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Finding equivalent impedience Zeq of circuit

  • Engineering
  • Thread starter asdf12312
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  • #1
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1

Homework Statement


1yad55.png



Homework Equations


w = 10^5 rad/s
Z(L)= jwL
Z(C) = -j/wC
Z(eq)= ??

Zeq(rectangular form) = a+jb
Zeq(polar form)= p<(theta)
p = sqrt(a^2 + b^2)
theta = arctan(b/a)

The Attempt at a Solution


i calculated the impedences across the inductor/resistors:
across 0.5uF capacitor: -j/wC = -j/(10^5)(0.5 x 10^-6) = -j/0.05 = -20j
across 0.1mH inductor: jwL = j(10^5)(0.1x10^-3)= 10j
across 10uF capacitor: -j/wC = -j/(10^5)(10x10^-6)= -j

still not sure what to do with the resistances. would i have to find R(eq)? if i do this by 10+ 10||10 i get 15ohms.

to get to rectangular form before converting to polar form would be:
R+jX, where i think R would be the R(eq) and X would be the equivalent impedence across both inductors and capacitors? how would i find this by adding them together?
 

Answers and Replies

  • #2
818
67
For starters, what's the impedance of an ideal resistor?

Try drawing the circuit on a piece of paper where you replace all the circuit elements with generic elements with impedances Z1, Z2, ... and so on. If I had, for instance, a circuit with a bunch of circuit elements in parallel, I could draw the following instead:

200px-Impedances_in_parallel.svg.png


Then it's just a matter of using the rules for impedances in series/parallel to find the equivalent impedance.
 
  • #3
gneill
Mentor
20,781
2,759
Impedance is just a fancy version of resistance. You can apply the same formulas and techniques that you've learned for simplifying resistor networks, only each "resistance" happens to be specified as a complex number.

You've calculated the impedances of the reactive components (inductors, capacitors). Just treat them as though they were resistor values. Simplify series and parallel connections as you would normal resistance values, but do the math using complex numbers.
 
  • #4
199
1
well i do know that the impedence of resistor would be the resistance value itself.
so Z(10ohm)=10 + 0j = 10< 84.29 degrees (in polar form)
Z(0.5uF): 0 -20j = 20 < -87.14 degrees
Z(0.1mH): 0+10j = 10 < 84.29 degrees
Z(10uF): 0 -j = 1 < -45 degrees

if i replace the elements with their impedences, would i add them in rectangular or polar form? and i am guessing adding impedences is same rule as adding resistors?
 
  • #5
gneill
Mentor
20,781
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well i do know that the impedence of resistor would be the resistance value itself.
so Z(10ohm)=10 + 0j = 10< 84.29 degrees (in polar form)
Z(0.5uF): 0 -20j = 20 < -87.14 degrees
Z(0.1mH): 0+10j = 10 < 84.29 degrees
Z(10uF): 0 -j = 1 < -45 degrees

if i replace the elements with their impedences, would i add them in rectangular or polar form? and i am guessing adding impedences is same rule as adding resistors?
Your angles look wonky. How did you determine them? For pure components (resistance, inductance, capacitance), the impedances lie directly along the axes of the complex plane. So, 0,+90, and -90 degrees.

Yes, you combine them as you would pure resistances using the same formulas. Adding complex numbers is simpler in rectangular form. Multiplication and division is simpler in polar form. Easiest of all is to use a calculator or computer application that handles complex numbers automagically.
 
  • #6
199
1
i used the formula theta=arctan(b/a), but to be honest i wasn't sure what i was doing either.

so if i work with impedences in rect form, i can replace all elements with their impedances and find the equivalent impedence:
Z1 = Z(0.5uF) + Z(10ohm) = (0-20j) + (10+0j) = 10-20j
Z2 = Z(0.1mH) + Z(10ohm) = (0+10j) + (10+0j) = 10+10j
Z3 = Z(10uF) + Z(10ohm) = (0-j) + (10+0j) = 10-1j

in this particular circuit Z(eq) would be Z1+ (Z2 || Z3).
actually your right, i could use an online calculator to find the parallel combination of Z2||Z3, since i have no idea how to do this with complex numbers. however, it did say to calculate by hand so i'm not sure. if i were to use an online impedence calculator http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html#c4
then i get the value:
Z2||Z3 = 6.258 + 1.684j

then the Zeq (rectangular form) would be: Z1+ Z2||Z3= (10-20j)+(6.258+1.684j) = 16.258-18.316j
when converting to polar form this would be 24.5<-48.4 degrees
 
