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Complex impedance and phase angle of a circuit

  • Thread starter Pierson5
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Homework Statement:

Circuit contains 2 inductors, 2 resistors and 2 capacitors (I'll attach a photo).
L1 = 20mH, L2 = 50mH, C1 = 20uF, C2 = 50uF, R1 = 4 ohms, R2 = 8ohms, Vmax = 48V, f = 159.155Hz
Solve for the complex impedance (Zeq). Solve for Vmax and the phase 0eq of the current I(t) relative to V(t).

Relevant Equations:

Omega = 2(pi)f
Imax = Vmax / Zeq
0eq = tan^-1 (imaginary portion of equation (coefficient with i in Zeq) / real portion of equation)
I've attached my work below. The numbers seem odd to me though. Are my equations correct? Is the phase angle really (0/12)? If so, what are the implications of that?
 

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Answers and Replies

  • #2
cnh1995
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Why don't you compute the impedances one branch at a time and later add all together? Looks like you went straight for the total impedance by writing one big equation with a lot of symbols. It is a bit difficult to follow.

What is the "net" reactance of "each" branch?
 
  • #3
ehild
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I've attached my work below. The numbers seem odd to me though. Are my equations correct? Is the phase angle really (0/12)? If so, what are the implications of that?
Your derivation is correct, but you made a mistake when evaluating the parallel resultant of Z1 and Z2.
 
  • #4
gneill
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That the frequency is specified to three decimal places raises my suspicion that the circuit will turn out to be at or very close to resonance. It would be a good idea to compute the individual impedances of the parallel branches as suggested by @cnh1995. Keep lots of decimal places in intermediate computations.
 
  • #5
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That the frequency is specified to three decimal places raises my suspicion that the circuit will turn out to be at or very close to resonance. It would be a good idea to compute the individual impedances of the parallel branches as suggested by @cnh1995. Keep lots of decimal places in intermediate computations.
Like this (for Z1)?
XC = 1/(ω * C) = 1/(1000 * 20E-6) = 20
XL = ω * L = 1000 * 20E-3 = 50
R = 4

Z1 = √R2 + (XL + XC)2

Plugging in the values above, I get 30.27 for Z1
Doing the same thing for Z2 I get 31.05

Would Zeq be the sum of them both?
 
  • #6
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Why don't you compute the impedances one branch at a time and later add all together? Looks like you went straight for the total impedance by writing one big equation with a lot of symbols. It is a bit difficult to follow.

What is the "net" reactance of "each" branch?
What about:
Using: ZC + ZRL
Gives us: (1/iωL) + 1 / ((1/R) + (1/iωL))

To get equation for ZRLC

ZRLC = R - ω2RLC + iωL / iωRC - ω2LC

Plugging in the values for R1 (4), C1 (20E-6), L1 (20E-3), ω (1000) I get (2.4 + 20i) / (0.80i - 0.40)

For Z2 I get (-12 + 50i) / (4.0i - 2.50)

I'll start working on Zeq using
√(Z*)(Z) for Z1 and Z2 to see if anything turns up.

Z* being the complex conjugate of Z
 
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  • #7
gneill
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Like this (for Z1)?
XC = 1/(omega * C) = 1/(1000 * 20E-6) = 20
XL = omega * L = 1000 * 20E-3 = 50
R = 4

Z1 = √R2 + (XL + XC)2
No, there's no resistor in either branch. And you'll probably want to use complex impedance calculations rather than reactance values. That way you won't have to keep track of whether the net reactance is capacitive or inductive in nature -- in one the current leads the voltage and in the other the voltage leads the current. Complex number representation for the impedances and the complex mathematics automatically takes care of this.

Once you've got the individual branch impedances you can combine them into a single impedance which will then be in series with the two resistors.
 
  • #8
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No, there's no resistor in either branch. And you'll probably want to use complex impedance calculations rather than reactance values. That way you won't have to keep track of whether the net reactance is capacitive or inductive in nature -- in one the current leads the voltage and in the other the voltage leads the current. Complex number representation for the impedances and the complex mathematics automatically takes care of this.

Once you've got the individual branch impedances you can combine them into a single impedance which will then be in series with the two resistors.
Ah, that makes sense. I'll keep working with the complex impedance values. Is the reply above on the right track?

Calculating Zeq for branch one, nothing jumps out that demonstrates I would be on the right track. This is what I got so far (using values for Z in my previous post):

Zeq1 = √(Z*)(Z)
(Z)(Z*) = (2.4 + 20i) / (-0.40 + 0.80i) * (2.4 - 20i) / (-0.40 - 0.80i) = 405/0.80, taking square root = 22.52

Zeq2 = √(Z*)(Z)
(Z)(Z*) = (-12 + 50i) / (-2.5 + 0.40i) * (-12 - 50i) / (-2.5 - 40i) = 2644/6.41, taking square root = 51.36

This doesn't seem right to me given the precision of the starting f value, I think I should be dealing with simpler/rounded numbers.
 
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  • #9
cnh1995
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Forget about the resistors for now.
You have two parallel branches, containing one L and one C in series in each branch.

What is the individual reactance of each branch?
 
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  • #10
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Forget about the resistors for now.
You have two parallel branches, containing one L and one C in series in each branch.

What is the individual reactance of each branch?
I think I see what your saying, instead of computing everything algebraically to find Zeq, plug in values for L and C and compute those first?

ZL = iωL
ZC = 1/iωC

Plugging in the values from my original post for ZL + ZC I get:
1. ZL + ZC = -30i
2. Same as above = 30i

I'm not sure where to go from here.
 
