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From: ankleand@athena.mit.edu (Andy Karanicolas)
Newsgroups: sci.physics,sci.electronics
Subject: Re: capacitors
Keywords: electromagnetism, capacitors, circuits
Message-ID: <5578@bloom-beacon.MIT.EDU>
Date: 31 May 88 06:05:14 GMT
References: <2992@phoenix.Princeton.EDU>
Sender: daemon@bloom-beacon.MIT.EDU
Reply-To: ankleand@athena.mit.edu (Andy Karanicolas)
Organization: Massachusetts Institute of Technology
Lines: 106

In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes:
>The following problem appears in many books on elementary physics,
>elementary electromagnetism, and elementary circuit theory:
>
>Two capacitors, with capacitances C1 and C2 respectively, carry
>charges Q1 and Q2 respectively.  Find the total energy stored.
>Now connect the positive plates and connect the negative plates.
>Show that the total energy stored decreases.  Where has the
>energy gone? 
>...
>I hate long .signature files.        -- T. Chow

This problem is a "classic" alright.   Many answers are possible, each
depending on the circuit model used.  It is necessary to look one level
higher to see the fundamental problem of asking "Where has the energy gone"

Define:  u(t)      unit step function
         delta(t)  Dirac delta function; unit impulse
	 integral(a, b, whatever-kernel, dx)  integral of kernel from
	  	a to b over dx

Now, 	 	v1 = q1/c1
		v2 = q2/c2


Since the total charge will be conserved,  q1 + q2 = (c1)(v) + (c2)(v)
where  v  is the final voltage.  Thus,
		     
	v1(t) = v1 + [v - v1] u(t)
		     
	v2(t) = v2 + [v - v2] u(t)

with the following convention for constituent relations

   +i(t)  ---->-----[device]------    	v(t) = f( i(t) )
		  +          -
		      v(t)

	i1(t) = c1 dv1/dt
	i2(t) = c2 dv2/dt

Note we will be taking the derivative of a unit step function.

Operationally, du(t)/dt = delta(t); as generalized functions, delta(t) and
du(t)/dt behave equivalently:

   integral(-inf, inf, delta(t), dt) = integral(-inf, inf, du(t)/dt, dt) = 1

The function delta(t) has the "sampling" property:

	integral(-inf, inf, g(t)delta(t), dt) = g(0)

as long as g(t) is a "smooth" function for t = 0.  As we will see, it is
this requirement that g(t) is smooth at t = 0 that is violated when the 
value of the energy change is asked for.

It is very important to realize that delta(t) is not an ordinary function; 
an ordinary function can be assigned a value or it can be graphed.  A
common, and treacherous, definition of delta(t) is that it is infinity when
t = 0 but it is 0 for any other t.  This is convenient for visualizing 
such functions that are impulsive.  However, delta(t) can be consistently
defined only in terms of the above integrals and conditions; that is, 
delta(t) is defined in terms of what it does in these integrals.

As will be seen, asking for the energy "lost" violates the definition of the
dirac delta function.

Taking the time derivatives above,

		      
	i1(t) =    c1 [v - v1] delta(t)

	i2(t) =    c2 [v - v2] delta(t)

The energy of the system is the time integral of the power:

 E = integral(-inf, inf, p1(t), dt) + integral(-inf, inf, p2(t), dt)
 
 p1(t) = c1 v1 (v - v1) delta(t)   +   c1 (v - v1)^2 delta(t) u(t)

 p2(t) = c2 v2 (v - v2) delta(t)   +   c2 (v - v2)^2 delta(t) u(t)

Now, we see that

 E = c1 v1 (v - v1)  +  c2 v2 (v - v2)  +
     [c1 (v - v1)^2  +  c2 (v - v2)^2] integral(-inf, inf, delta(t)u(t), dt)

The problem is now evident: we are attempting to assign a value to 
integral(-inf, inf, delta(t)u(t), dt).  The unit step u(t) is not a smooth
function at t = 0; u(t) is not defined for t = 0.

That is, the DETAILS of how the capacitor plates are connected determine what
the energy of the system will be.  If the plates are suddenly connected in a
step function manner ( u(t) ) in a lossless capacitive circuit then the 
energy  E  CANNOT be determined uniquely.

The energy can be determined if, for instance, there is finite resistance
in the circuit or if the switches connecting the plates take finite time to
close.  The particular model used for the switches will determine the outcome.

If Kirchoff's voltage and current laws hold, then the system is 
electro-quasi-static; that is curl(E) = -dB/dt = 0.  In this limiting case,
energy is not "lost" to EM radiation.  However, if EM radiation is to be
worked into the system, then the structure of the capacitor, its dielectric
constant and the circuit layout, among many other factors, will, in addition,
determine what the energy E will be.