# Need Help With This Question. Circuit Impedance. Much Appreciated

#### JohnKid5

Joined Nov 2, 2017
19
Hi
I'm struggling with this question and have attempted it several times but unsure what formula to proceed forward with.
If someone could help me determine the formula or equation for each a-e I will attempt the question and post my results for review.

I have attached the Circuit as a PDF file.

For the circuit shown in Figure Q8, calculate:

a) the total circuit impedance in polar form.
b) the supply current Is.
c) the branch current I1.
d) the active and reactive powers for the complete circuit.
e) the circuit power factor.

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#### shteii01

Joined Feb 19, 2010
4,644

#### The Electrician

Joined Oct 9, 2007
2,952
Hi
I'm struggling with this question and have attempted it several times......

#### WBahn

Joined Mar 31, 2012
29,497
Hi
I'm struggling with this question and have attempted it several times
Then it should be a fairly simple matter to show what you think is probably your best attempt.

Imagine how you would answer the questions with just normal resistors. Now treat the complex impedances the same way.

#### JohnKid5

Joined Nov 2, 2017
19
Hi still struggling with this question.

Not even sure where to start. could someone please help me or put me on the correct track to attempting it?

do I find the total impedance first using : Zt= Zr +Zl+ Zc
then AxB= (10+j10) x (-j12)

really confused.

#### shteii01

Joined Feb 19, 2010
4,644
Hi still struggling with this question.

Not even sure where to start. could someone please help me or put me on the correct track to attempting it?

do I find the total impedance first using : Zt= Zr +Zl+ Zc
then AxB= (10+j10) x (-j12)

really confused.
First you get rid of the resistors in parallel.
That will result in two resistors in series.
Then you combine the two resistors in series.
That will result in total impedance.

• JohnKid5

#### RBR1317

Joined Nov 13, 2010
712
• JohnKid5

#### JohnKid5

Joined Nov 2, 2017
19
For the following resistor network, can you find the total resistance?
View attachment 139178
Now take the formula for total resistance and substitute R1=(12+j16), R2=(10+j10), R3=(-j12). Then reduce the complex expression to polar form, and you have the answer to part a).
Thanks, will have a go and repost my results

#### WBahn

Joined Mar 31, 2012
29,497
Hi still struggling with this question.

Not even sure where to start. could someone please help me or put me on the correct track to attempting it?

do I find the total impedance first using : Zt= Zr +Zl+ Zc
then AxB= (10+j10) x (-j12)

really confused.
First, consider the case of a simpler problem in which all of the impedances are simple resistors. Could you solve that one?

If yes, then the thing to remember is that one of the primary motivations for using complex impedances for capacitors and inductors is that it allows you to treat impedances just like resistors except that the numbers involved happen to be complex.

If no, then your problem is more fundamental and you need to take a step back and really get a firm handle on circuit analysis principles and techniques on purely resistive circuits without the complicating factor of complex impedances.

#### JohnKid5

Joined Nov 2, 2017
19
ok i have had a few cracks at this question and the (-j12) is throwing me off every attempt.

R1=(12+j16)
R2=(10+j10)
R3=(-j12)

If it was a simple Resistor i would do:
1/Rt = 1/R2 + 1/R3. Get the Rt for the resistors in parallel.
Then, Rt = R1 + (sum of 1/Rt) to find the series.

How can i find the Impedance for the parallel? after that I think its just Polar Multiplication between the two in series and i should have my answer?

Thanks

#### The Electrician

Joined Oct 9, 2007
2,952
You just do the same math you would with just resistors, but in complex arithmetic. You could get yourself calculator that can do complex arithmetic, or find an app on the web that can do it.
Here's a page that discusses complex arithmetic: https://www.mathportal.org/algebra/complex-numbers/complex-numbers-arithmetic.php

Here's a calculator that can do the complex arithmetic for you: https://www.mathportal.org/calculat...tor/complex-numbers-operations-calculator.php

To do reciprocals on the that calculator, just make the first number 1 + j0, then divide by the number whose reciprocal you want.

