IEEE Solid-State Circuits Magazine - Fall 2016 - 9

before and after we open the valve,
but we have now created an effective glass with twice the cross-section
area. This will result in a new water
height that is half of the initial height
V/2. This situation is identical to that
of two capacitors. The total amount
of charge (CV ) will remain the same
before and after we close the switch.
However, the equivalent capacitance
is 2C after we close the switch. Therefore, the voltage across the capacitors
will settle to V/2.
What about the energy before and
after the valve is turned ON? Before
turning the valve ON, all the potential energy is stored in the left glass.
We can write
E before = C # V # V/2 + 0 = CV 2 /2.
After turning the valve ON, we
can write
E after = 2C # V/2 # V/4 = CV 2 /4,
where E before and E after represent the
total potential energy stored in the
system before and long after we turn
on the valve. We have also assumed
unity water density (t) and gravity
(g). These same equations are also
valid for the case of two capacitors. In
both cases (see Figure 3), we lose half
of the initial energy. The question is
where the energy is lost. To answer
this question, the reader is encouraged to think of an answer for the
case of two glasses first and see if the
answer can be extended to the case of
two capacitors.
Both the glass and the pipe walls
present friction to the flow of water
from one glass to the other. When we
first turn ON the valve, part of the potential energy is turned to kinetic energy,
moving water from the left glass to the
right one. The kinetic energy, however,
is turned to heat as the water flow will
have friction with the walls of the pipe
and the glass. Once the water level is
settled to its final value, all the kinetic
energy is turned to heat, warming up
the water and the glass accordingly.
What is interesting in this case is that
the total energy lost to heat is always

Two Capacitors
(Charge Sharing)

Equation

Two Glasses
(Water Sharing)

Total Charge at t < 0

CV + 0

Total Weight at t < 0

Total Charge at t > 0

CV /2 + CV /2

Total Weight at t > 0

Final Voltage at t > 0

V /2

Water Height at t > 0

Total Energy at t < 0

½ CV 2

Potential Energy at t < 0

Total Energy at t > 0

¼ CV 2

Potential Energy at t > 0

Figure 3: The equations governing the charge sharing between two capacitors are identical
to those governing the water sharing between two glasses of water. We have assumed unity
water density (t) and gravity (g) in writing these equations.

half of the total initial energy, independent of the pipe diameter and the glass
surface roughness (which contributes
to its resistance to water movement).
If the overall resistance to water movement is larger, it will take longer for the
water to settle to its final height, but
the energy will be lost at a lower rate.
If the resistance is smaller, the water
will move faster, but the rate of energy
loss will be higher. Again, in all cases,
exactly half of the initial energy is lost
in the process of water sharing.
Let us now return to the case of
two capacitors. Similar to the valve
and the pipe, the switch and its associated resistance will impede the flow of
charge from one capacitor to the other.
The energy lost in this process is the
energy dissipated as heat in the resistor. To see this quantitatively, let us
write an equation for the current flowing from the left capacitor to the right.
For t 2 0, we have
-2t
i R ^ t h = V e RC .
R

And hence the power being dissipated in the resistor can be written as
2
-4t
p R ^ t h = V e RC .
R

These equations show that the lower the R, the larger the initial current
and the initial power (proportional to
1/R), but the smaller the time constant
(proportional to R). Since the area under a decaying exponential curve is
the product of its initial value and its

time constant, the energy dissipated,
which is the area under p R ^ t h, will be
independent of R:
2
E R = V # RC = 1 CV 2 .
R
4
4

This equation confirms that the
energy wasted in the process of
charge sharing is independent of the
switch resistance.
An interesting exercise to contemplate here is the case where R is exactly
zero. Clearly, there will be no heat dissipation in this case as there is no resistance. What happens then? Will half of
the initial stored energy be wasted again
in the process of charge sharing? If yes,
where is it wasted and in what form?
The reader is encouraged to resort back
to our analogy again and see if a similar
case can be built in the analogous world
and if an answer may emerge. In the
interest of giving the readers a chance
to explore these questions on their own,
we will answer these questions in a
future article.
In summary, the process of charge
sharing of two capacitors using a switch
is analogous to the process of water
sharing among two adjacent glasses
using a valve and a pipe. In both cases,
where the two capacitors (glasses)
have equal capacitances (cross-section
areas), exactly half of the initial stored
(potential) energy is lost in the process
of charge (water) sharing. This loss is
independent of the switch resistance
(pipe diameter and friction).

IEEE SOLID-STATE CIRCUITS MAGAZINE

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Table of Contents for the Digital Edition of IEEE Solid-State Circuits Magazine - Fall 2016