Chapter 21
Electric Current and Direct-Current Circuit
Outline
21-1
Electric Current
21-2
21-3
Resistance and Ohm’s Law
Energy and Power in Electric Circuit
21-4
Resistance in Series and Parallel
21-5
Kirchhoff’s Rules
21-6
Circuits containing Capacitors
21-7
RC Circuits
21-6
Circuits containing Capacitors
Capacitors in Parallel
Capacitor are connected in parallel, and they have the same
voltage.
Figure 21-16 shows three capacitors connected in parallel.
Figure 21-16
Capacitors in Parallel
Deriving equivalent capacitance
In fig. 21-16 (a), the magnitudes of the charges on each capacitor
are
Q1  C1 ,
Q2  C2 ,
Q3  C3 ,
The total charge on the three capacitors is
Q  Q1  Q2  Q3   C1   C2   C3   (C1  C2  C3 )
(1)
In fig. 21-16 (b), the magnitude of the charge on an equivalent
capacitor is
Q   Ceq
(2)
Compared the above two equations (1) and (2), we have
Ceq  C1  C2  C3
Equivalent capacitance for Capacitors in Parallel
Ceq  C1  C2  C3  ...   C
SI unit: farad, F
(21  14)
Problem 21-56
capacitor in parallel
Two capacitors, one 7.5 µF, and the 15µF, are connected in
parallel across a 12-V battery.
(a) Find the equivalent capacitance of the two capacitors.
(b) Find the charge stored in each capacitor.
Solution
1) Since the 2 capacitors in parallel:
Ceq  C1  C2
 Ceq  7.5F  15F
 22.5F
2) Both capacitors have the same voltage 12 V:
Q1  C
 Q1  12(V )  (7.5 10 6 F )  9.0 10 5 C
Q2  C
 Q2  12(V )  (15 10 6 F )  1.8 10  4 C
Capacitors in Series
Capacitor are connected in series (one after the other), and they
have the same charge.
Figure 21-17 shows three capacitors connected in series.
Figure 21-17
Capacitors in Series
Deriving equivalent capacitance
In figure (a), since all the capacitors have the same charge Q,
we have
V1 
Q
,
C1
V2 
Q
,
C2
V3 
Q
,
C3
And the total potential difference across the three capacitors
must equal the battery emf,
  V1  V2  V3 
Q Q Q
1
1
1


 Q( 
 )
C1 C2 C3
C1 C2 C3
In figure (a), since Q = Ceqε, we have
1
 Q
(21  16)
Ceq
Compare (21-15) and (21-16), one has
1
1
1
1
 

Ceq C1 C2 C3
(21  15)
Equivalent capacitance for Capacitors in Series
1
1
1
1
1
 
  ...  
Ceq C1 C2 C3
C
SI unit: farad, F
(21  17)
Problem 21-56
capacitor in series
Two capacitors, one 7.5 µF, and the 15µF, are connected in series
across a 12-V battery.
(a) Find the equivalent capacitance of the two capacitors.
(b) Find the charge stored in each capacitor.
Solution
1) Since the 2 capacitors in series:
1
1
1


Ceq C1 C2

1
1
1


 0.20
Ceq 7.5F 15F
 Ceq  5.0
( F ) 1
F
2) Both capacitors have the same voltage 12 V:
Q1  Q2  Q
 12(V )  (5 10 6 F )  6.0 10 5 C
Active Example 21-3 Find Equivalent Capacitance
A electric circuit consists of a 12.0 V battery and three capacitors
connected partly in series and partly in parallel. Find the total energy
store in the capacitors.
Active Example 21-3
Find the Equivalent
Capacitance and the Stored
Energy
Solution
1) Find the equivalent capacitance of a 10.0 uf and 5.00 uf in series,
3.33 uF
2) Find the equivalent capacitance of circuit (3.33 uf and 20.0 uf in
parallel)
Ceq = 23.3 uF
3) Find the stored energy
1
U  CeqV 2  1.68 103 J
2
21-7 RC Circuit
A RC circuit consists of a resistor R and capacitor C.
Figure 21-18
A Typical RC Circuit
Charging the circuit !!
The formula that can be used to calculate the charge on the
capacitor with time t as a parameter is,
q(t )  C (1  e t / )
(21  18),
where, e is exponential.  = RC is called Time Constant (s: second),
which determines the property of the CR circuit. When t = ,
q = 0.632 Cε (63.2% of the final charge).
Figure 21-19
Charge Versus Time for an RC Circuit
The current as a function of time t can be expressed as

I (t )  ( )e t /
R
(21  19),
 = RC is called Time Constant (s: second),
Figure 21-20
Current Versus Time in an RC
Circuit
Example 21-9
Charging a Capacitor
A circuit consists of a 126-Ω resistor, a 275-Ω resistor, a 182-uF
capacitor, a switch, and a 3.00-V battery all in series. Initially the
capacitor is uncharged and the switch is open. At time t = 0, the
switch is closed.
(a) What charge will the capacitor have a long time after the switch is
closed?
(b) At what time will the charge on the capacitor be 80.0% of the value
found in part (a)?
Picture the problem
Example 21-9
Charging a Capacitor
Solution Time constant   RC  (126  275)(182 106 F )  73.0  103 s
Part (a)
1) After a long time ( t  ),
q(t )  C (1  et / )  C
 (182 106 F )(3.00V )  546 106 C
2) Set q(t) = 0.800 Cε, and solving for t,
q(t )  C (1  et / )  0.8C
We have
1  et /  0.8
t   ln(0.200)
so,
t   ln(0.200)  (73.0 103 s) ln(0.200)  118 103 s
Summary
Equivalent capacitance for Capacitors in Parallel
Ceq  C1  C2  C3  ...   C
(21  14)
Equivalent capacitance for Capacitors in Series
1
1
1
1
1
 
  ...  
Ceq C1 C2 C3
C
(21  17)
RC Circuit
q(t )  C (1  et / )
(21  18),
Example 21-8
Energy in Parallel
Two capacitors, one 12.0 uf, and the other of unknown capacitance
C, are connected in parallel across a battery with an emf 9.00 V.
The total energy stored in the two capacitors is 0.0115J. What is
the value of the capacitance C.
Example 21-8
Energy in Parallel
Solution
1) Since the Capacitor energy is
1
1
U  QV  CeqV 2
2
2
We have
1
0.0115 J  Ceq (9.00V ) 2
2
Ceq  284 106 F  284  F
2) Solve for C in terms of equivalent capacitance:
Ceq  C1  C
284 F  12.0  F  C
C  284 F  12.0  F  272  F