Electromagnetic Oscillations

11.2  Alternating Current

11.2.1  LC Circuits

The easiest-to-understand system that demonstrates electromagnetic oscillations is the combination of a capacitor and an inductor. Typically, we begin with a fully charged capacitor which is then linked to an inductor by a switch. The basic circuit is shown in figure 11.6.


Figure 11.6: A combination of an inductor and a capacitor forms an LC circuit.

For this circuit, the behavior is characterized by oscillating current, charge, and voltages. This is in sharp contrast to the exponential behavior of the current, voltages, and charges for the case of RL or RC circuits. If we assume we have initially charged the capacitor to a voltage Vmax before connecting it to the inductor, we can ask what the behavior will be.

While it is possible to give a qualitative picture of what happens when the capacitor and inductor are conjoined, it might be insightful to go through the quantitative determination first, then see if that adequately meets what we would naively expect. Before starting even the quantitative approach though, we need to state that this is a case in which the energy approach for looking at the circuit is more immediately obvious than the voltage approach. The main reason is that we can clearly see that energy is conserved for this case since capacitors and inductors involve only electric and magnetic fields which are conservative. We can use Maple to handle the mathematical details.

In dealing with specific cases of circuits, we now need to deal with one of the ever-present "difficult" issues of physics, namely: how do you decide what approach to use for solving any particular problem? This is especially true for circuits involving inductors, capacitors, and resistors because the formulas which describe the behavior of the individual circuit elements are difficult to memorize. In such cases, which do arise quite often in physics, you must remember the mantra: memorize behaviors, not formulas! Typically, you will remember enough about a specific formula to be able to reconstruct it either through a re-derivation of the formula from first principles, or through consideration of what mathematically is needed to get a correct description of the behavior combined with dimensional analysis. The physics is in the behavior!

As a way of remembering the behavior, it sometimes helps to see an animation of the process. This Java animation shows this for the case of an LC circuit (with a negligibly small resistance added to make the current well-defined) if you click on the "Start the Animation" button.

In applying this idea to the case at hand, remember that the behavior is determined by the same considerations that were important for physics problem in Mechanics: if there are no dissipative elements, use energy conservation! Conservation rules are always simpler to use for most applications than complex formulas describing intermediate behavior (i.e. what happens between the initial and final states being considered in the problem). In cases where dissipative elements, resistors in this case, are involved, think about the behavior in terms of the initial and long-time aspects of inductor or capacitor behavior. As you have seen in previous lectures, these kinds of considerations are essential to deriving the exact formulas anyway. But, for a large class of problems, these same considerations allow us to dispense with the exact formulas. Let's consider two contrasting cases.

Problem 1:
In the LC circuit shown in figure 11.6, C = 4 m F and L = 2.5 mH. At time t = 0, a current of 1 ampere flows through the inductor and is decreasing, and the charge on the capacitor is 57.7 m C and increasing with time.

  1. What is the maximum charge that will appear on the capacitor?
  2. What is the maximum current that will flow through the inductor?
  3. Derive a quantitative expression for the charge as a function of time. Be sure to specify the angular frequency w and the phase angle f.
  4. At what time after t = 0 will the charge on the capacitor next be zero?

Solution:
To solve this, we need to consider energy conservation predominantly as the formulas, while they can be used, are not as effective in terms of time and effort for the first part of the problem.

  1. Since energy is conserved in this system, we have, at any given time t,
    UC, max
    =
    UC(t) + UL(t)
    1
    2
    		 Qmax2
    C
    =
    1
    2
    		 q2
    C
    		 + 1
    2
    		 Li2 Þ
    Qmax
    =
      ________
    Öq2 + LCi2
     
    =

    Ö
     
    (57.7×10-6 C)2 +(2.5×10-3 H)(4×10-6 F)(1 A)2
     
    Qmax
    =
    1.2×10-4 C = 120 mC
    		(11.2.1.16)
  2. The same approach works for the inductor
    UL, max
    =
    UC, max
    1
    2
    		 Limax2
    =
    1
    2
    		 Qmax2
    C
    		 Þ
    imax
    =
      æ
     ú
    Ö
    Qmax2
    LC
     
    =
      æ
     ú
    Ö
    (1.2×10-6 C)2
    (2.5×10-3 H)(4×10-6 F)
     
    imax
    =
    0.012 A
    		(11.2.1.17)
  3. The formula for the charge as a function of time is given as shown below.
    q(t) = Q cos(wt + f)
    		(11.2.1.18)
    Therefore, we need only identify the values for each of the terms in the equation. The maximum charge we have determined as 120 mC. The angular frequency is 1/ÖLC, so
    w = q
      __
    ÖLC
    		 = 1

