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Physics_documet_with_circurit_diagram_tables_and_equations.pdf-Chapter 28 Direct Current cshowing page 1-5 out of 7

Page 1
Chapter 28 Direct Current circuits
28.1 Electromotive Force (emf)
Sources of emf: A battery or any other device that provides
electrical energy.
“charge pump”
r
:
internal resistance
Ir
: potential decrease by r
IR = V
: potential on R, or
terminal voltage of the battery
V = - Ir, or
= IR +Ir
is the open-circuit voltage. (if I = 0, V =
= V
max
)
If r is small (a good battery), V
.
If r is large (a bad battery), V <<
.
Power dissipation
I
=
I
2
R
+
I
2
r
+
_
Resistor
Example
: A battery has an emf of 12 V, and internal
resistance of 0.05
.
Its terminals are connected to a load
resistance of 3
.
(a)Find the current.
I
=
R
+
r
=
12
V
3.05
=
3.93
A
(b) Find the terminal voltage:
V
=
Ir
=
12
V
(3.93
A
)(0.05
)
=
11.8
V
Find the equivalent resistance
6
8
4
3
12
2
14


Page 2
28.2 Resistors in Series and in Parallel
In Series
:
The current is the same through each resistor in series
connection:
I = I
1
= I
2
The total voltage = sum of voltages of each resistor:
V
=
V
1
+
V
2
=
IR
1
+
IR
2
=
I
(
R
1
+
R
2
)
=
IR
eq
Generally
,
R
eq
=
R
1
+
R
2
+
R
3
+
...
R1
I
In Parallel:
The voltage (potential difference) across each resistor in a
parallel circuit is the same:
V = V
1
= V
2
The total current = sum of currents passing through each
resistor in a parallel circuit
I
=
I
1
+
I
2
=
V
R
1
+
V
R
2
=
V
1
R
1
+
1
R
2
=
v
R
eq
Generally,
1
R
eq
=
1
R
1
+
1
R
2
+
1
R
3
+
...


Page 4
28.3 Kirchhoff’s Rules
(1)
Junction
Rule
: "The sum of the current entering any
junction must equal the sum the currents leaving that
junction."
I
in
= I
out
I
1
= I
2
+ I
3
Conservation of Charge
(2)
Loop Rule
: "The sum of potential differences across all
the elements around any closed circuit loop must be
zero."
--> “
Conservation of Energy
Strategy to solve complex circuits:
Label all quantities and assign directions of the currents.
Apply the junction rule (Kirchhoff’s
1
st
rule).
Apply Kirchhoff’s loop rule to loops. Pay attention to signs.
Solve the equations.
I
V
=
V
b
V
a
=
IR
b
a
I
V
=
V
b
V
a
=+
IR
b
a
V
=
V
b
V
a
=+
b
a
+
_
+
_
V
=
V
b
V
a
=
b
a
I
2
I
1
I
3
Example
: Find the current
I
1
, I
2
, and
I
3
(1)
Assign directions of currents.
Just make a reasonable guess.
(2)
Junction rule:
I
1
+ I
2
= I
3
(3) Identify loops (3):
Apply Kirchhoff’s 2
nd
rule 3-1= 2 times.
Lower loop:
10 V – (6
)
I
1
- (2
)
I
3
= 0
Upper loop: -14 V – 10 V + (6
)
I
1
- (4
)
I
2
= 0
(4) 3 unknowns, and 3 independent equations.
I
1
= 2 A,
I
2
= -3 A, and
I
3
= -1 A
Negative value indicates that the actual direction of current is
opposite to the labeled direction.
4
6
2
14V
10V
I
3
I
2
I
1
+
+
_
_


Page 5
Water in the bucket
Charge in the capacitor
Water flow in the pipe
Electric current
q
t
C
0.632
I
t
I
0
0.368I
0
I
0
=
R
R
S
I
+q
-q
t
flow
rate
0.368F
0
Water in bucket
t
W
max
0.632
Bucket
pipe
Reservoir


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