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Transcript
```Chapter 28
Direct Current Circuits
1
Direct Current

When the current in a circuit has a constant
direction, the current is called direct current



Most of the circuits analyzed will be assumed to
be in steady state, with constant magnitude and
direction
Because the potential difference between the
terminals of a battery is constant, the battery
produces direct current
The battery is known as a source of emf
2
Electromotive Force

The electromotive force (emf), e, of a battery
is the maximum possible voltage that the
battery can provide between its terminals




The emf supplies energy, it does not apply a force
The battery will normally be the source of
energy in the circuit
The positive terminal of the battery is at a
higher potential than the negative terminal
We consider the wires to have no resistance
3
Internal Battery Resistance



If the internal
resistance is zero, the
terminal voltage equals
the emf
In a real battery, there
is internal resistance, r
The terminal voltage,
DV = e – Ir
4
EMF, cont

The emf is equivalent to the open-circuit
voltage



This is the terminal voltage when no current is in
the circuit
This is the voltage labeled on the battery
The actual potential difference between the
terminals of the battery depends on the
current in the circuit
5

The terminal voltage also equals the voltage
across the external resistance



This external resistor is called the load resistance
In the previous circuit, the load resistance is just
the external resistor
In general, the load resistance could be any
electrical device

These resistances represent loads on the battery
since it supplies the energy to operate the device
containing the resistance
6
Power

The total power output of the battery is
  I DV  Ie


This power is delivered to the external
resistor (I 2 R) and to the internal resistor (I2 r)
 I R I r
2
2
7
Example 28.1 Terminal Voltage
of a Battery




A battery has an emf of 12V and an internal
resistance of 0.05
Its terminals are connected to a load resistance of 3

Find the current in the circuit and the terminal
voltage of the battery
Calculate the power delivered to the load resistor,
the power delivered to the internal resistance of the
battery and the power delivered by the battery
8
Example 28.1 Terminal Voltage
of a Battery

Find the current in the circuit and the terminal
voltage of the battery
e
12V
I

 3.93 A
R  r (3  0.05)
DV  e  Ir  12V  (3.93 A)(0.05)  11.8V
9
Example 28.1 Terminal Voltage
of a Battery

Calculate the power delivered to the load resistor, the
power delivered to the internal resistance of the battery
and the power delivered by the battery
PR  I 2 R  (3.93 A) 2 (3)  46.3W


2
2
P

I
r

(
3
.
93
A
)
(0.05)  0.772W
Internal
r

battery
P  PR  Pr  46.3W  0.772W  47.1W
10
Example 28.2 Matching the

Find the load resistance R for which the maximum
power is delivered to the load resistance
e
 e 
PR  I R  
 R
Rr
Rr
2

dPR
d  e  

 R  0

dR dR  R  r  
2
e (r  R)
0
Rr
3
(R  r)
I
2
2
11
Resistors in Series



When two or more resistors are connected end-toend, they are said to be in series
For a series combination of resistors, the currents
are the same in all the resistors because the amount
of charge that passes through one resistor must
also pass through the other resistors in the same
time interval
The potential difference will divide among the
resistors such that the sum of the potential
differences across the resistors is equal to the total
potential difference across the combination
12
Resistors in Series, cont




ΔV = IR1 + IR2
= I (R1+R2)
Consequence of
Conservation of Energy
The equivalent
resistance has the
same effect on the
circuit as the original
combination of resistors
13
Resistors in Series – Example

Observe the effect on the currents and voltages
of the individual resistors
14
Equivalent Resistance – Series



Req = R1 + R2 + R3 + …
The equivalent resistance of a series
combination of resistors is the algebraic
sum of the individual resistances and is
always greater than any individual
resistance
If one device in the series circuit creates
an open circuit, all devices are inoperative
15
Equivalent Resistance –
Series – An Example

Two resistors are replaced with their equivalent
resistance
16
Some Circuit Notes

A local change in one part of a circuit may
result in a global change throughout the
circuit



For example, changing one resistor will affect the
currents and voltages in all the other resistors and
the terminal voltage of the battery
In a series circuit, there is one path for the
current to take
In a parallel circuit, there are multiple paths
for the current to take
17
Resistors in Parallel



The potential difference across each resistor is the
same because each is connected directly across the
battery terminals
A junction is a point where the current can split
The current, I, that enters a point must be equal to
the total current leaving that point



I=I1+I2
The currents are generally not the same
Consequence of Conservation of Charge
18
Equivalent Resistance –
Parallel, Examples

Equivalent resistance replaces the two original
resistances
I  I1  I 2
DV
I
Req
DV  DV1  DV2
DV2
DV1
I2 
I1 
R2
R1
1
1 1
 
Req R1 R2
19
Equivalent Resistance –
Parallel

Equivalent Resistance
1
1
1
1




Req R1 R2 R3

The inverse of the
equivalent resistance of two
or more resistors connected
in parallel is the algebraic
sum of the inverses of the
individual resistance
 The equivalent is always
less than the smallest
resistor in the group
20
Resistors in Parallel – Example

Observe the effect on the currents and voltages of
the individual resistors
21
Resistors in Parallel, Final



In parallel, each device operates independently of
the others so that if one is switched off, the others
remain on
In parallel, all of the devices operate on the same
voltage
The current takes all the paths



The lower resistance will have higher currents
Even very high resistances will have some currents
Household circuits are wired so that electrical
devices are connected in parallel
22
Example 28.4 Combinations of
Resistors



The 8.0- and 4.0-
resistors are in series and
can be replaced with their
equivalent, 12.0 
The 6.0- and 3.0-
resistors are in parallel and
can be replaced with their
equivalent, 2.0 
These equivalent
resistances are in series
and can be replaced with
their equivalent resistance,
14.0 
23
Example 28.4 Combinations of
Resistors

What is the current in
each resistor if a potential
difference of 42V is
maintained between a
and c?
