# Finding total impedance of the circuit

• Engineering

## Homework Statement

here is the circuit I am trying to figure out

## Homework Equations

I have to find the total impedance for the circuit.

## The Attempt at a Solution

I know how to find the total impedance in a series circuit and in a paralell circuit when I have just the Resistor and the capacitor. But I have no Idea how to figure this out. I been at it for 2 days and i can't seem to find anywhere on line or the book on how to do this.

I have tried to add up the resistors first in every form and then try to find the total Z = RXc/sqr(R+Xc), -Tan^-1(XC/R)

can someone point me in the right direction.

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Zryn
Gold Member
I know how to find the total impedance in a series circuit and in a paralell circuit when I have just the Resistor and the capacitor. But I have no Idea how to figure this out
Could you demonstrate this knowledge so someone can help figure out exactly where you went wrong?

Alternatively if you can answer the following questions, you should be well on your way to solving this problem:

What is the impedance of a resistor with a resistance of 40R.

What is the impedance of a capacitor with a reactance of 50R.

How do you combine impedances in series or in parallel?

Could you demonstrate this knowledge so someone can help figure out exactly where you went wrong?

Alternatively if you can answer the following questions, you should be well on your way to solving this problem:

What is the impedance of a resistor with a resistance of 40R.

What is the impedance of a capacitor with a reactance of 50R.

How do you combine impedances in series or in parallel?

well the impedance of a series circuit is found by Zt=SQr(R^2+Xc^2)
You add impedances in series to get the total Impedance. and in paralell you use, 1/((1/Z)+(1/Zn)

I am still stuck though ?

Zryn
Gold Member
well the impedance of a series circuit is found by Zt=SQr(R^2+Xc^2)
An impedance is made up of either a Resistance R and Reactance X in "rectangular form" and looks like Z = R + jX (where X(L) = wL and X(C) = -1/(wC) ).

An impedance can also be made up of a Magnitude M and Angle A in "polar form" and looks like Z = M / A ( where M = SQr(R^2+X^2) and A = arctan (X/R) ).

Are you familiar with both of these forms of impedance?

A 15R resistance will thus have an impedance of 15 + j0 OR 15 / 0, and a 40R capacitor will have an impedance of 0 - j40 OR 40 / -90.

You add impedances in series to get the total Impedance. and in paralell you use, 1/((1/Z)+(1/Zn)
Correct! Can you write out this math starting with the numbers I have shown you?

Last edited:
An impedance is made up of either a Resistance R and Reactance X in "rectangular form" and looks like Z = R + jX (where X(L) = wL and X(C) = -1/(wC) ).

An impedance can also be made up of a Magnitude M and Angle A in "polar form" and looks like Z = M / A ( where M = SQr(R^2+X^2) and A = arctan (X/R) ).

Are you familiar with both of these forms of impedance?

A 15R resistance will thus have an impedance of 15 + j0 OR 15 / 0, and a 40R capacitor will have an impedance of 0 - j40 OR 40 / -90.

Correct! Can you write out this math starting with the numbers I have shown you?
Yes I am familiar with both forms but not as in depth. so please tell me if I am right till this point

35-j0
15-j0
40-j0
0-j50

now where do I go from here ? add all of them together ???

Zryn
Gold Member
The numbers look good. Keep in mind that the sign for the imaginary component (reactive component, the one with the 'j') will be a + for inductors and - for capacitors, but when there is no reactive component then you can have +/- as j * 0 = 0.

You only add them together when they are in series! So the 15R resistor and 50R capacitor could add together to give you 15 + j0 + (0 - j50) = 15 - j50 as the equivalent impedance for that branch.

You could then redraw the circuit to look something like what I have attached. In this case you have two impedance's in parallel and you know how to handle that! (Keep in mind that j^2 = -1 also).

After that, you can redraw the circuit again and you will have one final resistance to combine into your impedance conglomerate.

#### Attachments

• 8.8 KB Views: 369
ok so I got this far from reading some stuff in the web.
15-j50
35-j0
40-j0

now I have to find the Z in parallel 1/(1/z+1/z) now how do I do this with these numbers

ok so using this formula Ztot= Z1 Z2 / (Z1 + Z2 ) my answer was (600-j2000)/55-j50
am I way off ?

Zryn
Gold Member
Treat them like normal algebra with 'j' as an unknown, except that j^2 = -1.

1 / ( 1/(15-j50) + 1/(40) ) = ...

As an alternative, first perhaps figure out algebraically what the parallel combination of two resistances A & B is using that formula. This is a shortcut that is quite useful to know and always works for any two resistance / impedance in parallel. Show your working though, just in case

*Edit: beat me to the punch, your answer is correct, but unsimplified! Z1*Z2 / (Z1 + Z2 ) is the easy shorthand method for 1/(1/z+1/z) that always works with two resistance / impedance in parallel.

Treat them like normal algebra with 'j' as an unknown, except that j^2 = -1.