  • #11
cnh1995
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Plugging in the values from my original post for ZL + ZC I get:
1. ZL + ZC = -30i
2. Same as above = 30i
That's correct.
What is the parallel equivalent of these two?
 
  • #12
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That's correct.
What is the parallel equivalent of these two?
Combining them would give us zero.
 
  • #13
cnh1995
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Combining them would give us zero.
They are in parallel.
 
  • #14
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They are in parallel.
Isn't Ztotal = 1/z1 + 1/z2?

Or is it: (Z1)(Z2) / Z1 + Z2 (undefined?)
 
  • #15
cnh1995
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Or is it: (Z1)(Z2) / Z1 + Z2 (undefined?)
It's not undefined, it is infinite.
Look up parallel resonance.
The parallel L-C network forms a resonant circuit which makes the impedance seen by the source infinite.
 
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  • #16
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It's not undefined, it is infinite.
Look up parallel resonance.
The parallel L-C network forms a resonant circuit which makes the impedance seen by the source infinite.
I see, so that would explain why the phase angle is 0? The inductor/capacitor reactances have the same magnitude but cancel each other out because they are 180 degrees apart in phase?

Does that mean Imax approaches zero as it's denominator approaches infinity? I'm not sure what this means for the rest of the problem or how to demonstrate it mathematically.
 
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  • #17
gneill
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Does that mean Imax approaches zero as it's denominator approaches infinity? I'm not sure what this means for the rest of the problem or how to demonstrate it mathematically.
The real component of the impedance remains finite (##12 \Omega##) while the imaginary component has gone to infinity. You can use some calculus (limits) to show how arctan(y/x) behaves in such a case, where y →∞. Or maybe just look at a graph of atan() and draw your conclusion by inspection.
 
  • #18
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The real component of the impedance remains finite (##12 \Omega##) while the imaginary component has gone to infinity. You can use some calculus (limits) to show how arctan(y/x) behaves in such a case, where y →∞. Or maybe just look at a graph of atan() and draw your conclusion by inspection.
So, when calculating Zeq, I can show that the parallel components have infinite reactance, therefore I would not need to continue any other calculation except (R1 + R2) + imaginary component (infinity), and the rest of the problem is the same as my original attempt?

When calculating the phase angle, wouldn't that change the answer from zero to pi/2 (or 90 degrees)?
 
  • #19
gneill
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So, when calculating Zeq, I can show that the parallel components have infinite reactance, therefore I would not need to continue any other calculation except (R1 + R2) + imaginary component (infinity), and the rest of the problem is the same as my original attempt?
Yep, pretty much.
When calculating the phase angle, wouldn't that change the answer from zero to pi/2 (or 90 degrees)?
Yes, but it's questionable what a phase angle means for zero current :smile:

If you assume that the given frequency is exact and the components ideal, then theoretically the reactive part of the impedance would not be infinite, simply very large. That's because ##\omega = 2 \pi f## would not be precisely 1000 rad/sec, and the branch impedances would not precisely match in magnitude. In such a case you could calculate an actual current (likely ridiculously small) and a phase angle of either + or - 90° depending upon whether the net reactance turned out to be capacitive or inductive in nature.
 
  • #20
gneill
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I've used Mathcad to plot how the phase angle behaves around the ##\omega = 1000 \frac{rad}{sec}## point for the circuit under consideration. I 've assumed that the components are ideal. Here's how it looks:
243926


If the angular frequency of the source were exactly 1000 rad/sec, then one might argue that the phase angle was zero, but I think it would be better to say that it was undefined for a practical situation where component values and frequency specification are subject to uncertainty.
 
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  • #21
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I've used Mathcad to plot how the phase angle behaves around the ##\omega = 1000 \frac{rad}{sec}## point for the circuit under consideration. I 've assumed that the components are ideal. Here's how it looks:

If the angular frequency of the source were exactly 1000 rad/sec, then one might argue that the phase angle was zero, but I think it would be better to say that it was undefined for a practical situation where component values and frequency specification are subject to uncertainty.
Ah, now I'm not sure what to conclude. It makes since if it was tan-1(infinity) = pi/2 or 90 degrees. This is consistent with the rest of the problem we've worked through. How would tan-1(infinity) = zero or undefined?
 
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  • #22
gneill
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In a practical circuit it would be undefined, since component tolerances and accuracy of the source frequency would influence the result.

In the present case, if the components and frequency are considered ideal values, a calculation keeping enough digits would show that the current phase would be ##-\frac{\pi}{2}##. But the current would have the incredibly small magnitude of ##I = 3.79\times 10^{-16} \text{A}##.

It's a tricky question that's designed to make you think about what can happen near resonance.
 
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  • #23
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In a practical circuit it would be undefined, since component tolerances and accuracy of the source frequency would influence the result.

In the present case, if the components and frequency are considered ideal values, a calculation keeping enough digits would show that the current phase would be ##-\frac{\pi}{2}##. But the current would have the incredibly small magnitude of ##I = 3.79\times 10^{-16} \text{A}##.

It's a tricky question that's designed to make you think about what can happen near resonance.

Talked to my professor, he said Imax is 48/infinity and therefore goes to zero. There is no current and therefore no phase. Phase is undefined. Thankyou all for your help. Really helped solidify my understanding of this problem!
 
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  • #24
gneill
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I'm very happy that I could be of help! Good luck in your studies.
 
  • #25
cnh1995
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Thankyou all for your help. Really helped solidify my understanding of this problem!
Glad to hear your feedback!
Plugging in the values from my original post for ZL + ZC I get:
1. ZL + ZC = -30i
2. Same as above = 30i
As a side note, when dealing with complex quantities in electric circuits, the complex operator 'i' is replaced by 'j' as 'i' is reserved for current.
 
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