Phasor analysis is convenient with AC circuits when looking for simple quantities like impedance. After converting capacitive and inductive quantities to reactances, I then treat them like resistors. It's a convenient way to make AC circuit analysis relatively simple. However, I found out today that I may have been doing it wrong, and I need help understanding why. For example, the circuit below: enter image description here

To find the impedance in this circuit, I would first find XL, then Xc:

\begin{equation*} {X_L}=2{\pi}fL=52.779{\quad}{\quad}{X_C}=-{\frac{1}{2{\pi}fC}}=-16.579 \end{equation*}

Then I would treat the reactances as resistances, in this case using the simple parallel resistor formula:

\begin{equation*} Z = \frac{1}{\frac{1}{R}+\frac{1}{X_L}+\frac{1}{X_C}} \end{equation*}

The result I get this way is Z = 24.75Ω

In studying for the FE exam, I've come across a method which relies on the impedance triangle:

                                     enter image description here

Through some trig on the relations on this diagram, the formula for impedance in a parallel AC circuit with reactive components becomes:

\begin{equation*} Z = \frac{1}{ \sqrt{\left(\frac{1}{R}\right)^2+\left(\frac{1}{X_L}+\frac{1}{X_C}\right)^2} } \end{equation*}

The result I get this way is Z = 24.17Ω

As you can see, they both produce a very similar result, but not the same. One is an approximation, and I have a suspicion based on the impedance triangle that the latter method is correct.

Can someone please elaborate as to why that is or isn't the case?

Thank you.


Oops. They are the same. I was using a calculator and was failing to insert the imaginary 'i' (or 'j', depending on your preference) in the reactances when using the first method. After doing so and extracting the magnitude of the complex result, the answers turn out identically.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.