•
•
•
•
•
•
Chapter 28 Direct current circuit
Electromotive force
Resistors in series and in parallel
Kirchoff’s rules
RC circuits
(Optional) Electrical instruments
(Optional) Household wiring and electrical
safety
p = eI = I 2 r + I 2 R
e - Ir = IR, e = I( r + R)
Electromotive force
Vb -Va = e - Ir
Vb -Va = Vc -Vd = IR
Matching the load
e2R
p=
(r + R)2
e
p = I 2 R, I =
r+R
Show that the maximum power delivered to the load resistance
occurs when the load resistance matches the internal resistance,
that is, when R = r.
†
dp
(r + R)2 - 2R(r
† + R)
= e 2[
]= 0
dR
(r + R)4
R=r
Resistors in series
Req = R1 + R2 + ...
DV = IR1 + IR2 = I (R1 + R2 )
Req = R1 + R2
For a series combination of resistance, the currents
in all the resistors are the same.
†
Resistors in parallel
1 1
1 1
Req = + , Req = + + ...
R1 R2
R1 R2
DV DV
1 1
DV
I = I1 + I 2 =
+
= DV ( + ) =
R1 R2
R1 R2
Req
When resistors are connected in parallel, the
potential differences across them are the same.
†
Rpara
RR
= 1 2
R1 + R2
Find the equivalent resistance
†
Find the equivalent resistance by
symmetry
Kirchhoff’s rules
1. The sum of the currents entering any junction in
a circuit must equal the sum of the currents
leaving that junction:
 I in =  I out
2. The sum of the potential differences across all
†
elements
around any closed loop must be zero:
 DV = 0
closed
loop
Sign conventions
Each circuit element is traversed from left to right
22V = (11W)I1
I1 = 2A, I 2 = -3A, I 3 = -1A
10V - (6W)I1 - (2W)(I1 + I 2 ) = 0
(4) 10V = (8W)I1 + (2W)I 2
(5) -12V = -(3W)I1 + (2W)I 2
(2) abcda 10V - (6W)I1 - (2W)I 3 = 0
(3) befcb - (4W)I 2 -14V + (6W)I1 -10V = 0
(1) I1 + I 2 = I 3
Example 28.8 Applying Kirchhoff’s rules
†
†
DVcap = 11.0V, Q = CDVcap = 66.0mC
-8.00V + DVcap - 3.00V = 0
(1) I1 + I 2 = I 3
(2) defcd 4.00V - (3.00W)I 2 - (5.00W)I 3 = 0
(3) cfgbc (3.00W)I 2 - (5.00W)I1 + 8.0V = 0
I1 = 1.38A, I 2 = -0.364 A, I 3 = 1.02A
Example 28.9 A multiloop circuit
†
†
q
e - - IR = 0
C
e
t = 0, q = 0 so I =
0
R
t = •, I = 0 so Qmax = Ce
Charging a capacitor
†
1
t
RC
)
1 -1t e -1t
I = dq / dt = Ce ( )e RC = e RC
RC
R
q = Ce (1- e
q dq
dq
q - Ce
e - - R = 0,
=C dt
dt
RC
q
dq
1
dq
1
=dt, Ú
=t
q - Ce
RC
RC
0 q - Ce
q - Ce
1
ln(
)=t
-Ce
RC
Charging a capacitor
When t = t = RC
I = e-1 I 0
†
-
1
t
RC
Q -1t
I = dq / dt = e RC
RC
q = Qe
q dq
dq
q
- - R = 0,
=C dt
dt
RC
q
dq
1
dq
1
=dt, Ú
=t
q
RC
RC
Q q
q
1
ln( ) = t
Q
RC
Discharging a capacitor
†
Discharging a capacitor in an RC circuit
-
1
t
RC
1
t
1
t
RC
-
2
U0 / 4 = U0e
t
-
1
t
RC
1
t
q2
U=
, q = Qe RC
2C
2
2
t
Q2 - t
U=
e RC = U 0 e RC
2C
Consider an RC circuit. (a) After how many time constants is the
charge on a capacitor one-fourth its initial value? (b) After how
many time constants is this stored energy one-fourth its initial
value?
q = Qe
Q / 4 = Qe
-
-1 / 4 = e RC
-ln 4 = -t / RC
t = RC(ln 4) = 1.39RC = 1.39t
-1 / 4 = e RC
-ln 4 = -2t / RC
t = RC(ln 4) / 2 = 0.693RC = 0.693t