Aljalal-Phys.102-3 May 2004-Ch28-page 1
Chapter 28
Circuits
Aljalal-Phys.102-3 May 2004-Ch28-page 2
28-1 Pumping Charges
Battery
Electric generator
Solar cell
devices that maintain potential
difference between their terminals by
doing work on charge carriers
These devices are called emf devices
Emf is an outdated
phrase which stands
for electromotive force
An emf device provides an emf E
Aljalal-Phys.102-3 May 2004-Ch28-page 3
28-2 Work, Energy, and EMF
Positive terminal
Higher potential
Internal chemistry of the
battery causes a net
flow of positive carriers
from the negative
terminal to the positive
terminal
E
Negative terminal
Lower potential
i
Battery
(Emf device)
i
R
Aljalal-Phys.102-3 May 2004-Ch28-page 4
28-2 Work, Energy, and EMF
Motion is
opposite to the
electric field
direction
→
E
There must be some
source of energy within
the source that enables
it to do work on the
charge carriers
Battery
Electric generator
Solar cell
Æ chemical energy
Æ mechanical energy
Æ light energy
Aljalal-Phys.102-3 May 2004-Ch28-page 5
28-2 Work, Energy, and EMF
dW
E =
dq
E
dq
The emf E of an emf device is the work
per unit charge that an emf device does
in moving charge from its lower-potential
terminal to its higher-potential terminal
Joule
E is measured in
= Volt
Coulomb
Aljalal-Phys.102-3 May 2004-Ch28-page 6
28-2 Work, Energy, and EMF
Ideal emf devices
have no internal resistance to the movement of charge
V
E
E
V
R
Not connected to a circuit
connected to a circuit
V=E
V=E
Potential difference between the terminals = the emf of the device
Aljalal-Phys.102-3 May 2004-Ch28-page 7
28-2 Work, Energy, and EMF
Real emf devices
have internal resistance to the movement of charge
V
E
E
V
R
Not connected to a circuit
connected to a circuit
V=E
V<E
Aljalal-Phys.102-3 May 2004-Ch28-page 8
28-2 Work, Energy, and EMF
Chemical energy is
increasing as energy is
transferred form the charge
carriers passing though it
EA
Ideal
rechargeable
battery
Electrical energy is
converted into work
Motor
Mass
i
Current direction is
determined by the
battery with the
larger emf
EB > EA
R
EB
Ideal
rechargeable
battery
Chemical energy is decreasing
as energy is transferred to the
charge carriers passing though it
Electrical energy
is dissipated as
heat
Aljalal-Phys.102-3 May 2004-Ch28-page 9
28-2 Work, Energy, and EMF
EA
i
Motor
Mass m
EB > EA
EB
Chemical energy lost by B
R
Thermal energy produced by
resistance R
Chemical energy stored in A
Work done by motor on
mass m
Aljalal-Phys.102-3 May 2004-Ch28-page 10
28-3 Calculating the Current in a Single-Loop Circuit
The energy method
i
E
R
In time dt, thermal
energy of
dE = i2 R dt = i R dq will
appear in the resistor
In time dt,
The battery will do
work of dW = E dq
on dq
Energy is conserved
dW = dE
E dq = i R dq
E =iR
We will not use this method to analyze circuits
Aljalal-Phys.102-3 May 2004-Ch28-page 11
28-3 Calculating the Current in a Single-Loop Circuit
b i
The Potential method
E
Higher
potential
R
a
Lower
potential
Find the change in the electric potential as you move around
the circuit
We will choose the clockwise direction
Let potential
at a
Va
Potential
at b
Va + E = Vb
Potential
at a
Va + E - i R = Va
+E -iR=0
E=iR
Aljalal-Phys.102-3 May 2004-Ch28-page 12
28-3 Calculating the Current in a Single-Loop Circuit
