Chapter 27 Subjects
Î Emf
devices (review)
Î Resistors
Î Power
in series and parallel
(examples)
Î Multiloop
circuits
Î Ammeters
Î RC
and voltmeters
circuits and time constant
PHY2049: Chapter 27
1
Emf devices
ÎWhat
do they do?
‹Maintain
potential difference by pumping
charge, which requires work. (Total
charge remains 0.)
ΓCharge
‹Don’t
up” a battery
be misled by this expression
PHY2049: Chapter 27
2
Simple Circuits
ÎOnly
one battery in the
circuit
ℇ
ℇ=V
ÎTwo
V
batteries in series
ℇ1+ ℇ2 = V
ÎBatteries
in parallel
ℇ2
ℇ1
To be useful and safe,
their ℇ must be the same. ℇ
PHY2049: Chapter 27
V
V
3
Resistors in Series
ÎLike
additional constriction. It
becomes harder for current to
flow.
ℇ
ÎMeasure of difficulty for
current to flow: R
ÎThus
R1
R2
Req = R1 + R2
Req = equivalent resistance
Req = R1 + R2 + R3 + ⋅ ⋅ ⋅
ℇ
Req
(in general)
(Read Section 27-5 for rigorous derivation)
PHY2049: Chapter 27
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Resistors in Parallel
ÎAdds
a bypass. It becomes
easier for current to flow.
ÎMeasure
of easiness for current
flow: 1/R (‘conductance’)
ÎTherefore,
R1
R2
1
1
1
=
+
Req R1 R2
Req = equivalent resistance
1
1
1
1
=
+
+
+L
Req R1 R2 R3
ÎReq
ℇ
ℇ
Req
(in general)
(Read Section 27-7 for rigorous derivation)
is smaller than the smallest
PHY2049: Chapter 27
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Resistors in Parallel
ÎAs
more resistors R are added in parallel to the circuit, how
does total resistance between points P and Q change?
‹(a) increases
‹(b) remains the same
‹(c) decreases
ÎIf
the voltage between P & Q is
held constant, and more resistors
are added, what happens to
the current through each resistor?
‹(a) increases
Overall current increases,
‹(b) remains the same
but current through each
branch is still V/R.
‹(c) decreases
PHY2049: Chapter 27
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Resistors in Parallel (continued)
1
1
1
=
+
Req R1 R2
ℇ
R1
R2
R1 R2
Req =
R1 + R2
(Engineers and physicists
memorize this formula.) ℇ
PHY2049: Chapter 27
Req
7
Parallel-Series Combo
ÎWhat
is the net resistance of the circuit
connected to the battery? Each resistance
has R = 3 kΩ
2
‹1&2
‹3
in
‹4 in
‹5 in
‹6 in
in series ⇒ 6 kΩ
parallel with 1&2 ⇒ 2 kΩ
1
series ⇒ 5 kΩ
parallel ⇒ 15/8 kΩ = 1.875 kΩ
series ⇒ 4.875 kΩ
R1 R2
Req =
R1 + R2
1
1
1
=
+
is faster than
Req R1 R2
PHY2049: Chapter 27
Req =
R1 R2
R1 + R2
3
4
5
6
8
Currents
ÎIf
the light bulbs are all the same in each of these two
circuits, which circuit has the higher current?
‹(a) circuit A
‹(b) circuit B
‹(c) both the same
ÎIn
which case is each light bulb
brighter?
‹(a) circuit A
‹(b) circuit B
‹(c) both the same
Current through each
branch is unchanged (V/R)
A
B
PHY2049: Chapter 27
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Currents
ÎCurrent
flows through a light bulb. If a wire is
now connected across the bulb as shown, what
happens?
‹(a)
All the current continues to flow through the bulb
‹(b) Current splits 50-50 into wire and bulb
‹(c) All the current flows through the wire
‹(d) None of the above
The wire “shunt” has almost no
resistance and it is in parallel with a
bulb having resistance.
Therefore all the current follows the
zero (or extremely low) resistance
path.
PHY2049: Chapter 27
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Currents
ÎThe
light bulbs in the circuits below are identical.
Which configuration produces more light?
‹(a) circuit I
‹(b) circuit II
‹(c) both the same
Circuit II has ½ current of
each branch of circuit I, so
each bulb is ¼ as bright.
The total light in circuit I is
thus 4x that of circuit II.
PHY2049: Chapter 27
11
Currents
ÎTwo
light bulbs A and B are connected in series
to a constant emf source. When a wire is
connected across B, what will happen to bulb A?
‹(a)
burns more brightly than before
‹(b) burns as brightly as before
‹(c) burns more dimly than before
‹(d) goes out
The wire shunt effectively eliminates the
second resistance, hence increasing the
current in the circuit by 2x. The first
bulb burns 4x brighter (I2 R).
PHY2049: Chapter 27
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Currents
ÎThe
three light bulbs in the circuit are identical.
The current flowing through bulb B, compared to
the current flowing through bulb A, is
‹(a) 4 times as much
‹(b) twice as much
‹(c) the same
‹(d) half as much
‹(e) 1/4 as much
Branch of circuit A has ½ resistance, thus it
has 2x current.
