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© The scientific sentence. 2010

Series RLC circuits




Capacitors and inductors are energy-storage devices. A capacitor stores electric energy and an inductor stores magnetic energy. Resistors just cause energy to be dissipated as a heat.



1. LC circuit: Oscillations

we examine the behavior a real circuit (resistance is significant), which contains an inductance L, a capacitance C, and a resistor R, an RLC circuit.



The capacitor in the circuit is charged by an external battery and then the battery is taken away.

When the switch is closed (position 1), the capacitor will begin to discharge through the inductor and the resistor.

During the discharging, there is a current I(t) crossing the circuit and charge Q(t) on the plates of the capacitor.

The current I is the rate at which charge is transferred from a plate to the other. In the figure, the sense of the current (positive) is counterclockwise. The potential difference between the plates is positive.
The potential through the inductor decreases then the difference between the terminals of the inductor is negative.
The potential difference between the terminals of the resistor is negative.

Kirchhoff's loop rule gives:

(Vb - Vc) + (Va - Vb) + (Vc - Va) = 0 . That is

+ Q/C - L (dI/dt) - R I = 0     (Eq. 1)

We have I = ± dQ/dt

During the discharging of the capacitor, Q decreases, hence dQ is negative so is I = dQ/dt and d(dQ/dt)/dt = d2Q/dt2.

Therefore

I = - dQ/dt. The equation (Eq. 1) becomes:

Q/C + L d2Q/dt2 + R I = 0 . Rearranging, we find

d2Q/dt2 + (1/LC)Q + (R/L) dQ/dt = 0


Equation of RLC oscillation circuit:

d2Q/dt2 + (R/L) dQ/dt + (1/LC)Q = 0



2. Expression of the charge Q(t)

This differential equation is the same as the differential equation of a damped harmonic oscillator, like the mass-spring with friction system. The mass-spring-friction harmonic oscillator provides a mechanical analog to the RLC circuit

Using the solutions of a second order differential equations, we have:

a) Δ = (R/L)2 - 4/LC > 0:
Q(t) = C1 exp{r1t} + C2 exp{r2t}
With
r1 = (- R/L + Δ1/2)/2
r2 = (- R/L - Δ1/2)/2

b) Δ = (R/L)2 - 4/LC = 0:
Q(t) = = C1 exp{r t} + C2 t exp{r t}
With
r = - R/2L

c) Δ = (R/L)2 - 4/LC < 0:
Q(t)= C1 exp{r1t} + C2 exp{r2t} = exp{αt} (c1 cos βt + c2 sin βt)

With
α = - R/2L
β = (- Δ)1/2/2

The electromagnetic energy of an LC circuit is constant because its resistance is negligible. The electromagnetic energy of the RLC circuit is not constant. It decreases with time because energy is dissipated as heat from the resistor.

Among the three cases given by the value of Δ the real solution is the third case in which Δ < 0. In an RLC circuit, the charge on the capacitor and the current in the circuit will tend to approach zero as time goes by, and the oscillations in the charge and current my occur while they are dying out.



RLC under-damped circuit: Oscillating charge with damping
ωo = [(1/LC)]1/2, τ = 2L/R,
damped angular frequency: ωd = [(1/LC) - (R/2L)2]1/2 =
o2 - (1/τ)2]1/2


Q(t) = Qm exp{- t/τ} cos (ωdt + φ)



If R > (4L/C)1/2, the circuit is over-damped
If R = (4L/C)1/2, the circuit is critically over-damped
If R < (4L/C)1/2, the circuit is under-damped



3. Example

R = 200.00 Ω , L = 1.00 mH, and C = 1.00 nF

4L/C = 4 x - 3 / 10- 9 = 4 x 106
(4L/C)1/2 = 2 x 103 200 < 2000, hence
R < (4L/C)1/2, the circuit is under-damped

τ = 2L/R = 2 x 10- 3/200 = 10- 5 /second

R/2L = 1/τ = 105 second

1/LC = 1/1 x 10- 3 x 10- 9 = 1012

ωo = [(1/LC)]1/2 = 106

ωd = [(1/LC) - (R/2L)2]1/2 = [1012 - 1010 ]1/2 =
105 [99]1/2 ≈ 106/second

For the initial conditions such that
Qm = Qo, and φ = 0, we have:

Q(t) = Qo exp{- 105 t} cos (106t )

Q(t µs) = Qo exp{- t/10} cos(t)






 


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