For example, the calculator gives the reciprocal of 12 + j16 as 3/100 - j (1/25)

Using this calculator you shouldn't have any trouble carrying the needed calculations

#### JohnKid5

Joined Nov 2, 2017
19
(10+j10) +(j12)
= (10+0) + j(10+12)
=10 - j2

(12+j16) + (10 - j2)
= (12+10) + j(16-2)
= 22+ j14

Something like this ?

#### JohnKid5

Joined Nov 2, 2017
19
Polar form : square root (22 squared + 14 squared = 26.07

Arc tan 14/22
= 32.47degrees

Polar Form : 26<32.47Degrees

#### The Electrician

Joined Oct 9, 2007
2,952
(10+j10) +(j12)
= (10+0) + j(10+12)
=10 - j2

(12+j16) + (10 - j2)
= (12+10) + j(16-2)
= 22+ j14

Something like this ?
"(10+j10) +(j12)
= (10+0) + j(10+12)
=10 - j2"

This is what you would do if those two impedances were in series, but they are not in series--they're in parallel. In post #10 you said:

"If it was a simple Resistor i would do:
1/Rt = 1/R2 + 1/R3. Get the Rt for the resistors in parallel."

You have to do that same thing with your two impedances in parallel.

Also, what happened to the minus sign associated with j12?

Try again. Use the web calculator I linked to do the necessary reciprocals.

#### WBahn

Joined Mar 31, 2012
29,497
ok i have had a few cracks at this question and the (-j12) is throwing me off every attempt.

R1=(12+j16)
R2=(10+j10)
R3=(-j12)

If it was a simple Resistor i would do:
1/Rt = 1/R2 + 1/R3. Get the Rt for the resistors in parallel.
Then, Rt = R1 + (sum of 1/Rt) to find the series.

How can i find the Impedance for the parallel? after that I think its just Polar Multiplication between the two in series and i should have my answer?

Thanks
Remember that -j12 a number, like any other number. It just happens to be what we have chosen to call "imaginary".

If the -j12 is throwing you off but the 10+j10 isn't, then just write the former as 0-j12.

You also need to get into habit of tracking units. If you just use the numbers for the various R values, then when you did your last step and computed

Rt = R1 + (sum of 1/Rt)

You would have gotten a number for the answer, tacked on an Ω symbol to the answer, and gotten little or no credit and would have been wondering why. But if you tracked units throughout your work, your units would have gotten messed up and you would have known that you made a mistake.

#### JohnKid5

Joined Nov 2, 2017
19
Thanks for the help, finally got my head around the complex numbers for impedance.

ok, I've managed to get the Total Impedance in Polar form. its 20.2<84.9degrees.

for the supply current, I've used I=V/Z and have worked it out to be 11.38<-84.9degrees A

To calculate the Branch Current at I1. What approach should I take for this?

#### The Electrician

Joined Oct 9, 2007
2,952
Thanks for the help, finally got my head around the complex numbers for impedance.

ok, I've managed to get the Total Impedance in Polar form. its 20.2<84.9degrees.

for the supply current, I've used I=V/Z and have worked it out to be 11.38<-84.9degrees A

To calculate the Branch Current at I1. What approach should I take for this?
You don't have the correct result for the total impedance. You will need to show every step in your calculation if you want help in tracking down your errors.

#### JohnKid5

Joined Nov 2, 2017
19

#### The Electrician

Joined Oct 9, 2007
2,952
You're doing it wrong. You first have to calculate the equivalent of the parallel combination of R2 (which is (10+j10)) and R3 (which is (0-j12)).

You could use the equivalence you showed in an earlier post: 1/Zeq = 1/Z2 + 1/Z3

You could also use the well known product over the sum method: Zeq = (Z2*Z3)/(Z2+Z3)

Having found Zeq then Ztotal = R1 + Zeq which is (12+j16) + Zeq