W9_parallel_resonance.eps

Week 9: Alternating Current Circuits 307

If we substitute this relation in for theI’s and divide byL, we get the following second order,

linear, homogeneous ordinary differential equation:

d^2 Q

dt^2

+

R

L

dQ

dt

+

Q

LC

= 0 (651)

We recognize this as the differential equation for adamped harmonic oscillator. To solve it,

we “guess”^86 :

Q(t) =Q 0 eαt (652)

and substitute this into the ODE to get the characteristic:

α^2 +R

L

α+^1

LC

= 0 (653)

We solve for:

α = −

R

2 L±

√(

R

L

) 2

−LC^4

2

= −R

2 L

±iω 0

√

1 −R

(^2) C
4 L
= −R
2 L
±iω 0

√

1 −τC

4 τL

= −

R

2 L±iω

′

(654)

whereτL=L/R τC=RC,ω′= 0

√

1 − 4 ττCL, and our final solution looks like:

Q(t) =Q 0 e

− 2 RtL

cos(ω′t) (655)

(after we choose the real part of the complex exponential and use the initial conditions).

From this we can easily find the current through and voltage acrossall of the elements of the

circuit. Finally, given the current and voltages it is easy to show thatenergy is conserved, that

the initial energy stored in the capacitor exactly balances the energy consumed in the resistor

ast→∞.

• AC voltage across a resistanceR:

V(t) R

I(t)

Figure 114: AC voltage acrossR

We use Kirchhoff’s voltage rule and Ohm’s Law to get:

V 0 sin(ωt)−IR= 0 (656)

or

IR(t) =

V 0

R

sin(ωt) (657)

and we see that the current isin phasewith the voltage drop across a resistor.