Direction does not affect the final result
b i
E
b i
Higher
potential
E
R
Lower
potential
a
Higher
potential
R
Lower
potential
a
Change in Potential
for clockwise
direction
Va + E - i R = Va
Change in Potential
for counterclockwise
direction
Va + i R - E = Va
+E -iR=0
+iR-E =0
E=iR
E=iR
Aljalal-Phys.102-3 May 2004-Ch28-page 13
28-3 Calculating the Current in a Single-Loop Circuit
b i
E
R
a
Loop Rule (Kirchhoff’s loop rule):
The algebraic sum of the changes in potential encountered
in a complete traversal of any loop of a circuit must be zero
∑ ∆V = 0
loop
Conservation of energy
Aljalal-Phys.102-3 May 2004-Ch28-page 14
28-3 Calculating the Current in a Single-Loop Circuit
i
Vi
R
Vf
i
Vi
Vi
Vi
R
E
E
Vf
Vf
Vf
Vf - Vi = - i R
Vf - Vi = + i R
V f - Vi = - E
Vf - Vi = + E
Aljalal-Phys.102-3 May 2004-Ch28-page 15
28-3 Calculating the Current in a Single-Loop Circuit
Checkpoint 1
i
a
c
B
R
b
Draw the emf arrow
E
At points a, b, and c rank …
The magnitude of current
All tie
The electric potential
Vb > Va = Vc
The electric potential energy
Ub > Ua = Uc
Aljalal-Phys.102-3 May 2004-Ch28-page 16
b
28-4 Other Single-Loop Circuits
i
Internal Resistance
r
R
E
Real battery
a
Potential (V)
E
a
r
i
R
b
a
E - i r - iR = 0
ir
E
iR
E
i=
r+R
Aljalal-Phys.102-3 May 2004-Ch28-page 17
28-5 Potential Differences
The potential difference between any two points in a circuit
does not depend on the path
i
b
i
b
r
R
r
E
R
E
a
a
Vb + i r - E = Va
Vb - i R = Va
Vb - Va = i R
E
Vb - Va =
R
r+R
Vb - Va = - i r + E
E
Vb - Va = r+E
r+R
E
Vb - Va =
R
r+R
Aljalal-Phys.102-3 May 2004-Ch28-page 18
28-4 Other Single-Loop Circuits
r=2Ω
E = 12 V
R = 10 Ω
b
b i
r
r
V
V
E
E
a
a
connected to a circuit
E
12
i=
=
=1A
r+R
2 + 10
V = E- i r
=12 - 1 (2) = 10 V < E
Not connected to a circuit
i=0
V = E - i r = E = 12 V
Aljalal-Phys.102-3 May 2004-Ch28-page 19
28-5 Potential Differences
Checkpoint 3
Compare E and V
b i
r
b
r
V
E
b
i
r
V
E
V
E
a
connected
a
connected
V = + E- i r
V =+E +ir
a
Not connected
V = + E + 0(r)
V<E
V>E
V=E
Aljalal-Phys.102-3 May 2004-Ch28-page 20
28-5 Potential Differences
Power, Potential, and Emf
Real emf
device
Rate of energy transfer form the
emf device to the charge carriers
b i
r
V
R
E
P=iV
P = i (E - i r)
P = i E - i2 r
a
Rate of energy transfer
form the emf device to the
charge carriers and to
internal thermal energy
P = Pemf - Pr
Rate of energy transfer to
internal thermal energy
Aljalal-Phys.102-3 May 2004-Ch28-page 21
28-5 Potential Differences
Sample Problem 28-1
a
i
battery 1
E1 = 4.4 V, E2= 2.1 V,
r1 = 2.3 Ω, r2 = 1.8 Ω, R = 5.5 Ω.
battery 2
r1
r2
E1
What is current i in the circuit?
Pick any direction for the current
Pick any direction for the loop rule
i r1 - E1 + i R + E2 + i r2 = 0
E1 - E2
i=
= 0.240 A = 240 mA
r1 + R + r2
E2
b
i
R c
If your guess is wrong,
your current will be
negative
Aljalal-Phys.102-3 May 2004-Ch28-page 22
28-5 Potential Differences
Sample Problem 28-1
i
a
battery 1
E1 = 4.4 V, E2= 2.1 V,
r1 = 2.3 Ω, r2 = 1.8 Ω, R = 5.5 Ω,
I = 240 mA.
E1
battery 2
r1
b
What is the potential difference
between the terminals of battery 1?
Va + i r1 - E1 = Vb
Va - Vb = - i r1 + E1 = 3.84 V
r2
E2
i
R c
Aljalal-Phys.102-3 May 2004-Ch28-page 23
28-5 Potential Differences
i
Sample Problem 28-1
battery 1
E1 = 4.4 V, E2= 2.1 V,
r1 = 2.3 Ω, r2 = 1.8 Ω, R = 5.5 Ω,
i = 240 mA.
