Handout Transients in the LRC Circuit PDF

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Handout Transients in the LRC Circuit...

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George Mason University Physics 263 Transients in the LRC Circuit 1

Introduction

When the applied voltage in an LRC circuit is a rapid step from one constant level to another, a short-term “transient” current in produced. This report gives a derivation of the equations for the current and various voltages in the circuit for this case. For reference, here is the circuit diagram:

Figure 1: The LRC series circuit. The circuit equation is

di q + Ri + = Vs C dt where Vs is the voltage of the source. For the special case which we are considering here, Vs is a constant, so we simplify the solution with the substitution L

q = Q + CVs Then the equation becomes, in terms of Q L

d2 Q dQ Q +R + =0 2 C dt dt

(1)

We use the common technique of assuming a “trial” solution. In this case it is: Q = Aest

(2)

When this is substituted into equation 1, and common terms are cancelled out, the result is Ls2 + Rs + 1/C = 0

This implies that the trial solution has the correct form.1 This equation can be solved for s, with the result s 1 R2 R − (3) s=− ± 2 4L LC 2L which can be re-written s 1 R2 R ±j − 2 s=− 2L LC 4L √ where j = −1.

1.1

Defining the constants α, ω, and ω0

The above equation can be written, in terms of some new constants α and ω , s = α ± jω and with the “natural” angular frequency defined via √ ω0 = 1/ LC then ω can be written: ω=

q

ω02 − α2

We can see that with ω0 > α, i.e. the case in which the resistance, R, is relatively small, that since Q will have the form Q = Ae−αte±j ωt the solution will be oscillatory, since ejωt = cos ωt + j sin ωt

2

Solution for the Oscillatory Case

So let’s assume a real, damped oscillatory solution of the form: Q = Ae−αt sin (ωt + δ)

(4)

Or, reverting back to q, the charge on the capacitor, q = CVs + Ae−αt sin (ωt + δ)

(5)

This equation has 2 constants, A and δ, which must be determined by the initial conditions. This is as expected, since the original differential equation is 2nd order. Differentiating q with respect to time gives the current i: i = Ae−αt(ω cos (ωt + δ) − α sin (ωt + δ))

(6)

Note that the current still oscillates “sinusoidally”, since the two terms in the sum both have the same frequency, ω . 1 We ignore here the problem of “uniqueness” i.e. whether or not there is another, different, correct form.

2

2.1

Applying the Initial Conditions

Let’s assume that the voltage jumps from −V0 to +V0 at time t = 0. These conditions can apply when driving an LRC circuit with a rectangular waveform from a signal generator, if the period of the waveform is much greater than the decay time, τ = 1/α. Just at time 0, no current has yet developed; the inductance prevents a sudden change. And the charge, q, on the capacitor is still −CV 0 . So the initial (t = 0) conditions are:

Using equation 7 in equation 6 gives

i=0

(7)

q = −CV0

(8)

ω cos δ − α sin δ = 0 or

ω α Using some trigonometric identities we can also write this as: tan δ =

sin δ =

s

cos δ =

s

or

ω2 α2 + ω 2 α2 α2 + ω 2

(9)

(10)

(11)

From equation 5 and 8, we also get −CV0 = CV0 + A sin δ This gives A=

−2CV0 sin δ

or A = −2CV0

s

α2 + ω 2 ω2

So the solution for q is q = CV0 − 2CV0

s

α2 + ω 2 −αt e sin (ωt + δ) ω2

(12)

Differentiating gives the current: i = −2CV0

s

α2 + ω 2 −αt e (ω cos (ωt + δ) − α sin (ωt + δ)) ω2

The equation for q may be simplified first rewriting it as: q = CV0 − (2CV0 / sin δ)e−αt sin (ωt + δ) 3

(13)

and then expanding the last term sin (ωt + δ) = sin ωt cos δ + cos ωt sin δ Using equation 9 then yields α sin ωt) (14) ω Making the same kind of expansion of the cos and sin terms in equation 13 results in the cancellation of the cos terms, giving the result q = CV0 − 2CV0 e−αt (cos ωt +

i = +2CV0

α2 + ω 2 −αt e sin ωt ω

(15)

For some typical parameters, here is a plot of V R = iR and VC = q/C vs. time:

Figure 2: Underdamped solution for voltage across R and C vs. time for -V 0 to + V0 step, for V0 = 5V . Note that sometimes VC exceeds V0 !

3 3.1

Non-oscillatory Solutions Overdamping

What happens when R is large enough to make α > ω? When that occurs, the solutions of equation 2 can be written: q s1 = −α + α2 − ω02 s2 = −α −

4

q

α2 − ω02

Note that both s1 and s2 are < 0. Instead of equation 4 the solution will look like: Q = A1 es1 t + A2 es2 t

(16)

i = s1 A1 es1 t + s2 A2 es2 t

(17)

So The initial conditions, given by equations 7 and 8 then allow determination of the constants A1 and A2 . After some algebra, the result for q is es1 t es2 t + 1 − ss21 1 − ss21

q = CV0 − 2CV0

3.2

!

Critical Damping

Finally, there is the interesting case when α = ω 0 For this case only, it turns out that there is another allowed form of the solution. It is: Q = Bte−αt as can be verified by substution into equation 1. So the general form of the solution is: Q = (A + Bt)e−αt Again, using the intial conditions for q and i to determine A and B, the result for q is q = CV0 − 2CV0 (1 + αt)e−αt The plot below shows q for the overdamped and critically damped cases.

5

Figure 3: VC vs. time for overdamped and critically damped cases, for a -V 0 to + V0 step, for V0 = 5V .

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