# Half power frequency divides by sqrt(2) instead of 2 why?

1. Aug 16, 2016

### fahraynk

• Member advised to use the homework template for posts in the homework sections of PF.
1. The problem statement, all variables and given/known data
Problem is to find the half power frequencies of a RLC circuit.
Z = √(R2 + (WL-1/WC)2)

2. Relevant equations

Power(P) = V2/Z
Pmax = V2/R <---because Z min = R

3. The attempt at a solution
P/2 = V2/2R = V2/Z
Z=2R
(R2 + (WL-1/WC)2) = 4R2
(WL-1/WC)2 - 3R2 = 0
W2-3WR/L-1/LC = 0
Solve for w to get half power frequencies.

Book gets : W2+-WR/L -1/LC = 0

Cant figure out why Pmax is divided by sqrt(2) instead of 2

Last edited: Aug 16, 2016
2. Aug 16, 2016

### Twigg

Did you make sure to convert from amplitudes to RMS values of voltage and/or current?

3. Aug 16, 2016

### fahraynk

To convert V2/R to VRMS2/R you divide V by 1/√2
It would square, and become power = V2/2R
P/2 = V2/4R
Trying to get P/2 = V2/R√2
I dunno... interesting tip though. 10 min messing with it and I am still not seeing it though.

4. Aug 16, 2016

### Twigg

The issue isn't RMS/peak conversions, sorry for the confusion. I missed this the first time I read your post, but $P \neq \frac{V^{2}}{Z}$. The correct relationship is $P = RI^{2} = R\frac{V^{2}}{Z^{2}} = R\frac{V^{2}}{R^{2} + (X_{L} - X_{C})^{2}}$. I can't tell what the book is doing from the information given though. Can you show their steps and refer us to the book and page?

5. Aug 16, 2016

### cnh1995

Is there resonance involved? Is that the complete problem statement? Check out this thread if it is any help.
https://www.physicsforums.com/posts/5372988/

6. Aug 17, 2016

### Staff: Mentor

Only if Z is a pure resistance AND the full voltage V is across that pure resistance Z.

For your series RLC circuit, at resonance, Z is purely real, and equal to R Ohms.
Power = power in the resistance = I2⋅R

Off resonance, at a frequency where X = R, the value of Z has increased to magnitude $\sqrt{2}{R}$ so current is reduced by a factor $\frac 1 {\sqrt2}$
At this reduced current, what now is the power in the circuit's series resistance?

7. Aug 17, 2016

### fahraynk

Schaums outlines Electric Circuits 5th edition Q12.5

Here is the complete information :
For the series RLC circuit shown in fig. 12-36, find the resonant frequency $W_{0}=2πf_{0}$. Also obtain the half power frequencies and the bandwidth β. V(w), R=100, L = .5H, C=0.4u (V(w) is a variable. I dont care about the numbers I only care about the derivation because I want to understand this...)

At resonance, $Z_{in}(w)=R$ and $w_{0} = \frac{1}{\sqrt{LC}}$
$w_{0} = \frac{1}{\sqrt{0.5(0.4x10^-6)}}=2236.1rads$ $f_{0}=\frac{w_{0}}{2π}=355.9 Hz$

The power formula $P = I_{eff}^{2}R = \frac{V_{eff}R}{|Z_{in}^{2}|}$ shows that $P_{max} = \frac{V_{eff}^{2}}{R}$, which is achieved at $w=w_{0}$, and that $P= \frac{1}{2}P_{max}$ when $|Z_{in}|^{2}=2R^{2}$; that is, when $wL - \frac{1}{WC} = +-R$ or $w^{2} +- \frac{R}{L}w - \frac{1}{LC} = 0$

corresponding to the upper sign, there is a single real positive root $w_{h} = \frac{R}{2L} + \sqrt{\frac{R}{2L}^{2} + \frac{1}{LC}} = 2338.3$ rad/s or $f_{h} = 372.1Hz$

and corresponding to the lower sign, the single real positive root $w_{l} = -\frac{R}{2L} + \sqrt{\frac{R}{2L}^{2} + \frac{1}{LC}} = 2138.3$ rad/s or $f_{l}=340.3$ Hz

Another thing I dont get is why the +- in the quadratic formula is $+-B+\frac{\sqrt{B^{2}-4AC}}{2A}$ instead of $B+-\frac{\sqrt{B^{2}-4AC}}{2A}$ I know it has to do with that +- R but I guess it just becomes arbitrary when you have a +- B term? even though doesnt it change the answers? Still, my main question is figuring out once and for all that blasted $\sqrt{2}$ ! This took like 25 minutes to type up! So ill read the other replys after a break!

