A plot of i(t) for the charging process is given in Fig. 27-16b. Note that the current
has the initial value Ᏹ/R and that it decreases to zero as the capacitor becomes
fully charged.
790
CHAPTE R 27 CI RCU ITS
A capacitor that is being charged initially acts like ordinary connecting wire
relative to the charging current. A long time later, it acts like a broken wire.
By combining Eq. 25-1 (q ϭ CV ) and Eq. 27-33, we find that the potential
difference V C (t) across the capacitor during the charging process is
(charging a capacitor).
(27-35)
This tells us that V C ϭ 0 at t ϭ 0 and that V C ϭ Ᏹ when the capacitor becomes
fully charged as t : ϱ.
The Time Constant
The product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensions
of time (both because the argument of an exponential must be dimensionless and
because, in fact, 1.0 ⍀ ϫ 1.0 F ϭ 1.0 s). The product RC is called the capacitive
time constant of the circuit and is represented with the symbol t:
t ϭ RC (time constant).
(27-36)
From Eq. 27-33, we can now see that at time t ϭ t (ϭ RC), the charge on the
initially uncharged capacitor of Fig. 27-15 has increased from zero to
q ϭ C Ᏹ(1 Ϫ e
Ϫ1
) ϭ 0.63C Ᏹ.
( 2 7 - 3 7 )
In words, during the first time constant t the charge has increased from zero to
63% of its final value C Ᏹ. In Fig. 27-16, the small triangles along the time axes
mark successive intervals of one time constant during the charging of the capacitor. The charging times for RC circuits are often stated in terms of t. For example,
a circuit with t ϭ 1 ms charges quickly while one with t ϭ 100 s charges much
more slowly,
Discharging a Capacitor
Assume now that the capacitor of Fig. 27-15 is fully charged to a potential V 0
equal to the emf Ᏹ of the battery. At a new time t ϭ 0, switch S is thrown from a to
b so that the capacitor can discharge through resistance R. How do the charge
q(t) on the capacitor and the current i(t) through the discharge loop of capacitor
and resistance now vary with time?
The differential equation describing q(t) is like Eq. 27-32 except that now,
with no battery in the discharge loop, Ᏹ ϭ 0. Thus,
(discharging equation).
(27-38)
The solution to this differential equation is
q ϭ q 0 e
Ϫt/RC
(discharging a capacitor),
(27-39)
where q 0 (ϭ CV 0 ) is the initial charge on the capacitor. You can verify by substitution that Eq. 27-39 is indeed a solution of Eq. 27-38.
R
dq
dt
ϩ
q
C
ϭ 0
V C ϭ
q
C
ϭ Ᏹ(1 Ϫ e
Ϫt/RC
)
has the initial value Ᏹ/R and that it decreases to zero as the capacitor becomes
fully charged.
790
CHAPTE R 27 CI RCU ITS
A capacitor that is being charged initially acts like ordinary connecting wire
relative to the charging current. A long time later, it acts like a broken wire.
By combining Eq. 25-1 (q ϭ CV ) and Eq. 27-33, we find that the potential
difference V C (t) across the capacitor during the charging process is
(charging a capacitor).
(27-35)
This tells us that V C ϭ 0 at t ϭ 0 and that V C ϭ Ᏹ when the capacitor becomes
fully charged as t : ϱ.
The Time Constant
The product RC that appears in Eqs. 27-33, 27-34, and 27-35 has the dimensions
of time (both because the argument of an exponential must be dimensionless and
because, in fact, 1.0 ⍀ ϫ 1.0 F ϭ 1.0 s). The product RC is called the capacitive
time constant of the circuit and is represented with the symbol t:
t ϭ RC (time constant).
(27-36)
From Eq. 27-33, we can now see that at time t ϭ t (ϭ RC), the charge on the
initially uncharged capacitor of Fig. 27-15 has increased from zero to
q ϭ C Ᏹ(1 Ϫ e
Ϫ1
) ϭ 0.63C Ᏹ.
( 2 7 - 3 7 )
In words, during the first time constant t the charge has increased from zero to
63% of its final value C Ᏹ. In Fig. 27-16, the small triangles along the time axes
mark successive intervals of one time constant during the charging of the capacitor. The charging times for RC circuits are often stated in terms of t. For example,
a circuit with t ϭ 1 ms charges quickly while one with t ϭ 100 s charges much
more slowly,
Discharging a Capacitor
Assume now that the capacitor of Fig. 27-15 is fully charged to a potential V 0
equal to the emf Ᏹ of the battery. At a new time t ϭ 0, switch S is thrown from a to
b so that the capacitor can discharge through resistance R. How do the charge
q(t) on the capacitor and the current i(t) through the discharge loop of capacitor
and resistance now vary with time?
The differential equation describing q(t) is like Eq. 27-32 except that now,
with no battery in the discharge loop, Ᏹ ϭ 0. Thus,
(discharging equation).
(27-38)
The solution to this differential equation is
q ϭ q 0 e
Ϫt/RC
(discharging a capacitor),
(27-39)
where q 0 (ϭ CV 0 ) is the initial charge on the capacitor. You can verify by substitution that Eq. 27-39 is indeed a solution of Eq. 27-38.
R
dq
dt
ϩ
q
C
ϭ 0
V C ϭ
q
C
ϭ Ᏹ(1 Ϫ e
Ϫt/RC
)