Question

A 120V rms potential at f=1000Hz is applied to a 30 mH inductor, a 40 microF capacitor and a resistor (R). Irms=400mA. What is the value of R?
What is the power factor phase angle phi?
The only equation I know that I could possibly use to find R is Vrms=IrmsR, when I plugged in my data, I got R=300 ohms.
Using R=300 ohms, I rearranged the equation PF = cos phi = R/Z, where Z=sqrt(R^2+[XL-Xc]^2) to solve for R and I got phi=90 degrees. Which makes me suspicious.
The data never specified that the circuit was in resonance, so was my original equation invalid?

Answer 1

300\(\Omega\) is the total impedance Z where \(Z = \sqrt{R^2 + (X_L - X_C)^2)}\)

the circuit will resonate when \(X_L = X_C \). i make that about 45 Hz but do it yourself and do not rely on my owlet packet calcs :p

Answer 2

@radar

Answer 3

I'm not sure what you are saying. Is my equation wrong?

Answer 4

OK

you said V = IR
I am saying V = IZ

Z is as described above

make sense?

Answer 5

are you familiar with phasor diagrams?

that is how you should "add" the equivalent "resistances" of the resistor inductor and capacitor

Answer 6

So I solve for Z using Vrms and Irms and then solve for R by plugging Z, L, C, and f into the long equation? Where XL = 2 pi f L and Xc = 1/2 pi f C ?

Answer 7

I know what a phasor diagram is but I don't know how to use it in this instance.

Answer 8

Imaginary parts: \(X_L = j \omega L\) and \(X_C = -j /\omega C\)
Real part : R

use \(w = 2 \pi f\) to get \(\omega\) from 1000Hz frequency



\(X_{net} = X_L - X_C = j \ ( \omega L - 1/ \omega C )\)

Answer 9

I have never taken calculus. I am assuming that you are saying that Z is the hypotenuse of a triangle where R gives the X value and Xnet gives the Y? So I convert f to omega, say XL = omega L and Xc = 1/omega C, subtract Xc from XL to get Xnet and then use pythagorean theorem to get R?

Answer 10

sounds good.

Answer 11

i would try to be more helpful but i got it in the ear recently for being too helpful so i must tread carefully :p

Answer 12

Thank you, I am just trying to make sure I understand what you are saying.
Is my original equation wrong because it is not in resonance?

Answer 13

to be in resonance \(X_L\) and \(X_C\) cancel each other out

so \(f_{res} = \frac{1}{2 \pi \sqrt{LC}}\)

you can calculate the resonant frequency. i think it's 145Hz but check for yourself

Answer 14

and yes you are right to say that, if it was resonating, you could just say V = IR because at that frequency R = Z.

Answer 15

You do not state whether the 120 V was applied to those components connected in series or parallel Would it make a difference?