1 / ( 1/(15-j50) + 1/(40) ) = ...

As an alternative, first perhaps figure out algebraically what the parallel combination of two resistances A & B is using that formula. This is a shortcut that is quite useful to know and always works for any two resistance / impedance in parallel. Show your working though, just in case

*Edit: beat me to the punch, your answer is correct, but unsimplified! Z1*Z2 / (Z1 + Z2 ) is the easy shorthand method for 1/(1/z+1/z) that always works with two resistance / impedance in parallel.
ok finally I am getting somewhere so now with the Z= Z1+Z2 I got the answer of (2225-j3750)/55-j50

is this correct ?

Now do i just cover it into polar form and then just divide in polar form?

i

Zryn
Gold Member
Z = Z1 + Z2 since the last two impedance's are in series, correct.

Z1 = 35 + j0
Z2 = (600 - j2000) / (55 - j50)

Could you write the maths you used to get (2225-j3750)/55-j50?

*Edit: Simple mistakes are easy, and if you don't show a bunch of working on these steps, people may mark you down in exams.

Z = Z1 + Z2 since the last two impedance's are in series, correct.

Z1 = 35 + j0
Z2 = (600 - j2000) / (55 - j50)

Could you write the maths you used to get (2225-j3750)/55-j50?

*Edit: Simple mistakes are easy, and if you don't show a bunch of working on these steps, people may mark you down in exams.

(600 - j2000)*********35
---------------+ ----------------=
(55 - j50)************1

(600 - j2000)*****35(55 - j50)
---------------+ ----------------=
(55 - j50)**********(55 - j50)

Zryn
Gold Member
35 * 55 = 1925
35 * -j50 = -j1750

600 - j2000 + 1925 - j1750 = 2525 - j3750

-1 mark for simple math mistake, given you wrote down what you intended (showed workings), OR -2 marks for the wrong answer with no explanation as to how you got it (did not show workings).

Yes, now you have the number you can simplify.

This can be done by converting both to polar form and treating the Magnitudes like a normal fraction, and the Angles when divided will actually subtract e.g. [X / Y] / [A / B] = X/A / (Y - B) ... hopefully you can tell which '/' I mean as a division and which is the 'angle' signifier!

This can also be done by multiplying the top and bottom of the fraction by the complex conjugate of the denominator. e.g [X + jY] / [A + jB] * [A - jB] / [A - jB]. You will note that the second fraction [A - jB] / [A - jB] = 1, so this is a legal multiplication, and you may know that when any complex number (impedance) is multiplied by its complex conjugate (the same number but with the sign of the 'j' component changed) the result will be a real number, and the fraction will reduce to something along with the form of Z = X + jY.

If you're keen, do it both ways and you should get the same answer.

ok so I am a bit confused so is the answer
2525 - j3750

or

2525 - j3750
---------------
55-j50

Zryn
Gold Member
2525 - j3750
---------------
55-j50

My comments above were in regards to 600 - j2000 + 35 (55 - j50) != 2225 - j3750 and how the expansion on the top line was incorrect (it should be 2525 - j3750). You still have the 55 - j50 on the bottom to contend with.

35 * 55 = 1925
35 * -j50 = -j1750

600 - j2000 + 1925 - j1750 = 2525 - j3750

-1 mark for simple math mistake, given you wrote down what you intended (showed workings), OR -2 marks for the wrong answer with no explanation as to how you got it (did not show workings).

Yes, now you have the number you can simplify.

This can be done by converting both to polar form and treating the Magnitudes like a normal fraction, and the Angles when divided will actually subtract e.g. [X / Y] / [A / B] = X/A / (Y - B) ... hopefully you can tell which '/' I mean as a division and which is the 'angle' signifier!

This can also be done by multiplying the top and bottom of the fraction by the complex conjugate of the denominator. e.g [X + jY] / [A + jB] * [A - jB] / [A - jB]. You will note that the second fraction [A - jB] / [A - jB] = 1, so this is a legal multiplication, and you may know that when any complex number (impedance) is multiplied by its complex conjugate (the same number but with the sign of the 'j' component changed) the result will be a real number, and the fraction will reduce to something along with the form of Z = X + jY.

If you're keen, do it both ways and you should get the same answer.
hey Zryn thank you for all of your help, specially having patience with walking me through this. I was looking for online tutorials for days and could not figure it out. My AC circuit books does not really explain this that good. I have learned more in 2 days in this forum than the last week from my online class.

Zryn
Gold Member
No worries, its difficult for everyone and just a matter of working through it and getting some practice.

Feel free to keep asking more stuff if any of it isn't clear.

2525 - j3750
---------------
55-j50

My comments above were in regards to 600 - j2000 + 35 (55 - j50) != 2225 - j3750 and how the expansion on the top line was incorrect (it should be 2525 - j3750). You still have the 55 - j50 on the bottom to contend with.
ok great I am back in track, thank you so much for the help. I am sure you will be hearing from me much more .