  • #7
gneill
Mentor
20,781
2,759
i used the formula theta=arctan(b/a), but to be honest i wasn't sure what i was doing either.

so if i work with impedences in rect form, i can replace all elements with their impedances and find the equivalent impedence:
Z1 = Z(0.5uF) + Z(10ohm) = (0-20j) + (10+0j) = 10-20j
Z2 = Z(0.1mH) + Z(10ohm) = (0+10j) + (10+0j) = 10+10j
Z3 = Z(10uF) + Z(10ohm) = (0-j) + (10+0j) = 10-1j

in this particular circuit Z(eq) would be Z1+ (Z2 || Z3).
actually your right, i could use an online calculator to find the parallel combination of Z2||Z3, since i have no idea how to do this with complex numbers. however, it did say to calculate by hand so i'm not sure. if i were to use an online impedence calculator http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html#c4
then i get the value:
Z2||Z3 = 6.258 + 1.684j

then the Zeq (rectangular form) would be: Z1+ Z2||Z3= (10-20j)+(6.258+1.684j) = 16.258-18.316j
when converting to polar form this would be 24.5<-48.4 degrees
Yup. Looks good.

You should review the mechanics of complex number manipulation, just in case this sort of problem comes up on an exam. A web search on "complex number tutorial" should help.
 
  • #8
199
1
OK so i figured out how to do division/multiplication in rect form, but i know it's easier in polar form. the only problem is, in polar form there's only 1 number so i'm not sure how to get it back in the form R(eq) + jX, suppose i were to do the parallel method Z2*Z3/(Z2+Z3) in polar form.

given:
Z2=10+10j
Z3=10-1j

i got the same answer for the parallel combination as the calculator, doing it by hand:
Z2 || Z3 = Z2*Z3/(Z2+Z3)

where i got Z2*Z3 using the polynomial method, (10+10j)(10-1j) = 110+90j, and Z2+Z3 is of course 20+9j

when dividing these two:
Z2*Z3 = a+bj = 110+90j
Z2+Z3 = c+dj = 20+9j

formula for dividing complex numbers i found online is: (ac+bd)+ (bc-ad)j / (c^2+d^2)
=[((110*20)+(90*9)) + ((90*20)-(110*9))] / (20^2 + 9^2)
= (3010 + 810j)/481
= 6.258 + 1.684j

which is the same answer as i got. was wondeirng if there was an easier way to do this though?
 
  • #9
gneill
Mentor
20,781
2,759
Polar form consists of two values: magnitude and angle. Rectangular form consists of two values: Real and Imaginary. No information is lost when converting from one form to the other.

Polar form is easier for division and multiplication.
1. First convert both values to polar form (magnitude; angle).
2. Multiply or divide the magnitudes as desired.
3. If multiplication, add the two angles. If division, subtract the divisor's angle from the numerator's.
3. Convert back to rectangular form, if required.
 
  • #10
199
1
thanks, i understand it now. for the last part, our teacher wants us to calculate the value of angular frequency w that causes the imaginary part of the impedance (the j values) to equal zero, or find the resonant frequency. would apreciate any help.

Z(L)=jwL
Z(C)=-j/wC

i tried leaving out the resistor values and simply working with the impedences for the inductor and 2 capacitors in the circuit, since those are the only imaginary (j) values. therefore I get: Z(0.5uF) + ( Z(0.1mH) || Z(10uF) ) = 0

substituting equations above:
-j/wC + ( jw(0.1mH) * || -j/w(10uF)) = 0

working only the parallel combination, and using variables only:
(jwL*-j/wC) / (jwL + -j/wC)
=L/C / (w^2*L*C) since j^2=-1
=1/(w^2*C^2)

-j/w(C1) + 1/(w^2*(C2)^2) = 0 (C1 and C2, since capacitance values are different)

did i do this right so far?
 
  • #11
gneill
Mentor
20,781
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Don't leave the resistors out of the calculation; they will contribute to both the real and imaginary parts of the impedance (thanks to the multiplications and divisions involved in the calculation when things are paralleled).

Looks like there might be a heavy bit of algebra involved...

(By the way, I don't think you're going to find a real-valued frequency ω for which the imaginary component of the impedance goes to zero. That's my assessment based upon experience, looking at the component values .... but don't take my word for it :smile: )
 
  • #12
818
67
I think you should try to present the problem here exactly like it was presented to you, like you did in your original post.

You'll only be able to show that the imaginary part of the equivalent impedance tends to zero as ω→∞.
 