I  I1  I 2
42V  I 14  I  3 A
DVbc  I1  6  I 2  3
I 2  2I1
I1  1A I 2  2 A
24
Example 28.5 Three Resistors
in Parallel





Three resistors are connected
in parallel
A potential difference of 18V is
maintained between points a
and b
Calculated the equivalent
resistance of the circuit
Find the current in each
resistor
Calculate the power delivered
to each resistor and the total
power delivered to the
combination of resistors.
25
Example 28.5 Three Resistors
in Parallel

Calculated the equivalent
resistance of the circuit
1
1
1
1 1 1 1
 

  
Req R1 R2 R3 3 6 9
18
Req 
11

Find the current in each resistor
DV 18V
I1 

 6A
R1
3
DV 18V
I2 

 3A
R2
6
I3  2 A
26
Example 28.5 Three Resistors
in Parallel
• Calculate the power delivered to
each resistor and the total
power delivered to the
combination of resistors.
P1  I12 R1
P2  I 22 R2
P3  I R3
2
3
27
Gustav Kirchhoff




1824 – 1887
German physicist
Worked with Robert Bunsen
They



Invented the spectroscope
and founded the science of
spectroscopy
Discovered the elements
cesium and rubidium
Invented astronomical
spectroscopy
28
Kirchhoff’s Rules


There are ways in which resistors can be
connected so that the circuits formed cannot
be reduced to a single equivalent resistor
Two rules, called Kirchhoff’s rules, can be
29
Kirchhoff’s Junction Rule

Junction Rule

The sum of the currents at any junction must
equal zero



Currents directed into the junction are entered
into the -equation as +I and those leaving as -I
A statement of Conservation of Charge
Mathematically,  I  0
junction
30



I1 - I2 - I3 = 0
Required by
Conservation of Charge
Diagram (b) shows a
mechanical analog
31
Kirchhoff’s Loop Rule

Loop Rule


The sum of the potential differences across all
elements around any closed circuit loop must be
zero
 A statement of Conservation of Energy
Mathematically,
 DV  0
closed
loop
32



Traveling around the loop
from a to b
In (a), the resistor is
traversed in the direction of
the current, the potential
across the resistor is – IR
In (b), the resistor is
traversed in the direction
opposite of the current, the
potential across the resistor
is is + IR
33
Loop Rule, final


In (c), the source of emf is
traversed in the direction of
the emf (from – to +), and
the change in the electric
potential is +ε
In (d), the source of emf is
traversed in the direction
opposite of the emf (from +
to -), and the change in the
electric potential is -ε
34
Junction Equations from
Kirchhoff’s Rules

Use the junction rule as often as needed, so
long as each time you write an equation, you
include in it a current that has not been used
in a previous junction rule equation

In general, the number of times the junction rule
can be used is one fewer than the number of
junction points in the circuit
35
Loop Equations from
Kirchhoff’s Rules


The loop rule can be used as often as
needed so long as a new circuit element
(resistor or battery) or a new current appears
in each new equation
You need as many independent equations as
you have unknowns
36
Kirchhoff’s Rules Equations,
final


In order to solve a particular circuit problem,
the number of independent equations you
need to obtain from the two rules equals the
number of unknown currents
Any capacitor acts as an open branch in a
circuit

The current in the branch containing the capacitor is
37
Example 28.6 A Single-Loop
Circuit




A single-loop circuit contains
two resistors and two batteries
Find the current in the circuit
Kirchhoff’s loop rule:
Starting from a,
e1  IR1  e 2  IR2  0
6V  8I  12V  10 I  0
I  0.33 A
38
Example 28.7 A Multiloop
Circuit

Find the current in the circuit
Junction rule at junction c:
I1  I 2  I 3  0
Loop rule for abcda:
10V  (6) I1  (2) I 3  0
Loop rule for befcb:
 (4)I 2 14V  (6) I1  0
Solve I1, I2, and I3!