Last edited: Aug 17, 2016
8. Aug 19, 2016

### BvU

Re: "why the +- in the quadratic formula is $\ \ \displaystyle \frac {\pm B+\sqrt{B^{2}-4AC}}{2A}$."

You were looking at $|Z_{in}|^{2}=2R^{2}$ which is satisfied if $w^{2} +\frac{R}{L}w - \frac{1}{LC} = 0$ and also if $w^{2} - \frac{R}{L}w - \frac{1}{LC} = 0$.

That's two quadratic equations. If there is a resonance, each has two real solutions, one positive $\omega$ and one negative. You only want the positive.

9. Aug 22, 2016

### fahraynk

Interesting. Is there a theory behind why to only take the positive roots that I can google?

10. Aug 22, 2016

### fahraynk

When you say "Only if Z is a pure resistance AND the full voltage V is across that pure resistance Z.", Can you elaborate a bit... I am trying to find the Z where $P=\frac{1}{2}Pmax$. Max power is where the impedance is pure resistance and the frequency is resonance. So if power = $\frac{V^2}{R}$ at resonance, there should be a answer for Z where that power is 1/2. So voltage, resistance, everything is known but Z in the equation $\frac{V^2}{2R} = \frac{V^2}{Z}$. So if I am trying to find the value for Z where it would be equal to half of pure resistance. Can you elaborate a bit on why thats wrong because I am staring blankly lol

Then you said : "Off resonance, at a frequency where X = R" <--- (do you mean AT resonance, X = R ?), "the value of Z has increased to magnitude $\sqrt{2}R$"

At resonance, $Z=\sqrt{R^2 + (WL-\frac{1}{WC})^2} = R^2$. So, how does $Z=R^2$ become $\sqrt{2}R$ ?

I see though if I take for face value and plug $P=IV = \frac{I}{\sqrt{2}}V = \frac{V^2}{\sqrt{2}R}$ it works out...
But at Vmax P=ImaxVmax, Z=R. Why does $I = \frac{1}{\sqrt{2}}Imax$? One of my assumptions is probably wrong I know, since I am getting the problem wrong.

11. Aug 22, 2016

### BvU

No. Your driving frequency is positive, so the response frequency is positive as well.

12. Aug 22, 2016

### Staff: Mentor

I mean move away from resonance until you find a frequency where reactance X equals resistance R, in magnitude. As you know, precisely at resonance, in a series circuit the total reactance = 0 ohms.

The equation $\require{enclose}\enclose{box}[mathcolor="white",mathbackground="yellow"]{\color{black}{power=\frac{V^2}Z}}$ is correct only where Z is resistance (no part of Z can be reactance in that equation), and V must be the voltage measured across that resistance (and not across the whole RLC circuit).

13. Aug 23, 2016

### ehild

Your power formula is wrong. Z is the magnitude of the impedance. The power is P=V I cos(θ), where V and I mean the rms values of voltage and current, and θ is the phase lag of the current behind the voltage. Now, you know that I=V/Z, and the resistance is the real part of the impedance, R=Zcos(θ). Therefore $P=\left(\frac{V^2}{Z}\right)\left(\frac{R}{Z}\right)=\frac{V^2R}{Z^2}$
You get the same formula by using the fact that AC power is dissipated only on resistors. The power is P=I2R, and I=V/Z, so $P=\frac {V^2R}{Z^2}$. At resonance, Z=R, so Pres=V2/R.

14. Aug 27, 2016

### fahraynk

God thanks dood I am finally seeing it now.