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394 ⏐⏐⏐ CAPACITORS
40
5.0
40
v C (V)
0 1t 2t 3t 4t 5t t
i C (mA)
2
E
– v R +
R
t = 32 ms
0 1t 2t 3t 4t 5t t
v R (V)
t = 32 ms
t = 32 ms
0 1t 2t 3t 4t 5t t
FIG. 10.36
Waveforms for the network of Fig. 10.35.
1
+
v C
–
2
(a)
= E
i C = i R = i discharge
v R = 0 V
FIG. 10.38
Demonstrating the discharge behavior of a
capacitive network.
R
By Eq. (10.13),
iC � � E
R � e�t/t � e �t/(32�10�3 40 V
)
�
8 k�
By Eq. (10.17),
The curves appear in Fig. 10.36.
b. 5t � 5(32 ms) � 160 ms
� (5 � 10 �3 )e �t/(32�10�3 )
v R � Ee �t/t � 40e �t/(32�10�3 )
Once the voltage across the capacitor has reached the input voltage
E, the capacitor is fully charged and will remain in this state if no further
changes are made in the circuit.
If the switch of Fig. 10.24 is opened, as shown in Fig. 10.37(a), the
capacitor will retain its charge for a period of time determined by its
leakage current. For capacitors such as the mica and ceramic, the leakage
current (i leakage � v C/R leakage) is very small, enabling the capacitor
to retain its charge, and hence the potential difference across its plates,
for a long time. For electrolytic capacitors, which have very high leakage
currents, the capacitor will discharge more rapidly, as shown in Fig.
10.37(b). In any event, to ensure that they are completely discharged,
capacitors should be shorted by a lead or a screwdriver before they are
handled.
i = 0
+
vC E
–
2
v R = 0 V i = 0
+
v C Rleakage
(b)
–
i leakage
FIG. 10.37
Effect of the leakage current on the steady-state behavior of a capacitor.
10.8 DISCHARGE PHASE
The network of Fig. 10.24 is designed to both charge and discharge the
capacitor. When the switch is placed in position 1, the capacitor will
charge toward the supply voltage, as described in the last section. At
any point in the charging process, if the switch is moved to position 2,
the capacitor will begin to discharge at a rate sensitive to the same time
constant t � RC. The established voltage across the capacitor will create
a flow of charge in the closed path that will eventually discharge the
capacitor completely. In essence, the capacitor functions like a battery
with a decreasing terminal voltage. Note in particular that the current i C
has reversed direction, changing the polarity of the voltage across R.
If the capacitor had charged to the full battery voltage as indicated in
Fig. 10.38, the equation for the decaying voltage across the capacitor
would be the following:
v C � Ee �t/RC
discharging
(10.18)