Chapter 20, Electric Circuits
(direct current only)
Electric Current
The flow of electric charge
• Electromotive force, resistance, Ohm’s law, power
I=
• Series and parallel wiring
• Internal resistance
!q
amps (A), 1 A = 1 C/s
!t
V, the voltage of the battery
or power supply.
• Electric circuits – Kirchhoff’s rules
The maximum potential
difference appearing across
the terminals is termed the
“electromotive force”, or
emf (volts).
• Measurement of voltage and current
• Omit 20.5, 12, 13 (alternating current, capacitors, RC circuits)
V
The emf is measured when negligible current is being supplied by the
battery. It can be measured by a voltmeter with high resistance.
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20.1
2
Ohm’s Law
Prob. 20.102/1: A FAX machine uses 0.11 A of current in normal mode,
0.067 A in standby mode. The machine operates using a potential
difference of 120 V.
a) How much charge flows in 1 minute in normal and standby modes?
The current flowing around a circuit is
proportional to the voltage applied.
b) How much more energy is used in 1 minute in normal mode?
That is, I ! V
And, V = IR (Ohm’s Law)
R = resistance, in ohms (!)
• Current is the rate of flow of charge
If I = 0.4 A when V = 3 V, then
• How much PE does a charge lose in travelling from + terminal to – ?
“Conventional”
flow of current
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R = V/I = (3 V)/(0.4 A) = 7.5 !
4
Conduction of heat
Resistivity
Rate of flow of heat:
Q=
The resistance of a piece of wire of length, L, and cross-sectional
area, A, is
kA
∆T
L
R=
k = thermal conductivity
!L
A
! is the “resistivity” of the material of the wire
Conduction of charge
!=
Rate of flow of charge:
I=
RA ".m2
=
= ".m
L
m
A
∆V
ρL
! = electrical resistivity
Electrical conductivity = 1/!
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Table of Resistivities
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Prob. 20.11/13: A cylindrical copper cable carries a current of 1200 A.
There is a potential difference of 0.016 V between two points on the
cable that are 0.24 m apart.
What is the radius of the cable?
[Resistivity of Cu = 1.72 " 10-8 !.m]
• What is the resistance of 0.24 m of the cable?
#
V = IR...
The resistivity varies with temperature
20.13
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Test for deep vein thrombosis – how fast does leg
recover?
Variation of resistivity with temperature
Measure the resistance of
part of the calf:
!L
!L
R=
=
A
Vcalf/L
!L2
R=
Vcalf
! = !0[1 + "(T − T0)]
Resistivity = !0 when temperature = T0.
Volume V,
cross sectional
area A
" = temperature coefficient of resistivity (or resistance)
Vcalf = volume of calf of length L
=LA
Same relation holds for resistance:
R = R0[1 + !(T − T0)]
Inflate cuff to cut off blood flow from
the leg, but not to it – volume of calf
increases, R drops.
as R = !L/A
Release cuff, volume and resistance return to normal, but how quickly?
Should be fast.
Similar to variation of length or volume with temperature.
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20.15
9
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10
Prob. 20.105/15: A platinum resistance thermometer has a resistance
of 125 ! at 20º C. When immersed in boiling chlorine, its resistance
drops to 99.6 !.
Prob. 20.19: Two wires (W, Cu) have the same cross-sectional area.
They are joined end to end to form a single wire. The total resistance is
the sum of the resistances of the pieces.
The temperature coefficient of resistance of Pt is " = 0.00372 ºC-1.
The total resistance does not change with temperature. What is the ratio
of the lengths?
What is the temperature of boiling chlorine?
W: # !01 = 5.6 " 10-8 !.m, # "1 = 0.0045 ºC-1
• How does resistance change with temperature?
Cu:# !02 = 3.5 " 10-5 !.m,#
"2 = –0.0005 ºC-1
• What is the sum of the resistances?
• To make the total resistance independent of temperature, the sum of
# the terms involving temperature must be zero...
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Electrical Power
Prob. 20.-/100: A car battery is being charged at a voltage of 12 V and a
current of 19 A.
A charge #q in falling through a potential difference V loses
potential energy V#q.
How much power is being produced by the charger?
Power, P = VI = (12 V) " (19 A) = 228 W.
This energy is supplied by the battery or power supply.
I=
The current flow is:
!q
!t
Prob. 20.28: A piece of nichrome wire has a radius of 0.65 mm. It
dissipates 400 W of power when connected to a 120 V DC supply.
The rate at which energy is delivered by the supply is:
P=
Estimate the length of wire.
V ∆q
= VI
∆t
! = 10-6 !.m
which is the power supplied.
Ohm’s law: V = IR, so P = VI = V2/R = I2R
• From P and V, what is the resistance of the wire?