    Ö
    (2.5×10-3 H)(4×10-6 F)
    		 = 104 s-1
    		(11.2.1.19)
    The phase is given by the initial condition. We want q(t = 0) = 57.7 m C, but, in addition, we want the current in the inductor to be decreasing. The formula for current in the inductor is derived by taking the derivative of the charge w.r.t. time. So,
    i(t) = -wQmaxsin(wt + f)
    		(11.2.1.20)
    Therefore, we need to find the value of f that not only gives the correct initial charge, but has the property of decreasing the magnitude of the current. Thus, if we take just the initial charge for deciding on the phase, we would have
    q(t = 0)
    =
    Q cosfÞ
    =
    cos-1 q
    Q
    =
    cos-1 é
    ê
    ë     		57.7 mC
    120 mC
    		 ù
    ú
    û
    f
    =
    61.3° or 299°
    		(11.2.1.21)
    If we want current to decrease as time increases right after t = 0, we must choose f = 299° or f = 2.61 radians. Putting it all together gives
    q(t) = Q cos(wt + f) = (120 mC)cos((104 s-1)t + 2.61)
    		(11.2.1.22)
  4. We can get this by evaluating the next time cos(wt + f) = 0. Since f > p/2 or 3p/2, the next zero occurs when the argument of the cosine is at 5p/2, so
    q(t) = 0
    =
    Qmaxcos(wt + f) Þ
    cos(wt + f)
    =
    0 Þ
    wt + f
    =
    5p/2 Þ
    t
    =
    (5p/2 - f)
    w
    =
    (7.85 - 2.61)
    104 s-1
    t
    =
    0.52 ms
    		(11.2.1.23)

11.2.2  RLC Circuits

Including a non-negligible resistance in our calculation is straightforward. We simply need to account for the effect of resistance in terms of the power lost. In this case, we know that the time rate of change of potential energy in the system is dU/dt = Li di/dt + (q/C)dq/dt. Since energy is no longer conserved, we can't have dU/dt = 0, but rather dU/dt = PR = -i2R, i.e. the Joule heating in the resistor is the rate at which energy is lost from the system. The differential equation describing the charge as a function of time is then
L d2q
dt2
 + q
C
 ·q + R dq
dt
 = 0
(11.2.2.24)
We once again turn to Maple to get the solution of this equation and it's ramifications.

11.2.3  AC Circuits

We can also consider the case of resistors, inductors, and capacitors connected to an alternating current source. Such a source produces a current that varies with time, usually sinusoidally, that looks very much like the current in an LC circuit. We can easily see the result of introducing such a current into a resistor, capacitor, or inductor in terms of the voltage across each. Suppose that the current has the form of i(t) = I coswt. Then the voltage for each component connected as shown in figure  11.7 is as follows.


Figure 11.7: Plots of voltage and current for an alternating current source connected to a resistor, an inductor, or a capacitor. The magnitudes of the current and voltage across each component are I, VR, VL, and VC, respectively.


i(t)
=
Icoswt
vR
=
i(t)R = (IR)coswt = VRcoswt
vL
=
L di
dt
 = L d
dt
 (Icoswt) = -(IwL)sinwt = VLcos(wt + p/2)
vC
=
q(t)
C
 = 1
C
 ó
õ 		dq(t)
dt
  dt = 1
C
 ó
õ 		i(t) dt
=
1
C
 ó
õ 		Icoswt dt
=
I
C
 sinwt
w
=
I
wC
 cos(wt - p/2)
vC
=
VCcos(wt - p/2)
(11.2.3.25)
where we made use of the trigonometric identities
cos(q- p/2)
=
sinq
cos(q+ p/2)
=
-sinq
(11.2.3.26)
We describe the resistor as having a voltage which is in phase with the current source. The voltage across the inductor is out of phase because it leads the current source by a quarter-cycle. This is in keeping with our knowledge of the behavior of inductors in that any attempt to change the current leads to an EMF that opposes the change, hence the current itself must "lag" the change to the current. For a capacitor, the voltage is also out of phase with the current source but in this case it lags the current source by a quarter-cycle. It is the current that delivers the charge to the capacitor plates in order to create the potential difference that causes the current to change, hence the current in this circuit must lead the change in current.

Send comments to larryg@upenn5.hep.upenn.edu.
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