# Using complex impedance

Any help getting the answer shown here?

I have tried adding the complex impedance of L and R in series and then adding that to C in parallel. I'm not super sure about that last step, and then attempting to simplify that. I get annoyingly close, having the two terms shown but they are the denominator with a bunch of gibberish on top.

Am I correct in the assumption that the total complex impedance between Vin and V out is the sum in series of L and R added in parallel to C? Do I then just have B/Z = A?

Thank you

• Show your work. Maybe someone can check the issue from what you have. But you have to be specific what help you need, you don't have an answerable question about your homework. May 21 at 17:10

$$\frac{\frac{1}{j\omega C}}{\frac{1}{j\omega C}+R+j\omega L}$$ $$\frac{1}{1+jRC\omega-\omega^{2}LC}$$ Take the absolute value of this to get $$\frac{1}{\sqrt{(1-\omega^{2}LC)^{2}+(\omega RC)^2}}$$ Use voltage division.

• Thank you very much, can you clarify the physics behind the top expression? May 21 at 17:44
• You use voltage division. I also used phasor analysis. May 21 at 17:45
• The current through a capacitor is proportional to the change of the voltage across it. The voltage across an inductor is proportional to the change of the current through it. The voltage across a resistor is proportional to the current through it. The quantity that is the ratio between the voltage and current is called in general the impedance. May 21 at 17:46
• Complex exponentials are used, as they are used to build the sinusoidal functions. May 21 at 17:48

This should help give you a few hints (from my website): -

I'm sure if I'd have derived it without substituting for 1/LC and RCwn you'd see the expected answer in your question much more simply.

Am I correct in the assumption that the total complex impedance between Vin and V out is the sum in series of L and R added in parallel to C?

That is the complex impedance between the upper connection for $$\V_{in}\$$ and $$\V_{out}\$$. However that isn't useful to know here.

can you clarify the physics behind the top expression?

There is an unstated assumption that $$\V_{out}\$$ is being measured with an ideal voltmeter with infinite input resistance. So the impedance seen by the voltage source $$\V_{in}\$$ is the impedance of the capacitor, resistor and inductor in series, i.e. just the sum of their individual impedances:

$$\frac{1}{j\omega C}+R+j\omega L \ \ \Omega$$

(This is the denominator (bottom line) of Tahmid's first expression.)

Now we know the impedance of the LRC series combination we can use Ohm's Law ($$\I=V/Z\$$) to work out the current around the circuit:

$$\frac{V_{in}}{\frac{1}{j\omega C}+R+j\omega L} \ A$$

$$\V_{out}\$$ is the voltage across the capacitor. From Ohm's Law again this is equal to the impedance of the capacitor multiplied by the current through it ($$\V =ZI\$$):

$$V_{out} = {\frac{1}{j\omega C}} \cdot \frac{V_{in}}{\frac{1}{j\omega C}+R+j\omega L} \ V$$

Re-arranging a couple of terms we get:

$$\frac{V_{out}}{V_{in}} = \frac{\frac{1}{j\omega C}}{\frac{1}{j\omega C}+R+j\omega L}$$

We also note that the answer only needs to be expressed as $$\B/A\$$ where $$\B\$$ and $$\A\$$ represent the magnitude of $$\V_{out}\$$ and $$\V_{in}\$$ respectively (specifically their peak values). So as we go through the analysis we can later take the absolute magnitude and ignore the phase argument. Hence we get:

$$\frac {B}{A} = {\Bigg\lvert \frac {V_{out}}{V_{in}} \Bigg\rvert} = \Bigg\lvert \frac{\frac{1}{j\omega C}}{\frac{1}{j\omega C}+R+j\omega L }\Bigg\rvert$$

So that covers the physics background and the rest is algebraic manipulation. The above line is the first expression in Tahmid Hassan's answer so you can follow the rest there.

Hat tip to Tahmid Hassan for typesetting the algebra expressions, which I've re-used here.