Second order ODE for RLC circuit

Click For Summary
The discussion revolves around deriving the second-order ordinary differential equation (ODE) for an RLC circuit using Kirchhoff's voltage law, leading to the characteristic equation. The roots of the equation are calculated, revealing a dependence on the circuit's resistance (R), capacitance (C), and inductance (L). The expectation is to obtain a solution that includes decaying exponentials and sinusoids, indicating energy dissipation over time. However, the user encounters difficulty in deriving the charge function Q(t) as a function of time, specifically in obtaining the expected damped motion. The outcome hinges on the discriminant of the characteristic equation, which influences the nature of the solution.
icesalmon
Messages
270
Reaction score
13
if I consider a circuit consisting of a capacitor, an inductor and a resistor and using kirchhoffs voltage rule for the circuit i come up with the following

L(Q''(t)) + R(Q'(t)) + (Q(t))/C = 0 I solve for the roots using a characteristic equation of the form
LM2 +MR +(1/C) = 0
solving this for m I obtain
m = [-R/L +/- sqrt((r/l2) - 4(1)(1/LC))]/2
i'm expecting an equation using both decaying exponentials and sinusoids so that the energy will tend towards zero after a long time. But this depends on the values of R, C and L. I'm having trouble moving forward from this point in deriving the equation for the charge on the capacitor as a function of time.
Q(t) = Ae-Btcos(w't+θ)
B = R/2L
 
Last edited:
Engineering news on Phys.org
so obviously the characteristic equation doesn't directly lead to that damped charge motion then, it depends on the discriminant. I thought I was SUPPOSED to get the decaying exponential out of it. Anyways, thanks.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
6K
Replies
14
Views
2K
Replies
44
Views
6K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K