E1
i r1 - E1 + i R + E2 + i r2 = 0
Potential (V)
a
i
r1
i r1
a
battery 2
r1
r2
E2
i
R c
E2
r2
b
E1
b
R
c
E1
E2
iR
a
i r2
Aljalal-Phys.102-3 May 2004-Ch28-page 24
28-4 Other Single-Loop Circuits
Resistances in Series
b i
E
R1
V1
V R2
V2
R3
V3
a
For steady flow of charge
Resistances are wired one after
another and a potential difference
is applied to the two ends of the
series
The applied potential difference V
is equal to the sum of the
potential differences across all the
resistances
V = V1+V2+V3
Since the charge is conserved, all
the resistances have the same
current i
i = i1= i2 = i3
Aljalal-Phys.102-3 May 2004-Ch28-page 25
28-4 Other Single-Loop Circuits
Resistances in Series
b i
E
R1
V1
V R2
V2
R3
V3
Req has the same current
i = i1 = i 2 = i 3
and the same total potential
difference
V = V1+V2+V3
as the actual resistances
Req = R1 + R2 + R3
a
b i
E
V
a
Req
Req =
n
∑R
j=1
j
n resistances in series
Aljalal-Phys.102-3 May 2004-Ch28-page 26
28-4 Other Single-Loop Circuits
Derivation of Req = R1 + R2 + R3
b i
E
R1
V1
V R2
V2
R3
V3
a
b i
E
V
a
Req
V = V1 + V2 + V3
i Req = i R1 + i R2 + i R3
Req = R1 + R2 + R3
Aljalal-Phys.102-3 May 2004-Ch28-page 27
28-4 Other Single-Loop Circuits
Resistances in Parallel
b i
i1
i2
E
V R1
R2
i3
R3
Resistances are directly wired
together on one side and
directly wired together on the
other side and a potential
difference is applied across
the pair of connected sides
a
All the resistances have the same potential difference
V = V1= V2= V3
Since the charge is conserved, the total current passing
through the resistances is equal to the sum of the current
passing through each resistance
i = i1+ i2 + i3
For steady flow of charge
Aljalal-Phys.102-3 May 2004-Ch28-page 28
28-4 Other Single-Loop Circuits
Resistances in Parallel
b
i
i1
E
V R1
i2
R2
i3
R3
a
b
E
V
a
i
Req
Req has the same total
current
i = i1 + i2 + i3
and the same potential
difference
V = V1= V2 = V3
as the actual resistances
1
1
1
1
=
+
+
Req
R1 R2 R3
1
=
Req
n
∑
j=1
1
Rj
n resistances in series
Req is smaller than any of the actual resistances
Aljalal-Phys.102-3 May 2004-Ch28-page 29
28-4 Other Single-Loop Circuits
1
1
1
1
=
+
+
Derivation of
Req
R1 R2 R3
b
i
i1
E
V R1
i2
R2
a
b
E
V
a
i3
R3
i = i1 + i2 + i3
i
Req
V
V
V
V
=
+
+
Req
R1
R2
R3
1
1
1
1
=
+
+
Req
R1 R2 R3
Aljalal-Phys.102-3 May 2004-Ch28-page 30
28-4 Other Single-Loop Circuits
i
E
Resistances in Series
R1
V1
i = i1 = i2 = i3
V = V1 + V2 + V3
V R2
V2
Req = R1 + R2 + R3
R3
V3
i
i1
E
V R1
i2
R2
i3
R3
Resistances in Parallel
i = i1 + i 2 + i3
V = V1= V2 = V3
1
1
1
1
=
+
+
Req
R1 R2 R3
Aljalal-Phys.102-3 May 2004-Ch28-page 31
28-4 Other Single-Loop Circuits
i
Sample Problem 2
E= 12 V,
R1 = 20 Ω, R2 = 20 Ω,
R3 = 30 Ω, R4 = 8.0 Ω,
What is current i through
the battery?