39
RC Circuits

In direct current circuit containing capacitors, the
current may vary with time


The current is still in the same direction
An RC circuit will contain a series combination of a
resistor and a capacitor
40
Charging a Capacitor



When the circuit is completed, the capacitor
starts to charge
The capacitor continues to charge until it
reaches its maximum charge (Q = Cε)
Once the capacitor is fully charged, the
current in the circuit is zero
41
Charging a Capacitor

Kirchhoff’s loop rule:
q
e   IR  0
C
e

Initially, I i 

Finally, Q  Ce

dq
e
q
I  
dt
R CR
R
42
Charging a Capacitor





dq
e
q
I  
dt
R CR
q
t
dq
dt
dq q  Ce
 


q  Ce
RC
dt
RC
0
0
t
 q  Ce 
ln 

RC
  Ce 
q(t) = Ce(1 – e-t/RC)
= Q(1 – e-t/RC)
ε t RC
I( t )  e
R
Time constant =RC
43
Charging an RC Circuit, cont.



As the plates are being charged, the potential
difference across the capacitor increases
At the instant the switch is closed, the charge on the
capacitor is zero
Once the maximum charge is reached, the current in
the circuit is zero

The potential difference across the capacitor matches that
supplied by the battery
44
Charging a Capacitor in an RC
Circuit

The charge on the
capacitor varies with
time


q(t) = Ce(1 – e-t/RC)
= Q(1 – e-t/RC)
The current can be
found
ε t RC
I( t )  e
R
  is the time constant

 = RC
45
Time Constant, Charging



The time constant represents the time
required for the charge to increase from zero
to 63.2% of its maximum
 has units of time
The energy stored in the charged capacitor is
½ Qe = ½ Ce2
46
Discharging a
Capacitor



Kirchhoff’s loop rule:
q
  IR  0
C
dq
q
I 
dt
CR
q
t
dq
dt
Q q  0 RC

q(t) = Qe-t/RC

dq
Q t RC
I t  

e
dt
RC
47
Discharging a Capacitor in an
RC Circuit

When a charged
capacitor is placed in
the circuit, it can be
discharged


q(t) = Qe-t/RC
The charge decreases
exponentially
48
Discharging Capacitor

At t =  = RC, the charge decreases to 0.368 Qmax



In other words, in one time constant, the capacitor loses
63.2% of its initial charge
The current can be found
dq
Q t RC
I t  

e
dt
RC
Both charge and current decay exponentially at a
rate characterized by  = RC
49
Example 28.9 Charging a
Capacitor in an RC Circuit



An uncharged capacitor and a
resistor are connected in series
to a battery where e=12V,
C=5F, and R=8105
The switch is thrown to position
a
Find the time constant of the
circuit, the maximum charge on
the capacitor, the maximum
current, and the charge and
current as functions of time
50
Example 28.9 Charging a
Capacitor in an RC Circuit


e=12V, C=5F, and R=8105
time constant:
=RC= 8105510-6 =4s

the maximum charge on the
capacitor:
Q=Ce= 5F  12V =60 C

the maximum current:
Ii= e/R=12V/8105=15 A
51
Example 28.10 Discharging a
Capacitor in an RC Circuit



Consider a capacitor of
capacitance C that is being
discharged through a resistor of
resistance R
After how many time constants
is the charge on the capacitor
one-fourth its initial value?
Q
 Qe t / RC
4
t  RC ln 4   ln 4
52
Example 28.10 Discharging a
Capacitor in an RC Circuit



The energy stored in the
capacitor decreases with time as
the capacitor discharges
After how many time constants
is this stored energy one-fourth
its initial value?
Equation 26.11 and 28.18:
q 2 Q 2 2t / RC
U (t ) 

e
2C 2C
1
ln 4
1 Q 2 Q 2  2t / RC
t  RC ln 4  

e
2
2
4 2C 2C
53
Example 28.11 Energy
Delivered to a Resistor


A 5-F capacitor is charged to a potential difference
of 800V and then discharged through a resistor
How much energy is delivered to the resistor in the
time interval required to fully discharge the capacitor?

Energy types: Electric potential and internal energy

1 2
DU+Eint=0  E R  U C  Ce
2

Solve ER!
54
```
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