Electronic – need to convert impedance to admittance in this problem

circuit analysisimpedance

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?

Best Answer

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.

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