𝐼1 = 18 A, 𝐼2 = 15 A , 𝐼 = 30 A and R2 = 4 Ω. determine 𝑅1 and 𝑋L .

This is the picture of the circuit:

Image 1

And this my solution all the way to the point where I have uncertainties:

Image 2

Now I can solve this in two ways:

  1. I can solve it without converting impedance to admittance (gives the wrong answer)

  2. Or I can solve it with converting impedance to admittance (gives the correct answer)

Image 3

My question is why will I get the wrong answer when not converting impedance into admittance, it seems mathematically correct, but it is the wrong solution, can someone explain?

  • \$\begingroup\$ Can you try re-writing your attempted solution(s) in MathJax? Trying to read your faded pencil marks is making it more trouble than necessary to try to understand your problem. \$\endgroup\$
    – The Photon
    Jul 24, 2019 at 22:03
  • \$\begingroup\$ Yeah. Some of your 9s look exactly like 4s as well. I managed to get through it but half my time was spent trying to read the writing. \$\endgroup\$
    – DKNguyen
    Jul 24, 2019 at 22:42
  • \$\begingroup\$ Ok, im gonna try to fix that, is there any tutorial on how to write in MathJax? \$\endgroup\$ Jul 24, 2019 at 22:56
  • \$\begingroup\$ It's too bad you don't have high enough permissions to try and edit my answer so you can view the mathjax code. I pretty much used everything you need to use normally. \$ to start and stop. {} to group command arguments together, and each command is preceded by a \. Common commands are \times, \frac{}{}, \sqrt{}, \angle{}. Use your preview when you type. It will let you know whether you did it right or not. Underscore for subscripts: X_L or X_{L}. The {} is used if your subscript is multiple characters. \$\endgroup\$
    – DKNguyen
    Jul 24, 2019 at 22:59
  • \$\begingroup\$ Can you send me code , I don't know if you can send messages on stackExchange \$\endgroup\$ Jul 24, 2019 at 23:01

2 Answers 2


I think there are two things at work here:

1. You are not taking into account that \$R_1\$ and \$X_L\$ are in parallel and not in series.

When you write \$ Z_1 = R_1 + X_L \$, you are stating that they are in series, not in parallel.

The lumped impedance of \$Z_1\$ does have a form: \$Z_1 = Re(Z_1) + Im(Z_1)j\$

But these real and imaginary terms are not \$R_1\$ or \$X_L\$

You should have written \$ Z_1 = R_1 || jX_L \$ and expanded that out. Then your math will accurately reflect that they are in parallel. The real terms in this expanded expression will be \$R_1\$ and the imaginary terms will be \$X_L\$.

2. You cannot just invert the real and imaginary components of a complex impedance/admittance to find the admittance/impedance.

For example:

\$ Z = 2 \angle{45} = \sqrt{2} + j\sqrt{2}\$

\$ Y = \frac{1}{Z} = \frac{1}{2}\angle{-45} = \frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}} \$

We agree on those right?

But then in your second solution, you try and calculate Y by individually finding the reciprocal of the real and imaginary components of Z:

\$Y_{wrong} = \frac{1}{Re(Z)} + j\frac{1}{Im(Z)} =\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}\$

Or maybe:

\$Y_{wrong} = \frac{1}{Re(Z)} + \frac{1}{Im(Z)j} =\frac{1}{\sqrt{2}} - j\frac{1}{\sqrt{2}}\$

if you thought that the \$j\$ should be included in the reciprocal.

You might start feeling that something is wrong and doesn't make sense here because it's inconsistent. Look at the \$j\$. It only makes sense that you would have to include it as part of the reciprocal so it ends up in the denominator (or in the numerator as a \$-j\$)...but at the same time, if you do that then it's obviously wrong since there is an inductance, not a capacitance so \$-j\$ is obviously wrong. It doesn't feel right both ways and that's because it is wrong.

Either way, obviously \$Y \ne Y_{wrong}\$ so it does not work. It does not work because the real and imaginary components are tied together and so you can't just break them apart and invert them individually.

Here's an interesting exercise: What happens if you try to find the admittance of a resistor \$R\$ by doing what you tried to do, except now think of it as \$ R + 0j \$?

You get a divide by zero! We both know that in the end you do get \$ Y = \frac{1}{R}\$ but math to actually get there is is different. The reciprocal of a the singular component (real or imaginary, as long as there is only one) is just a shortcut that only works in that circumstance. It cannot be applied to complex numbers in general.

So I think your second solution is wrong too. You just happened to accidentally account for the parallel \$ R_1\$ and \$X_L \$ when you incorrectly tried to calculate Y by individually inverting the real and imaginary components of Z.

  • \$\begingroup\$ Thank you! that makes sense I should've used Xp= U<0 / I <-49.46 -> 1 / R + 1 / X_L = ( 2.17 + 2.53j) ^-1 and then I should was supposed to do 1 / R =0.195 and 1 / xl = - 0.223 j I's that correct \$\endgroup\$ Jul 24, 2019 at 23:24
  • \$\begingroup\$ Note that the reason you get the correct answer by reciprocating the \$R_1\$ and \$X_L\$ is that they are individual components with only a real or imaginary part, not both. If you tried to do that with the lump sum impedance \$Z_1\$, your answer would be wrong. So be wary when changing admittance to impedance. If it has both real and imaginary parts, change to phasor form and invert it there. \$\endgroup\$
    – DKNguyen
    Jul 24, 2019 at 23:28
  • \$\begingroup\$ Ok i will thanks \$\endgroup\$ Jul 24, 2019 at 23:31

Cannot possibly be 15 A, 18 A and 30 A - that would require \$\small I_1\$ and \$\small I_2\$ to be out of phase by \$\small 147^o\$, which is not possible. Maximum phase angle between these is \$\small <90^o\$.


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