20.100
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Chapter 20 so far...
14
Prob. 20.26/28: A piece of nichrome wire has a radius of 0.65 mm. It
dissipates 400 W of power when connected to a 120 V DC supply.
Ohm’s law
Estimate the length of wire.
V = IR
! = 10-6 !.m
Resistance, resistivity
R=
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!L
A
• From P and V, what is the resistance of the wire?
Temperature dependence
! = !0[1 + "(T − T0)]
R = R0[1 + !(T − T0)]
Power
P = VI = V2/R = I2R
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Experiment 2: Wheatstone Bridge
Experiment 2: Wheatstone Bridge
The object is to verify that the power radiated by a heated object is
proportional to T4, where T is the temperature of the object in Kelvin.
The filament of a lamp is heated by passing a current through it. The
filament radiates away power as electromagnetic radiation (visible
light + infrared). At a steady state, the power radiated is equal to the
electrical power supplied.
r
Va
Galvanometer. Reads zero when:
D
R1 R4 (then VA = VB,
Rx =
R2
theory later)
V2
Then, P =
Rx
Voltmeter
C
The electrical power supplied is P = V2/R. This is taken to be equal to
the radiated power.
V
The electrical resistance, R, of the filament is measured with a
Wheatstone bridge. V is measured with a voltmeter.
B
From Rx, find temperature of the
filament from graph provided.
Is P proportional to T4 ?
The temperature of the filament is determined from its resistance.
R = R0 [1 + $(T - T0)]
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A
es
er
l
iab
e
nc
a
t
is
17
Prob. 20.C6:
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Which bulb filament has the greater resistance, 75 W or 150 W?
Prob. 20.27: A tungsten wire is connected to a source of constant
voltage via a switch. At the instant the switch is closed, the temperature
of the wire is 28º C and the initial power dissipated in the wire is P0.
P = VI = V2/R
At what wire temperature has the power dissipated in the wire fallen to
P0/2?
so large R $ small P
" = 0.0045 ºC-1
Prob. 20.C2: The filament of a lamp is made of a tungsten wire. As the
filament heats up, does the power increase or decrease? The temperature
coefficient of resistance is positive.
• How does the power dissipated in the wire vary with the resistance of
# the wire?
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• How does the resistance of the wire vary with temperature?
• How does the power vary with resistance for constant V?
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Resistors in Series
I
R1
V1 = IR1
Resistors in Parallel
R2
R1 + R2
V2 = IR2
I = 12/9 = 1.33 A
The potential difference between A and B is:
V = I1R1 = I2R2
I1 =
V=
V=
So, I =
Total potential difference:
V = V1 + V2 = I " (R1 + R2) = IRs
I = V/Rs, where Rs = R1 + R2
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Resistors in Parallel
I2
A
I
I1
B
"
1
1
V
+
=
R1
R2
Rp
A
B
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Prob. 20.54/52: A wire of resistance R is cut into three equally long pieces,
which are then connected in parallel.
In terms of R, what is the resistance of the parallel combination?
I=
V
6
=
= 2.25 A
R p 8/3
• What is the resistance of each piece?
I1 = 6/8 = 0.75 A
• What is the equivalent parallel resistance of the pieces?
I
I
!
A B
The circuits are equivalent
1
1
1
=
+
Rp
R1
R2
result in the same current.
I2 = 6/4 = 1.50 A
V
V
+
=V
R1
R2
Rp is the equivalent parallel resistance:
A single resistance Rs = R1 + R2 (equivalent series resistance) would
1
1 1 3
= + =
Rp 4 8 8
8
Rp = !
3
I1
V
V
, I2 =
R1
R2
and I = I1 + I2 (current conserved).
The same current I passes through both resistors (current is conserved
and has nowhere else to go).
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I2
and so,
I
I = I1 + I2 = 0.75 + 1.50 = 2.25 A
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Combination of Series and Parallel Resistors
I
I = V/R = 24/240 = 0.1 A
Original circuit
1
1
1
=
+
R p 180 470
I = 0.1 A
I2
I1
Potential drop across 110 ! resistor is:
V = IR = (0.1 A)(110 !) = 11 V.
Therefore VAB = 24 – 11 = 13 V.
VAB = I1 " 180, I1 = 13/180 = 0.0722 A
VAB = I2 " 470, I2 = 13/470 = 0.0277 A
I = I1 + I2 = 0.0999... = 0.1 A
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Series and Parallel Resistances
I = 0.1 A
I1
I2
I1 = 0.0722 A
R1
I2 = 0.0277 A
R2
=
Rs
Resistors in series: Rs = R1 + R2 + . . .
