Download Exponential Functions and RC circuits

Exponential Functions and RC circuits
The defining characteristic of an exponential function is:
Rate of change of a quantity is proportional to the quantity.
In equation form, this statement amounts to
dy
= ky ,
dt
where k is a constant of proportionality. It is easy to integrate this equation to obtain an
exponential function:
dy
= k dt
y
ln y = kt + C
y = Cekt
where C is a constant of integration. Note the following points:
•
The units of k must be the units of 1/t, since the argument of an exponential must be
dimensionless.
•
It is often convenient to define a new variable τ = 1/ k (Greek letter tau). In this form,
our exponential function is
y = Cet / τ .
In this case, of course, the units of t and τ must be the same, if the exponent is to be
dimensionless.
•
Exponential functions have characteristic times for growth and decay. For example, let’s
write an exponential equation as
y = et / τ
where I have set C = 1 for convenience.
At t = 0, y = 1.
At t = τ, y = e; at t = 2τ, y = e2 ; and so on. The constant τ is a measure of how quickly
or slowly the exponential grows, and in each interval τ, the function increases by a factor
of e. This behavior is characteristic of exponential functions. Note that for exponentially
decaying functions (negative exponents), we would be dividing by e instead of
multiplying.
•
Finally, note that we could use any base, not just e, the base for natural logarithms.
Functions of the form 10kt or 2kt have much the same properties. We could, for example,
define doubling times or halving times (“half-lives”), as we did in class.
Now let’s see how exponential functions apply to a circuit consisting of an emf, a resistor, and a
capacitor in series (refer to Fig. 28-13 in the text). The Kirchhoff loop equation is
1
E − iR −
q
= 0.
C
(Remember that C = q / V so that the voltage drop across a capacitor is V = q / C .)
If we recall that i = dq / dt , then on substituting we obtain
E −R
dq q
− = 0,
dt C
or, rearranging terms,
R
dq
q
=E−
or, dividing both sides by R ,
dt
C
dq E
q
= −
.
dt R RC
IMPORTANT: Note that we have the rate of change of q proportional to q. Therefore we
expect an exponential solution!
This equation is easier to solve than is apparent from the text. Consider
CASE 1: Charge the capacitor fully, and then throw the switch in Fig. 28-13 to “b”. The
capacitor will now discharge through the resistor. Since the battery is out of the circuit, E is zero,
and the above loop equation becomes
dq
q
=−
.
dt
RC
But this equation is easy to integrate:
dq
1
=−
dt
q
RC
t
+K
ln q = −
RC
q = Ke− t / RC
where K is a constant of integration. We can determine K from the initial conditions: At t = 0,
the capacitor is fully charged, and so holds a charge q0 = C E . Setting t = 0 in the above
equation, one sees that K = q0 = CE , and so our equation becomes
q(t ) = q0 e− t / RC = CE e− t / RC .
Compare with the text, Eq. 28-36. It is now straightforward to find the current in the circuit and
the voltage drop across the capacitor as functions of time, as we did in class. For example, the
voltage drop across the capacitor would be
VC (t ) =
q(t )
= E e− t / RC .
C
2
CASE 2: Assume that at t = 0, the capacitor is completely uncharged. Now throw the switch in
Fig. 28-13 to “a”. The capacitor begins to charge, according to the loop equation we wrote down
before:
E − iR −
q
= 0 , or
C
dq E
q
= −
dt R RC
This equation is more difficult to solve, but not much, if we use a simplifying substitution. We
define a new variable
u≡
q
E
−
.
R RC
If we take the derivative of this equation, since the first term is constant, we have
du
1 dq
dq
du
=−
or
= − RC .
dt
RC dt
dt
dt
Our loop equation above now becomes
− RC
du
= u or, simplifying,
dt
du
1
.
=−
dt
RC
But we have already solved this equation. The solution is just
u = Ke− t / RC ,
where as before, K is a constant of integration. If we now substitute back for u, we obtain
E
q
−
= Ke− t / RC or, moving terms around,
R RC
E
q
= − Ke− t / RC or, multiplying both sides by RC ,
RC R
q(t ) = CE − Ke− t / RC .
Exercise: See if you can find K from the initial condition, in this case q(t = 0) = 0 , and show
that K = C E , so that our result becomes
q(t ) = CE (1 − e− t / RC ) .
Compare this result with Eq. 28-30 in the text. As before, it is now straightforward to find the
current i and the voltage drop V across the capacitor as functions of time.
3