  • #13
199
1
sorry, i thought that explaining i needed to find 'resonant frequency' was clear enough, but pehaps it wasn't :eek:
this is exactly how my instructor worded it:

To maximize the current that is transmitted to the circuit (which also maximizes the power
transferred to the circuit), we want to make the equivalent impedance as small as possible. Since
we can’t modify the value of the resistors, we should modify the value of the inductive and
capacitive impedances. Specifically, we want the inductor and capacitor impedances to cancel
each other out.
Recall: Z(L) = jwL and Z(C)= -j/wC

Since the inductive impedance is positive and the capacitive impedance is negative, if we choose
the correct frequency (w), the equivalent impedance of the entire circuit will be all real (no
imaginary part).

Calculate the frequency that maximizes the current entering the circuit. To do this, find the
angular frequency that causes the imaginary part of the impedance to equal zero. This frequency is called the resonant frequency.

i am still confused by this problem, i tried getting the imaginary part contributed from the inductor/capacitors to equal 0. i might be wrong, but that's what i understood from it. i tried adding the parallel impedences of the 2nd capacitor and the inductor together (along with resistor values), then adding this value to the impedence of the 1st capacitor plus the resistor. since there will be 2 j values in the equation, one positive and one negative, i could seperate these out and set them equal to 0, and solve for resonant frequency w. however, my main problem i am having is that i am unable to add the parallel components without knowing w.

supposing Z2*Z3/(Z2+Z3) as i did the previous problem, the fastest way to multiply to get the numerator would be to convert Z2 and Z3 to polar form and then multiply them, however this would result in a really messy solution, for example sqrt(100+(w^2)(0.1mH)^2)) would be the polar form for Z2, simply because of the unknown w. i'm probably making this harder than it is, but i honestly dont know how to approach this.
 
Last edited:
  • #14
The Electrician
Gold Member
1,246
152
I wonder if you have the right schematic for this problem.

You say "my main problem i am having is that i am unable to add the parallel components without knowing w. "

What you should do is leave w as a variable and solve for the value of w which would make the imaginary part of Zeq zero.

However, for the schematic and component values you show in post #1, there is no solution at a finite value of w.

Plotting the magnitude of Zeq and its imaginary part can help you to see the problem. Here are such plots:

attachment.php?attachmentid=58180&stc=1&d=1366838856.png


The magnitude of Zeq is in blue and the imaginary part is in red.

Your problem says "Calculate the frequency that maximizes the current entering the circuit. To do this, find the angular frequency that causes the imaginary part of the impedance to equal zero." This suggests that the frequency of maximum current is also the frequency where the imaginary part is zero, but you can see in the plots that the frequency at which the magnitude of the impedance is a minimum (that's where the current would be maximum) is not the frequency where the imaginary part is zero. In fact, the imaginary part isn't zero until the frequency is ∞, so to speak.

This is why I think you may have the wrong circuit schematic.
 

Attachments

  • #15
gneill
Mentor
20,781
2,759
Yes, the algebra for this problem will be a bear. But you do need to leave ω as a variable, since that's the variable you're looking for.

As milesyoung intimated above, and as I also suggested (but did not prove), the imaginary part of the impedance for this circuit won't go to zero for any finite value of ω. This is not to say, however, that you can't find a value of ω that would maximize the magnitude of the current, or the real portion of it. That will take even more algebra and a bit of calculus.

It is possible that the component values of this particular circuit were not well chosen to provide a simple, clear solution where the reactive part of the impedance goes to zero at a real, finite frequency.
 
  • #16
818
67
Since the inductive impedance is positive and the capacitive impedance is negative, if we choose the correct frequency (w), the equivalent impedance of the entire circuit will be all real (no imaginary part).

Calculate the frequency that maximizes the current entering the circuit. To do this, find the angular frequency that causes the imaginary part of the impedance to equal zero. This frequency is called the resonant frequency.
This reads an awful lot like you're supposed to be looking at a series RLC circuit instead.

If Zeq = R + jX is the equivalent impedance of your circuit, X = 0 doesn't minimize |Zeq|, in general, unless R is constant, which is why this:
Calculate the frequency that maximizes the current entering the circuit. To do this, find the angular frequency that causes the imaginary part of the impedance to equal zero.
doesn't make any sense for the circuit you included in your original post. As gneill wrote in post #11:
Don't leave the resistors out of the calculation; they will contribute to both the real and imaginary parts of the impedance (thanks to the multiplications and divisions involved in the calculation when things are paralleled).
which is also true for the reactive circuit elements. The real part of the equivalent impedance will be a function of ω.
 

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