The same current passes through the resistors
Power dissipated in 180 ! resistance is I12R = 0.07222 × 180 = 0.94 W
=
Resistors in parallel:
Rp
1
1
1
=
+
+ ...
Rp
R1
R2
Same potential difference across the resistors
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Prob. 20.C12: In one of the circuits, none of the resistors is in series or in
parallel. Which one?
Prob. 20.58: What is the equivalent resistance between points A and B?
• Resistances must be connected directly end to end with nothing
# between for them to be in series or in parallel
2!
6!
4!
1!
3!
2!
3!
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Prob. 20.63: The five resistors are identical. The battery delivers 58 W of
power to the circuit. What is the resistance R of each resistor?
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Prob. 20.-/54: When two resistors are connected in series with a 12 V
battery the current from the battery is 2 A. When they are connected in
parallel the current is 9 A.
Determine the values of the two resistances, R1, R2.
R
I1 = 2 A
R
2R
R
V = 12 V
Series
1
1
1
1
5
= + + =
R p R 2R R 2R
• What is the equivalent resistance in terms of R?
• What does R need to be so that 58 W is dissipated in the equivalent resistance?
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I2 = 9 A
31
Rs = R1 + R2
Rs =
V 12
=
=6Ω
I1
2
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Rs
V = 12 V
Rp
Parallel
1
1
1
= +
R p R 1 R2
V 12 4
Rp = =
= Ω
I2
9
3
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Internal Resistance, Terminal Potential Difference
Terminal Potential Difference
Load
Load (lights, starter, etc)
Internal resistance, r, increases as a
battery gets older – corrosion, etc.
I
I
I
I
Suppose r = 0.01 ! and V = 12 V and
the starter motor draws 100 A of current.
TPD
Then, the terminal potential difference is
reduced by:
TPD
Internal resistance of battery, r
I ! r = 100 ! 0.01 = 1 V.
Terminal Potential Difference, TPD = IR and V = I(r + R),
V
R
so, I =
and TPD = V
emf = V
r+R
r+R
TPD
The internal resistance decreases the
voltage available when the battery is
supplying current.
The potential difference that appears across the terminals of the battery
is reduced by the internal resistance of the battery.
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The TPD is 12 – 1 = 11 V.
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Prob. 20.112: 75 ! and 45 ! resistors are connected in parallel. When
connected to a battery, the current delivered is 0.294 A.
A 1.2 ! resistor is connected across a battery. If the battery had no
internal resistance, the power dissipated in the resistor would be Po.
When the 45 ! resistor is removed, the current drops to 0.116 A.
The battery does have an internal resistance of 0.06 !, so the power
dissipated in the resistor is P.
Find the emf and the internal resistance of the battery.
Find P/Po.
• Draw diagrams!
• For an internal resistance r and an emf V, write expressions involving
# the current for the two cases
• The power dissipated in the resistor is I2R, so work out I for the two
# cases for an emf of V...
# $ simultaneous equations in V and r
• Solve for V and r.
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Kirchhoff’s Rules
What is the equivalent resistance?
1) Junction Rule:
# The sum of the currents entering a junction
# is equal to the sum of currents leaving it.
# 7=5+2
Conservation of charge and current
2) Loop Rule:
# The sum of potential changes
# around any closed loop is zero.
Potential drops as you go with the
flow of current through a resistor.
No series or parallel combinations
12 – 2"5 – 2"1 = 0
Conservation of energy
You need Kirchhoff’s rules!
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Friday, January 26, 2007
Midterm Test
Loop Rule:
The water apparently always
flows downhill, and arrives
back where it started and
flows forever in a loop with
no energy input.
Thursday, March 1, 7 - 9 pm
Armes Lecture Theatres
(seating by student number to follow)
This cannot be, as there is no
pump to raise the potential
energy of the water back to
its initial value.
Chapters 18 – 25
20 multiple choice questions
Formula sheet provided
Likewise, electrons flowing
around a closed loop cannot
flow forever unless a power
source pumps them back up
to their initial potential.
There is NO deferred exam for this test
Final exam will count for 70% if you have to
miss the midterm (see me)
Thanks to Escher
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Kirchhoff’s Rules
Prob. 20.77: Determine the voltage across the 5 ! resistor. Which end
is at higher potential?
1) Junction Rule:
! The sum of currents entering a junction is equal to the sum of
# currents leaving it (conservation of charge and current).
I1 B I1 + I2
A
2) Loop Rule:
# Around any closed loop, the sum of potential changes is equal
# to zero (conservation of energy).
C
I2
I1
I1 + I2
# ! a charge %q flowing around any closed loop returns to the
# # potential energy at which it started.
# ! you don’t get something (energy) for nothing.
The potential decreases when you go with the flow of current
through a resistor.
