Wheatstone Bridge
ÎAn
ammeter A is connected between points a and b in the
circuit below, in which the four resistors are identical.
What is the current through the ammeter?
‹ a)
I/2
‹ b) I / 4
‹ c) zero
‹ d) need more information
The parallel branches have the same
resistance, so equal currents flow in
each branch. Thus (a) and (b) are at
the same potential and there is no
current flow across the ammeter.
PHY2054: Chapter 18
34
Circuit Problem (1)
ÎThe
light bulbs in the circuit are identical. What happens
when the switch is closed?
‹ (a)
both bulbs go out
‹ (b) the intensity of both bulbs increases
‹ (c) the intensity of both bulbs decreases
‹ (d) A gets brighter and B gets dimmer
‹ (e) nothing changes
Before: Potential at (a) is 24V, but so
is potential at (b) because equal
resistance divides 48V in half. When the
switch is closed, nothing will change
since (a) and (b) are still at same potential.
PHY2054: Chapter 18
35
Circuit Problem (2)
ÎBulbs
A and B are identical. What happens when the
switch is closed?
‹ (a)
‹ (b)
‹ (c)
‹ (d)
‹ (e)
nothing happens
A gets brighter, B dimmer
A gets dimmer, B brighter
Both A and B get brighter
Both A and B get dimmer
Before: Bulb A and bulb B both have
9V across them.
After: Bulb A has 12V across it and
bulb B has 6V across it (these
potentials are forced by the batteries).
PHY2054: Chapter 18
36
Calculate Currents (Not Simplest Way)
i2 = i1 + i3
1
+12 − 3i2 = 0
2
+6 − 3i1 = 0
i2 = 4
i1 = 2
i3 = 4 − 2 = 2
PHY2054: Chapter 18
37
Res-Monster Maze
All batteries are 4V
All resistors are 4Ω
Find current in R. (Hint: Find voltages
along path not connected by resistors)
PHY2054: Chapter 18
38
Res-Monster Maze
2
All batteries are 4V
All resistors are 4Ω
Find current in R. (Hint: Find voltages
along path not connected by resistors)
PHY2054: Chapter 18
39
Res-Monster Maze (Part 2)
All batteries are 4V
All resistors are 4Ω
Find currents across
these resistors
PHY2054: Chapter 18
40
First find currents in resistors using
potentials established by batteries
1
1
1
0
1
1
0
3
1
1
2
1
2
1
1
1
0
0
0
0
1
PHY2054: Chapter 18
41
Then use junction rules to find other currents
8
7
7
6
4
1
1
1
0
1
1
2
3
1
1
0
1
1
1
1
2
1
1
3
1
1
0
8
1
2
4
0
0
0
0
1
PHY2054: Chapter 18
42
Continue using junction rules until finished
8
7
7
6
4
1
1
1
3
1
1
0
8
1
9
9
1
1
1
11
1
1
2
4
12
1
2
2
1
0
1
1
3
10
1
0
3
1
9
0
2
0
0
0
1
10
9
9
PHY2054: Chapter 18
43
Light Bulbs
ÎA
three-way light bulb contains two filaments that can be
connected to the 120 V either individually or in parallel.
‹A
three-way light bulb can produce 50 W, 100 W or 150 W, at the
usual household voltage of 120 V.
‹ What are the resistances of the filaments that can give the three
wattages quoted above?
Use P = V2/R
¾R1 = 1202/50 = 288Ω (50W)
¾R2 = 1202/100 = 144Ω (100W)
PHY2054: Chapter 18
44
Problem
ÎWhat
is the maximum number of 100 W light bulbs you
can connect in parallel in a 100 V circuit without tripping a
20 A circuit breaker?
‹ (a)
‹ (b)
‹ (c)
‹ (d)
‹ (e)
1
5
10
20
100
Each bulb draws a current of 1A. Thus
only 20 bulbs are allowed before the
circuit breaker is tripped.
PHY2054: Chapter 18
45
Find Currents Through All Resistors
60
Fig. 18-8
PHY2054: Chapter 18
46
Find Currents (2)
ÎFind
total current (need equivalent R)
‹ 10.00Ω
+ 5.00Ω (parallel)
‹ 3.33Ω + 4.00Ω (series)
‹ 7.33Ω + 3.00Ω (parallel)
‹ 2.13Ω + 3.00Ω (series)
‹ I = 60 / 5.13 = 11.70 A
Î 3.33Ω
Î 7.33Ω
Î 2.13Ω
Î 5.13Ω
ÎCurrent
through outside
3.00Ω resistor = 11.70 A.
