# Homework Help: Calculating the value of R for a resistor in an RLC series circuit

1. Nov 30, 2012

### quellcrist

1. An RLC series circuit has C= 4.8mF, L=0.520H, and source voltage amplitude V= 56.0V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0V, what is the vaule of R for the resistor?

2. $I = (ΔVc)/(Xc)$,
$Xc= (1)/(ωc)$,
$ω= (1)/(sqrt(LC))$

resonance frequency = $(1)/(sqrt(LC)) = 20Hz$

$Xc= 10.4$

$Ic = IL = Ir$ in an RLC series circuit

therefore $I= 80V/10.4 = 7.7$

At resonance frequency $X=R$
Z is at a minimum
R is at a minimum, VL and VC cancel each other other becoming 0,
$Z= sqrt((R)^2+(XL-Xc)^2)$

$Vr=sqrt((ΔVtotal)^2-(ΔVL-ΔVc)^2)$

3. I know this question shouldn't be that hard but I have hit a mental block and cannot proceed. I know how to calculate impedance but the value of R isn't given. My question is "How do I calculate impedance when R is not given?" I know the RLC series circuit becomes purely resistive at resonance and that Z is at a minimum and Z=R. Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2012

### aralbrec

At resonance, the impedance of the series C and L is zero, so the voltage of the source will only appear across R.

Maybe doublecheck your Xc calculation. Windows calculator gadget is giving me something different but it could be I misclicked something. I tend to get three different answers on three different attempts when I use this thing. EDIT: I was thrown by 20Hz for 1/√LC, it's 20 rad/s and then I get the same as you.

Last edited: Nov 30, 2012
3. Nov 30, 2012

### quellcrist

So does the value R across the resistor simply equal the value of the resistance across Vc and VL (Vc= 10.4Ω and VL=10.4Ω but since they are 180° out of phase they end up canceling each other out.) ???

I mean to say does R=10.4 ohms?

Last edited: Nov 30, 2012
4. Nov 30, 2012

### aralbrec

The impedances of the inductor and capacitor cancel out so that the full source voltage appears across R only. The voltage across the capacitor will be equal and opposite (ie 180 degrees out of phase) with the voltage across the inductor so that adding them results in zero.

You can see it like so:

Ic = I = Vc jwC

VL = I jwL = Vc jwC jwL = -Vc w2LC = -Vc

because at resonance, w2LC = 1

Or more directly:

Zl+Zc = 0 so (Vc+Vl) = IZ = 0; Vl = -Vc

The series current I flows through R, C and L but generates equal and opposite voltages across C and L so that Vs appears across R only.

This means R = Vs / I which you could also get from I = Vs/Z = Vs/(R + Zl + Zc) = Vs/R

It may seem a bit strange but that's because the problem is the reverse of what is normally asked. Usually you have the resistance from which you can calculate the series current I=Vs/R (since Zc and Zl cancel at resonance). And from there you can find the voltage across the capacitor as Vc=I jwC

Last edited: Nov 30, 2012
5. Nov 30, 2012

### quellcrist

Thank you for clarifying this for me. I noticed this problem was weird for the same reason you stated (usually they give you a value for R and then you proceed from there).

I rechecked and re-did all my calculations. Turns out that R does equal 10.4 ohms. I found this by finding the current across Ic (since Ic=IL=Ir). I took the current and set it equal to 56V/R and solved for R. I ended up with R= 10.4 ohms.