F
# – use the “conventional” flow of current
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I2
I1
41
I1 + I2
E
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D
42
What is the equivalent resistance, Requiv ?
Wheatstone Bridge
R1
R2
Rx
R4
Requiv
Requiv
=
I
I
V
V = IRequiv
So, Requiv
V
V
=
I
Find I by applying Kirchhoff’s rules to the original circuit
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It’s a variation on the Wheatstone bridge
43
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44
20.79: Wheatstone Bridge
Find potential difference between
B and D
I1
20.104: Find the current through the 2 ! resistor and the voltage V of
the battery to the left of it.
I2
I1 – I2
What is the equivalent
resistance between A and C?
A
I2
I1
B
I – I1
I1 – I2
I
I – I1
3A
F
I
I – I2
I – I2
I
F
3–I
3A
C
I
I
E
D
E
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46
Mode of operation of an ammeter to measure current
A galvanometer
Full scale deflection (fsd) = 0.1 mA
(for example)
Meter
The ammeter is inserted into the
circuit to measure the current
flowing around the circuit.
Rc
R
Ideally, the ammeter should have
negligible resistance so as not to
affect the current being measured.
Resistance of galvanometer coil
A torque is exerted on the coil when a current flows through it. The rotation
of the coil is resisted by a spring. The angle of rotation is proportional to
the current in the coil.
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48
Measurement of current –&switching scales
Ammeter – a bypass (shunt) resistor is placed in
parallel with a galvanometer to limit the current
through the galvanometer to no more than that for
full scale deflection (fsd)
Prob. 20.107/83: A galvanometer has a coil resistance of 12 ! and a full
scale deflection current of 0.15 mA. It is used with a shunt resistor to
make an ammeter that registers 4 mA at full scale deflection.
Find the equivalent resistance of the ammeter.
• Work out what is the potential difference across the galvanometer
# when 0.15 mA flows through it.
To measure 60 mA when the fsd is 0.1 mA:
VAB = 0.1RC = 59.9R (mV)
galvanometer
(fsd)
• Find what the shunt resistance needs to be to produce an equal
# potential difference when the current flowing through the shunt resistor
# is 4 – 0.15 mA.
So, for the shunt resistor:
0.1
R = RC
= 0.00167RC
59.9
To measure current I: R = RC
fsd
I − fsd
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Mode of operation of a voltmeter to measure voltage
50
Measurement of voltage
Coil resistance
The voltmeter is attached to two points
of a circuit between which the potential
difference is to be measured.
galvanometer
Ideally, the voltmeter should have very
large resistance so as to draw very little
current from the circuit being studied.
I f sd = 0.1 mA
V = 100 V
R
RC = 50 !
I = 0.1 mA
A resistor is put in series with the
galvanometer to limit the current to
that giving full scale deflection (fsd)
at the desired voltage.
To measure 100 V when fsd is at 0.1 mA and the coil resistance RC = 50 !:
(R = resistance put in series with the galvanometer
to convert it to a voltmeter)
V
100 V
That is, R =
− RC =
− (50 !) = 999, 950 !
I f sd
0.1 × 10−3 A
V = I f sd (R + RC )
R ! 1 M!
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20.82
52
Prob. 20.82: Voltmeter A has an equivalent resistance of 2.4 " 105 ! and
a full scale voltage of 50 V.
Prob. 20.85: Consider a circuit consisting of two 1550 ! resistors
connected in series across a 60 V battery.
Voltmeter B uses the same galvanometer. It has an equivalent resistance
of 1.44 " 105 !. What is its full scale voltage?
a) Find the voltage across one of the resistors.
b) A voltmeter has a fsd voltage of 60 V and uses a galvanometer with a
# full scale current of 5 mA.
# Determine the voltage this voltmeter registers when it is connected
# across the resistor in part a).
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54
Puzzle: what is the equivalent resistance between A and B?
Symmetry! The circuit looks the same backwards as forwards!
3R
I1
I
A
I1
R
R
R
I – I1
I – 2I1
I – I1
I1
Requiv =
VAB
I
3R
I
B
The hazard of an ungrounded power plug
I1
Apply loop rule around one loop to find I1 in terms of I
Calculate VAB in terms of the currents and resistances along one path
from A to B – all paths should give the same result
Can also ignore symmetry and use 2 loops to find the currents.
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56
Grounded three-prong plug
Chapter 20
• Electric current:
• Resistance:
#
Ohm’s law
#
resistivity, temperature coefficient of resistance
#
measurement of temperature
• Power:
#
P = VI and alternative forms from Ohm’s law
• Electric circuits:
#
series and parallel resistors
#
Kirchhoff’s rules
#
ammeters and voltmeters
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Friday, January 26, 2007
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