PHY2049: Chapter 27
13
Household Circuits
ÎAll
devices are added in
parallel.
‹Why?
ÎOverload:
too many devices
that require a lot of current
can draw more current than
wires can handle.
‹Overheating of wires
‹Fire hazard!
PHY2049: Chapter 27
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Power
ÎThe
three light bulbs in the circuit are identical.
What is the brightness of bulb B compared to bulb
A?
‹(a) 4 times as much
‹(b) twice as much
‹(c) the same
‹(d) half as much
‹(e) 1/4 as much
Use P = I2R. Thus 2x current in A means it is 4x brighter.
Use P = IV. 2x current and voltage. Same conclusion.
PHY2049: Chapter 27
15
Power (Problem solving)
ÎFind the value of R that maximizes
Very large R. Very small R. Find I or V
6Ω
I
2
I1
I2
12Ω
18V
1
R
I2
=
IT
1
R
P2 =
I 22 R
1
+ 12
=
=
2
power emitted by R.
P2 = I 22 R
12 R
R =
12 + R
18
12 + R
IT =
=
12 R
4+ R
6+
12 + R
R
12
12
⇒ I2 =
12 + R
4+ R
Maximize
144 R
(4 + R )
2
R = 4Ω P2 = 9 W
PHY2049: Chapter 27
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Power lines
ÎAt
large distances, the resistance of power lines
becomes significant. To transmit maximum
power, is it better to transmit high V, low I or
high I, low V?
‹(a)
high V, low I
‹(b) low V, high I
‹(c) makes no difference
Power loss is I2R, so want
to minimize current.
ÎWhy
do birds sitting on a high-voltage power line
survive?
‹They
are not touching high and low potential
simultaneously to form a circuit that can conduct
current
PHY2049: Chapter 27
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Concept Questions
Î Two
incandescent light bulbs—40 W and
100 W. Which has a smaller resistance?
Î 40
W light bulb made for US market. If
used in Argentina, will it be brighter,
dimmer, or no different?
Î Connect
a piece of copper wire to a
(constant) current source. What will
happen?
Î Connect
it to an emf source (battery or
power outlet). What will happen?
PHY2049: Chapter 27
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Circuits
ÎWhen
the switch S is closed, then:
‹What
happens to the voltage across each resistor?
‹What happens to the current through each resistor?
‹What happens to the total power output of the battery?
‹Let R1 = R2 = R3 = R4 = 90 Ω and V = 9 V. Find the
current through each resistor before and after closing
the switch.
Before
¾I1 = 9/135 = 1/15 A
¾I2 = 0
¾I3 = I4 = I1/2=1/30 A
After
¾I1 = 9/120 = 3/40 A
¾I2 = I3 = I4 = 1/40 A
PHY2049: Chapter 27
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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) nothing changes
Before: Potential at (a) is 12 V, but so
is potential at (b) because equal
resistance divides 24 V in half. When the
switch is closed, nothing will change
since (a) and (b) are still at same potential.
PHY2049: Chapter 27
(a)
(b)
20
Circuit Problem (2)
ÎThe
light bulbs in the circuit shown below are identical.
When the switch is closed, what happens to the intensity
of the light bulbs?
‹(a) bulb A increases
‹(b) bulb A decreases
‹(c) bulb B increases
(b)
‹(d) bulb B decreases
(a)
‹(e) nothing changes
Before: Potential at (a) is 12 V, but so
is potential at (b), because equal
resistance divides 24 V in half. When
the switch is closed, nothing will
change since (a) and (b) are still at
same potential.
PHY2049: Chapter 27
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Circuit Problem (3)
ÎThe
bulbs A and B have the same R. What happens when
the switch is closed?
‹a) nothing happens
‹b) A gets brighter, B dimmer
6V
‹c) B gets brighter, A dimmer
‹d) both go out
(a)
(b)
Before: Bulb A and bulb B both
have 9 V across them.
After: Bulb A has 6 V across it
and bulb B has 12 V across it
(these potential differences are
forced by the batteries).
PHY2049: Chapter 27
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Wheatstone Bridge
ÎAn
ammeter A is connected between points a and b in the
circuit below, in which two R1 are in series and so are two
R2. What is the current through the ammeter?
‹(a) I / 2
‹(b) I / 4
a
R1
‹(c) zero
R1
‹(d) need more information
c
Before the ammeter is connected, the
potential difference between points (a) and
(c) is 3 V, as is the potential difference
between points (b) and (c). Thus (a) and
(b) are at the same potential, and there
will be no current flow between them even
after the ammeter is connected.
PHY2049: Chapter 27
R2
b
R2
I
6V
23
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
¾R2 = 1202/100
= 288Ω (50W)
= 144Ω (100W)
PHY2049: Chapter 27
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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) 1
Each bulb draws a current of 1A.
Thus only 20 bulbs are allowed
‹(b) 5
before the circuit breaker is
‹(c) 10
tripped.
‹(d) 20
‹(e) 100
PHY2049: Chapter 27
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