# Q of RLC circuit, confused with provided answer

1. Sep 4, 2016

### fahraynk

1. The problem statement, all variables and given/known data
The main question is to derive the Q of the series RLC circuit. This book shows the whole solution but I have a question about a part of their answer. Basically I want to know how $\frac{1}{2}CV_{Cmax}^2 = \frac{1}{2}C(\frac{i_{max}^2}{wC}^2)$
to me it looks like they are saying $V=i_{max}/Z$ instead of $V=IZ$.

Also, at the end why is the energy dissipated per cycle $W_d=\frac{I_{max}^2R\pi}{w}$.
I think $I^2R\frac{2\pi}{w}$ should be dissipated the energy dissipated per cylce. Why are they leaving it as $I^2R\frac{\pi}{w}$ so that the energy dissipated per cycle is divided by 2? Also then they multiply Q by $2\pi$ at the end which I also found to be strange. Can anyone please shed some light?
2. Relevant equations

3. The attempt at a solution
The book gives the solution:

In the time domain, the instantaneous stored energy in the circuit is given by : $W_s=\frac{1}{2}Li^2 + \frac{q^2}{2C}$
For a maximum: $\frac{dW_s}{dt} = Li\frac{di}{dt}+\frac{q}{c}\frac{dq}{dt} = i(L\frac{di}{dt}+\frac{q}{C})=i(V_l+V_C)=0$
Thus, the maximum stored energy is $W_s$ at i=0 or $W_s$ at $V_L+V_C=0$, whichever is larger. Now the capacitor voltage, and therefore the charge, lags the current by $90^\circ$; hence, i=0 implies $q=\pm Q_{max}$ and :
$W_{s|i=0} = \frac{Q_{max}^2}{2C} = \frac{1}{2}CV_{Cmax}^2 = \frac{1}{2}C(\frac{I_{max}}{wC})^2=\frac{I_{max}^2}{2Cw^2}$

On the other hand, $V_L + V_C=0$ implies $V_L=V_C=0$ and $i=\pm I_{max}$ so that $W_{s|V_L+V_C=0}=\frac{1}{2}LI_{max}^2$
It follows that
$W_{s-max}=\frac{I_{max}^2}{2Cw^2}$ for $(w\leq w_0)$ and
$W_{s-max}=\frac{LI_{max}^2}{2}$ for $(w \geq w_0)$

The energy dissipated per cycle (in the resistor) is $W_d=\frac{I_{max}^2R\pi}{w}$. Consequently,
$Q = 2\pi\frac{W_{smax}}{W_d} = \frac{1}{wCR}, (w\leq w_0)$ and $\frac{wL}{R}, (w \geq w_0)$

2. Sep 5, 2016

### rude man

Z = 1/ωC for a capacitor so what they wrote (& which you copied incorrectly BTW) is correct.
If you integrate Rimax2sin2(ωt) dt over 1 period = 2π/ω you will get what they state, which is πRimax2/ω.