I
PHY2054: Chapter 18
60
47
Find Currents (3)
ÎTotal
current I = 11.70, find current I2
‹ One
approach, find VA – VB
‹ VB = 0, VA = 60 – 3.00*11.70 = 24.90
‹ I2 = 24.9 / 3.00 = 8.30 A
C
I1
A
I
PHY2054: Chapter 18
B
I2
60
48
Find Currents (4)
ÎFind I1
‹ I1
ÎFind
= I – I2 = 3.40
remaining currents
ΔV = VA - VC
‹ VC = 0 + 3.40*4.00 = 13.60
‹ ΔV = 24.90 – 13.60 = 11.30
‹ I10 = 11.30 / 10 = 1.13
‹ I5 = 11.30 / 5 = 2.26
‹ Find
C
I1
A
I
PHY2054: Chapter 18
B
I2
60
49
Find Currents (5)
ÎFind
remaining currents
‹ Use I1 = 3.40
‹ Find ΔV = VA -
VC
‹ VC = 0 + 3.40*4.00 = 13.60
‹ ΔV = 24.90 – 13.60 = 11.30
‹ I10 = 11.30 / 10 = 1.13
‹ I5 = 11.30 / 5 = 2.26
C
I1
A
I
PHY2054: Chapter 18
B
I2
60
50
RC Circuits
ÎCharging
a capacitor takes time in a real circuit
‹ Resistance
allows only a certain amount of current to flow
‹ Current takes time to charge a capacitor
ÎAssume
uncharged capacitor initially
‹ Close
switch at t = 0
‹ Initial current is i = V
ÎCurrent
/ R (no charge on capacitor)
flows, charging capacitor
‹ Generates
capacitor potential of q/C
ΔVresistor = V − q / C = Ri
ÎCurrent
‹ Goes
i
V
decreases continuously as capacitor charges!
to 0 when fully charged
PHY2054: Chapter 18
51
Analysis of RC Circuits
ÎCurrent
ÎSo
and charge are related
i = dq / dt
can recast previous equation as “differential equation“
dq
q
V
+
=
dt RC R
V − q / C = Ri
ÎGeneral
−t / RC
solution is q = CV + Ke
‹ (Check
and see!)
‹ K = −CV (necessary to make q = 0 at t = 0)
ÎSolve
for charge q and current i
(
q = CV 1 − e −t / RC
)
i=
dq V −t / RC
= e
dt R
PHY2054: Chapter 18
52
Charge and Current vs Time
(For Initially Uncharged Capacitor)
(
q ( t ) = q0 1 − e
i ( t ) = i0e
PHY2054: Chapter 18
−t / RC
)
−t / RC
53
RC Example
ÎLet
V = 12 volts, C = 6μF, R = 100Ω
‹τ
= RC = 600μs
(
)
(
−t / RC
−t / 600
q
=
CV
1
−
e
=
72
1
−
e
ÎCharge vs time
‹ Using
ÎCurrent
‹ Using
ÎThis
units of μC and μsec
vs time i =
)
V −t / RC
e
= 0.12e−t / 600
R
units of amps and μsec
circuit will fully charge in a few millisec
PHY2054: Chapter 18
54
Exponential Behavior
Ît
= RC is the “characteristic time” of any RC circuit
‹ Only
Ît
t / RC is meaningful
= RC
‹ Current
falls to 37% of maximum value
‹ Charge rises to 63% of maximum value
Ît
=2RC
‹ Current
falls to 13.5% of maximum value
‹ Charge rises to 86.5% of maximum value
Ît
=3RC
‹ Current
falls to 5% of maximum value
‹ Charge rises to 95% of maximum value
Ît
=5RC
‹ Current
falls to 0.7% of maximum value
‹ Charge rises to 99.3% of maximum value
PHY2054: Chapter 18
55
Discharging a Capacitor
ÎConnect
fully charged capacitor to a resistor at t = 0
q
−iR − = 0
S
C
R
C
dq
q
+
=0
dt RC
ÎSolution
ÎSolve
is q = q0e
−t / RC
= CVe −t / RC
for current i
q, i fall exponentially
dq V −t / RC
= e
= i0e−t / RC
i=
dt
R
PHY2054: Chapter 18
56
Charge and Current vs Time
(For Initially Charged Capacitor)
q ( t ) = q0e
−t / RC
i ( t ) = i0e
−t / RC
PHY2054: Chapter 18
57
Same RC Example
ÎC
= 6μF, R = 100Ω
‹τ
= RC = 600μs
ÎCharge
capacitor to 12 volts, then pull out battery and let
the RC circuit discharge
ÎCharge
‹ Using
vs time q = CVe−t / RC = 72e−t / 600
units of μC and μsec
V −t / RC
e
= 0.12e−t / 600
ÎCurrent vs time i =
R
‹ Using
ÎThis
units of amps and μsec
circuit will fully discharge in a few millisec
PHY2054: Chapter 18
58