Simplify the circuit
R2R3
R23 =
= 12 Ω
R2 +R3
E
R4
i=
R2314
= 0.3 A
R2
R3
parallel
i
R1
E
R4
R23
series
R2314 = R1+R 4 +R23 = 40 Ω
E
R1
i
E
R2314
Aljalal-Phys.102-3 May 2004-Ch28-page 32
28-4 Other Single-Loop Circuits
i
Sample Problem 2
E= 12 V,
R1 = 20 Ω, R2 = 20 Ω,
R3 = 30 Ω, R4 = 8.0 Ω,
R23 = 12 Ω, i = 0.3 A
What is current i2 through R2?
Vba
i2 =
R2
E
i2 R
3
R1
R2
R4
a
R1
E
parallel
b
i
R23
R4
R2 and R3 are in parallel, potential
differences across R2 and R23 are equal
a
series
R1, R23 and R4 are in series, currents
thorough R2314 and R23 are equal
Vba = R23 i = 3.6 V
Vba
i2 =
= 0.18 A
R2
b
i
E
R2314
Aljalal-Phys.102-3 May 2004-Ch28-page 33
28-4 Other Single-Loop Circuits
i
Sample Problem 2
E= 12 V,
R1 = 20 Ω, R2 = 20 Ω,
R3 = 30 Ω, R4 = 8.0 Ω,
R23 = 12 Ω, i = 0.3 A,
i = 0.18, A,Vab = 3.6 A
What is current i3 through R3?
E
i3
R2
a
R1
E
parallel
b
i
R23
R4
a
series
Another way
i3 = i - i2 = 0.12 A
i2 R
3
R1
R4
Vba
i3 =
= 0.12 A
R3
i = i2 + i3
b
i
E
R2314
Aljalal-Phys.102-3 May 2004-Ch28-page 34
28-4 Other Single-Loop Circuits
b i
Checkpoint 2
E
R1
V1
V R2
V2
R3
V3
Let R1 > R2 > R3
Rank …
a
the current through the resistances
All tie
the potential difference across them
R 1 > R2 > R 3
i R 1 > i R 2 > i R3
V 1 > V2 > V3
Aljalal-Phys.102-3 May 2004-Ch28-page 35
28-6 Multiloop Circuits
junction a
branch
i1
R1
R3
a
i3
i2
E3
R2
E1
E2
i1
branch
R4
branch
i3
R5
b
junction b
Pick any direction for the currents
If your guess is wrong,
your current will be
negative. The actual
current is moving
opposite to your guess
Any point in a branch has the same current.
Aljalal-Phys.102-3 May 2004-Ch28-page 36
28-6 Multiloop Circuits
junction a
i1
R1
R3
a
i3
i2
E3
R2
E1
E2
i1
R4
i3
R5
b
junction b
Junction Rule (Kirchhoff’s junction rule):
The sum of the currents entering any junction must be
equal to the sum or the currents leaving that junction
∑i=∑i
in
out
Conservation of charge
for steady flow of charge
Aljalal-Phys.102-3 May 2004-Ch28-page 37
28-6 Multiloop Circuits
i1
R1
R3
a
i3
i2
E3
R2
E1
E2
i1
R4
i3
R5
b
To analyze a circuit, we need to find as many independent
equations as the number of unknowns in the circuit
Suppose R’s and E’s are given. we need to find i1, i2, and i3
Number of unknowns = 3
Number of independent equations = 3
Aljalal-Phys.102-3 May 2004-Ch28-page 38
28-6 Multiloop Circuits
junction a
i1
R1
R3
a
i3
i2
Loop
E1
E3
R2
Loop
1
E2
i1
2
R4
i3
R5
b
Junction rule at junction a
i3 = i1 + i2
Loop rule on loop 1
- i1R1 - E1 - i1R5 + E2 + i2R2 = 0
Loop rule on loop 2
- i2R2 - E2 - i3R 4 + E3 - i3R3 = 0
Aljalal-Phys.102-3 May 2004-Ch28-page 39
junction a
28-6 Multiloop Circuits
Sample Problem 28-3
i1
E1 = 3.0 V,
E2= E3 = 6.0 V,
Junction rule at junction a
Loop rule on loop 1
R3
a
i3
i2
R1 = R3 = R4 = R5 = 2.0 Ω,
E1
R2 = 4.0 Ω.
Find the magnitude and the
direction of the current in
each branch.
