasor diagram.

Or: In complex notation we consider U and I in the Euler form

U=Uo exp(jωt), I=Io exp(jωt).

Across a capacitor, U=Q/C. The current is the time derivative of Q.

I=dQ/dt=C dU/dt = C (jω)Uo exp(jωt) --> I=C (jω) U.

The complex impedance is Z=U/I. For a capacitor, it is Zc=1/(jωC)=-j/(ωC).

ehild

Ah I get it now. I had some misconception which is cleared.
Now I know 3 methods to solve this question. Thanks to both of you