anonymous
  • anonymous
A group that I'm in for physics is planning on presenting the math of an RLC circuit and the voltage across the capacitor at some time t. I did research on the math for the RLC circuit and I wanted to see if you guys had anything to add:
Physics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
To begin, we should note some initial conditions. At t=0, the voltage across the circuit is equal to that of the voltage source and thus the voltage "across" the capacitor is also that of the voltage source. Additionally, the inductor current at t=0 is 0A. Therefore: \(V_c(0)=V_s\) \(I_L(0)=0\text{A}\) We can also note that the inductor current is the derivative of the capacitor voltage times the capacitance: \(I_L(t)=CV_c'(t) \rightarrow V_c'(t)=\frac{1}{C}I_L(t)\) Therefore: \(V_c'(0)=0 \text{V/s}\)
anonymous
  • anonymous
Now we can apply `Kirchoff's Voltage Law:` \(V_R+V_L+V_c=0\) \(IR+LI'+V_c=0\) \(\Omega I_L(t)+\text{H}I_L'(t)+V_c(t)=0\) Where \(\Omega\) is the resistance and \(\text{H}\) is the inductance We also note that the current is the rate of change of the charge (Q) with respect to time. Therefore: \(I_L(t)=CV_c'(t)\) \(I_L'(t)=CV_c''(t)\) Plugging this into our differential equation from Kirchoff's Law of Voltage, we can express the differential equation all in terms of voltage of the capacitor at some particular time t. Thus: \(\text{(H)(F)}V_c''(t)+(\Omega) (\text{F})V_c'(t)+V_c(t)=0\) Where \(\Omega\) is the resistance, \(\text{H}\) is the inductance, and \(\text{F}\) is the capacitance.

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anonymous
  • anonymous
In order to solve this second order homogeneous differential equation, we can re-express the equation as a characteristic equation given by: \(\text{HF}r^2+\Omega \text{F}r+1=0\) Now that we have a quadratic equation, we can apply the quadratic formula obtaining the values of: \[\large r_1,r_2=\frac{ -(\Omega \text{F}) \pm \sqrt{(\Omega \text{F})^2-4(\text{HF})} }{ 2(\text{HF}) }\] This will give us the general solution: \(\huge V_c(t)=K_1e^{-r_1t}+K_2e^{-r_2t}\) And just for the sake of making connections, taking the derivative will give us the inductor current: \(\huge I_L(t)=-r_1K_1e^{-r_1t}-r_2K_2e^{-r_2t}\) We can then solve for the constants using the initial conditions that we came up with from the beginning!
anonymous
  • anonymous
The last equation for the inductor current needs to be multiplied by the capacitance since \(I_L(t)=CV'(t)\)
anonymous
  • anonymous
@Michele_Laino @IrishBoy123 What do you guys think? Is there any other math that you guys think would be good to include for the presentation?
anonymous
  • anonymous
`Some other notes:` I also did not include impedance nor resonance. We haven't exactly covered that material yet in my physics course. Actually, most of what I just did was beyond the scope of the course. But my group really wanted to do this so we're giving it a big shot. This is information and math that I gathered from numerous papers and videos. We wanted to keep it in a very very very simple scenario.
IrishBoy123
  • IrishBoy123
as it's a presentation, you might wish to find something good on youtube. i have tried walter lewin and lasseviren1, 2 of my faves, but nothing yet. it's 1 thing knowing your stuff, it's another getting that across.
anonymous
  • anonymous
@IrishBoy123 I'm familiar with Walter Lewin! I watched his lectures for kinematics and forces. I didn't know he has public videos for electromagnetism?
IrishBoy123
  • IrishBoy123
https://www.youtube.com/channel/UCiEHVhv0SBMpP75JbzJShqw not the only collection!
IrishBoy123
  • IrishBoy123
https://www.youtube.com/channel/UCliSRiiRVQuDfgxI_QN_Fmw/playlists
anonymous
  • anonymous
@IrishBoy123 I'm not sure how I missed those. . . Thanks so much!!
IrishBoy123
  • IrishBoy123
as i understand it, MIT did their best to rid the airwaves of him, but his loyal following had made copies and put a lot of it back up. i am sure there are other collections too!
Michele_Laino
  • Michele_Laino
I think that the transient state, of your circuit can be modeled by this ODE: \[\Large \frac{{{d^2}I}}{{d{t^2}}} + \left( {\frac{R}{L}} \right)\frac{{dI}}{{dt}} + \left( {\frac{1}{{LC}}} \right)I = {E_0}\cos \left( {\omega t} \right)\] where \(\Large E_0 \cos(\omega t)\), is the voltage provided, by the source power of the circuit
anonymous
  • anonymous
@Michele_Laino They way we're modeling our RLC circuit is that the capacitor is first being charged and then we're observing the decay of the voltage after time. After t=0, there is no voltage source being constantly applied.
Michele_Laino
  • Michele_Laino
If I call with \(V\) the voltage across the capacitor, and there are no external voltage sources, then I can write: \[\Large I = - \frac{{dQ}}{{dt}},\quad Q = CV,\quad V = L\frac{{dI}}{{dt}} + RI\] where \(\Large Q\) is the charge of the capacitor. Please note that the first equation, expresses the conservation of charge. Now from the first and second equations, I get: \[\Large I = - C\frac{{dV}}{{dt}}\] Substituting into the third equation, I can write: \[\Large \frac{{{d^2}V}}{{d{t^2}}} + \left( {\frac{R}{L}} \right)\frac{{dV}}{{dt}} + \left( {\frac{1}{{LC}}} \right)V = 0\] Please solve that ODE, with the subsequent initial conditions: \[\Large V\left( 0 \right) = {V_0},\quad \left( {\frac{{dV}}{{dt}}} \right)\left( 0 \right) = 0\] since at \(t=0\) we have no current inside the circuit

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