R1
Loop
E3
R2
Loop
1
E2
i1
2
i3
R5
b
i3 = i1 + i2
- i1R1 - E1 - i1R5 + E2 + i2R2 = 0
- 4 i1 + 3 + 4 i2 = 0
Loop rule on loop 2
- i2R2 - E2 - i3R 4 + E3 - i3R3 = 0
- 4 i2 - 4 i3 = 0
R4
Aljalal-Phys.102-3 May 2004-Ch28-page 40
28-6 Multiloop Circuits
Sample Problem 28-3
i3 = i1 + i2
- 4 i1 + 3 + 4 i2 = 0
-4 i2 - 4 i3 = 0
Eliminate one variable, say i3
- 4 i1 + 3 + 4 i2 = 0
- 4 i1 + 3 + 4 i2 = 0
- 4 i1 + 3 + 4 i2 = 0
- 4 i2 - 4 ( i1+i2 ) = 0
-8 i2 - 4 i1 = 0
i1
i2 = 2
Eliminate another variable, say i2
i1
- 4 i1 + 3 + 4 (- )= 0
2
- 6 i1 + 3 = 0
i1= 0.5 A
i2 = - 0.25 A
i3 = 0.25 A
Aljalal-Phys.102-3 May 2004-Ch28-page 41
28-6 Multiloop Circuits
Sample Problem 28-3
E1 = 3.0 V,
E2= E3 = 6.0 V,
i1
R1
i1= 0.5 A
i2 = - 0.25 A
i3 = 0.25 A
i3
i2
E3
R2
R1 = R3 = R4 = R5 = 2.0 Ω,
E1
R2 = 4.0 Ω.
Find the magnitude and the
direction of the current in
each branch.
R3
E2
i1
i2 i3
R5
Our guess for the direction of i2 is wrong,
it should be in the opposite direction.
Only after you finish finding all currents in a circuit,
you can correct your guesses abut directions
R4
Aljalal-Phys.102-3 May 2004-Ch28-page 42
28-6 Multiloop Circuits
junction a
i1
R1
R3
a
i3
i2
E3
R2
E1
E2
i1
R4
i3
R5
b
junction b
Junction rule at junction a
i3 = i1 + i2
Junction rule at junction b
i1 + i2 = i3
Same
No new information
Aljalal-Phys.102-3 May 2004-Ch28-page 43
28-6 Multiloop Circuits
i1
a
R 3 i2
R1
E3
R2
Loop
E1
Loop
1
E2
Loop3
i1
Loop rule on loop 1
Loop rule on loop 2
Add the two
equations
Loop rule on loop 3
R5
i3
b
2
R4
i3
- i1R1 - E1 - i1R5 + E2 + i2R2 = 0
- i2R2 - E2 - i3R 4 + E3 - i3R3 = 0
- i1R1 - E1 - i1R5 - i3R 4 + E3 - i3R3 = 0
Same as loop1 + loop2
No new information
You can use only two of the three loops
Aljalal-Phys.102-3 May 2004-Ch28-page 44
Charging a capacitor
28-8 RC Circuits
a
E
S
R
i
b
C
E
a
S
R
b
C
No
charge
initial
At t = 0, switch to position a
What is the charge on the capacitor as a function of time?
