Engineering Problems about Zin in complex circuit analysis

[e0ne199]

1. Homework Statement

the problem is my answer for question (a) is not the same as the answer provided by the question, i get 2.81 - j4.49 Ω while the answer demands 2.81 + j4.49 Ω

Homework Equations

simplifying the circuit, details can be seen below

The Attempt at a Solution

X_L1=jωL₁=j*20e-3*1000=j20Ω
X_L2=jωL₂=j*5e-3*1000=j5Ω
X_C1=1/(-jωC₁)=1/(j*100e-6*1000)=-j10Ω
X_C2=1/(-jωC₂)=1/(j*200e-6*1000)=-j5Ω

solving Z_par1:
Z_par1 = X_L1 * X_C1 /(X_L1 + X_C1)
Z_par1 = -j10Ω * j20Ω /(-j10Ω + j20Ω) = -j20Ω

solving Z_ser1:
Z_ser1 = -j20Ω - j5Ω = -j25Ω

solving Z_in:
Z_in = 1/((1/10)+(1/(j5))-(1/(j25))) = 2.81 - j4.49 Ω

 

[gneill]

Zin = 1/((1/10)+(1/(j5))-(1/(j25))) = 2.81 - j4.49 Ω

I believe that you are making some arithmetical mistake in evaluating the above expression. When I type the same thing into Mathcad I see:

 

[e0ne199]

I believe that you are making some arithmetical mistake in evaluating the above expression. When I type the same thing into Mathcad I see:

View attachment 234677

i don't know if mathcad can give result like that because i calculated using ti-89 emulator and got result like what i posted above... you can see the screenshot on my attachmet

 

[e0ne199]

does the use of i or j symbol affect the calculation?

 

[gneill]

What you've just shown me is :

Not the same.

 

[gneill]

does the use of i or j symbol affect the calculation?

No.

 

[e0ne199]

What you've just shown me is :

View attachment 234700

Not the same.

ok i think i have the solution... i did not input the equation into my calculator properly, i have achieved the same result after adding more bracket symbol into my equation... thanks for the guidance

 

[gneill]

Glad to have helped.

 

[thomastuck11]

The answers provided for questions (a) and (b) in problem 10.9 are easily arrived at, however the answer given for question (c) doesn't make sense to me. Can someone clarify?

 

[magoo]

The part (c) result matches what I came up with.

You have a delta configuration to start, with the capacitance from a to b across the top.
The left leg of the delta is an L-R combination. The right leg is an L-C combo.

For the L-R combo, I got a value of 2 + j4. For the L-C combo, I got a value of -j20.

Keep in mind that the L-R part and the L-C part are in series with each other and this series (L-R + L-C) is in parallel with the capacitance.

As I said, the results agree with what is shown for part (c).

 

[scottdave]

I confirm the answer for (c). I took the 10Ω resistor in parallel with the j5Ω inductor as Z_left. Then took -j10Ω cap in parallel with the j20Ω inductor as Z_right.
Then Z_left in series with Z_right.
Then that result in parallel with the -j5Ω cap.

 

[scottdave]

I don't want to ignite a battle, but looking at all of those parenthesis on the calculator screenshot makes me glad I have an RPN calculator.

It's a good idea to break it up into smaller equivalent impedances, then parallel those.

 

[Babadag]

 

[The Electrician]

All the required impedances can be obtained in one go using a little linear algebra. Choose the bottom node (g) of the circuit to be the reference node. Label the other two nodes 1 and 2.

Write the admittance matrix (Y) by inspection.
Invert the Y matrix to obtain the impedance (Z) matrix. Two of the elements of the Z matrix are directly the impedances Z(ag) (shown in red) and Z(bg) (shown in magenta). The impedance Z(ab) is the sum of Z(ag) and Z(bg) minus the two transfer impedances:

 

[Babadag]

By the way, it is easier-in my opinion-to use Excel complex:
Define Name:
ZR=COMPLEX(10,0)
ZLa=COMPLEX(0,5)
ZLb=COMPLEX(0,20)
ZCa=COMPLEX(0,-5)
ZCb=COMPLEX(0,-10)
Z1=ZLa*ZR/(ZLa+ZR)=IMDIV(IMPRODUCT(ZLa,ZR),IMSUM(ZLa,ZR))
Z2=ZLb*ZCb/(ZLb+ZCb)=IMDIV(IMPRODUCT(ZLb,ZCb),IMSUM(ZLb,ZCb))
ZA=IMSUM(Z1,Z2); ZB=IMSUM(Z2,ZCa);ZC=IMSUM(Z1,ZCa)
Zab=ZA*ZCa/(ZA+ZCa)=IMDIV(IMPRODUCT(ZCa,ZA),IMSUM(ZCa,ZA))
Zag=Z1*ZB/(Z1+ZB)=IMDIV(IMPRODUCT(Z1,ZB),IMSUM(Z1,ZB)
Zbg=Z2*ZC/(ZC+Z2)=IMDIV(IMPRODUCT(Z2,ZC),IMSUM(Z2,ZC))

 

[scottdave]

Wow! @Babadag that was kind of impressive. I do remember using Excel to attempt complex number mathematics, but found other methods.

If you are familiar with Python, it easily handles complex numbers.
note that Python uses the letter j after a number to indicate the imaginary component.

# Here is a simple function in Python, using a lambda function.
# it will calculate the parallel equivalent of two impedances.
#
zpar2 = (lambda x,y: x*y /(x+y))
print(zpar2(1,2j))
# here is the output
(0.8+0.4j)

R is another popular interpreted programming language, which is also available for free.
It also handles complex and imaginary (using the traditional i )

# R code for parallel impedances.
par2 <- function(a,b) { return (a*b / (a+b))}
par2(1,1i)
# here is the output
[1] 0.5+0.5i

Note these are simple calculations. I didn't put in any handling for when the denominator to be zero.

 

[Babadag]

Thank you very much, scottdave! It is very interesting, indeed.