q
=0
E -iRLoop rule
C
dq
q
RE =0
Differential equation
dt
C
Aljalal-Phys.102-3 May 2004-Ch28-page 45
28-8 RC Circuits
Differential equation
dq
q
R=0
E dt
C
At t = 0, q = 0
Solution
q = CE (1-e-t/RC )
At t = ∞, q = CE
Charge q
q = CE
q = CE (1-e-1 )
= 0.63 CE
Time t
t =RC
Aljalal-Phys.102-3 May 2004-Ch28-page 46
28-8 RC Circuits
q
i=
q = CE (1-e-t/RC )
dq
dt
i
E -t/RC
i=
e
R
t
i
E
a
S
t
E
t= 0, i =
R
R
like connecting wire
b
C
t= ∞, i = 0
like broken wire
Aljalal-Phys.102-3 May 2004-Ch28-page 47
28-8 RC Circuits
q
V=
q = CE (1-e-t/RC )
q
C
VC
VC = E (1-e-t/RC )
t
i
E
a
S
t
t= 0, VC = 0
R
b
C
VC
t= ∞, VC = E
Aljalal-Phys.102-3 May 2004-Ch28-page 48
28-8 RC Circuits
RC has dimension of time
RC = τ
RC = capacitive time constant
RC
Dimension like
V Q
Q
Q
=
=
=t
Q
i V
i
t
q = CE (1-e-t/RC ) = CE (1-e-t/τ )
After one time constant
q = CE (1-e-1 ) = 0.63 CE
E -1
E
i=
e = 0.37
R
R
VC = E (1-e-1 ) = 0.63 E
Aljalal-Phys.102-3 May 2004-Ch28-page 49
28-8 RC Circuits
Show that
q = CE (1-e-t/RC )
dq
q
R=0
is a solution of E dt
C
Check the solution
At t = 0,
q = CE (1-e0/RC ) = CE (1-1) = 0
At t = ∞,
q = CE (1-e- ∞/RC ) = CE (1-0) = CE
dq
1 -t/RC
E -t/RC
= CE (
e
) =
e
dt
RC
R
dq
q
RE dt
C
E -t/RC
CE (1-e-t/RC )
=E e
RR
C
-t/RC
-t/RC
=E -Ee
- E (1-e ) = 0
Aljalal-Phys.102-3 May 2004-Ch28-page 50
Discharging a capacitor
28-8 RC Circuits
a
E
b
S
a
R
q0
C
E
R
S
b
C
i
initial
At t = 0, switch to position b
What is the charge on the capacitor as a function of time?
q
+iR =0
C
q dq
+
R=0
Differential equation
C dt
Aljalal-Phys.102-3 May 2004-Ch28-page 51
28-8 RC Circuits
Discharging a capacitor
a
E
R
S
b
C
i
At t = 0, switch to position b
Differential equation
dq
q
R+
=0
dt
C
At t = 0, q = q0
At t = ∞, q = 0
Solution
q = q0 e-t/RC
Aljalal-Phys.102-3 May 2004-Ch28-page 52
28-8 RC Circuits
Discharging a capacitor
a
E
R
S
b
C
i
i
q = q0 e-t/RC
dq
i=
dt
q0 -t/RC
i=e
RC
Our guess for the direction
of the current is wrong
Aljalal-Phys.102-3 May 2004-Ch28-page 53
28-8 RC Circuits
a
q = q0 e-t/RC
q0 -t/RC
i=
e
RC
E
S
R
b
C
i
After one time constant
q = q0 e-1 = 0.37 q 0
q
q
i = - 0 e-1 = 0.37 0
RC
RC
q0 -1
q0
VC =
e = 0.37
C
C
Aljalal-Phys.102-3 May 2004-Ch28-page 54
28-8 RC Circuits
Sample Problem 28-5
In terms of RC, when will the
charge on the capacitor be
half its initial value?
Discharging a capacitor
a
E
S
R
b
C
q = q0 e-t/RC
1
q0 = q0 e-t/RC
2
1
= e-t/RC
2
1
ln( ) = ln(e-t/RC )
2
t
ln(2) =
RC
At t = 0, switch to position b
t = ln(2)RC = 0.69 RC = 0.69 τ
Aljalal-Phys.102-3 May 2004-Ch28-page 55
28-8 RC Circuits
Sample Problem 28-5
Discharging a capacitor
a
R
S
In terms of RC, when will the
energy stored in the capacitor
b
E
be half its initial value?
C
2
q
q02 -2t/RC
U=
= U0 e-2t/RC
=
e
2C
2C
1
U0 = U0 e-2t/RC
At t = 0, switch to position b
2
1
= e-2t/RC
2
2t
ln(2) =
RC
ln(2)
t=
RC = 0.35 RC = 0.35 τ
2
Aljalal-Phys.102-3 May 2004-Ch28-page 56
28-8 RC Circuits
Charging a capacitor
a
R
i
Checkpoint 5
Rank the sets according to …
A - initial current
B - time required for the
current to decrease to its
half its initial value
E(V)
12
12
10
10
R(Ω)
2
3
10
5
S
b
E
C
At t = 0, switch to position a
C(µF)
3
2
0.5
2
A
1
2
4
3
E
i=
R
B
2
2
4
1
t = 0.69 RC