WileyPLUS Assignment 1
Chapters 18, 19
10 Questions, mostly “GO” with tutorial assistance
Due Wednesday, January 27 at midnight
Week of Jan 26 – 28
Experiment 2: Wheatstone Bridge
Wednesday, January 20, 2010
1
WileyPLUS Marking Scheme
Wednesday, January 20, 2010
2
Chapter 20, Electric Circuits
(direct current only)
• Electromotive force, resistance, Ohm’s law, power
• Series and parallel wiring
• Internal resistance
• Electric circuits – Kirchhoff’s rules
• Measurement of voltage and current
• Omit 20.5, 12, 13 (alternating current, capacitors, RC circuits)
Wednesday, January 20, 2010
3
Electric Current
Current is the rate of flow of
charge
1 A = 1 C/s
Conventional Flow of Current
Positive charges flowing from the
+ to the - terminal of a battery or
power supply
Current really is a flow of
electrons (negative charge) in the
opposite direction.
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electromotive force, emf = V
By convention, we use the
conventional flow of current
4
Prob. 20.3/102: 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?
b) How much more energy is used in 1 minute in normal mode?
• Current is the rate of flow of charge
• How much PE does a charge lose in travelling from + terminal to – ?
a) 6.6 C, 4.02 C
b) 312 J
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Ohm’s Law
The current flowing around a circuit is
proportional to the voltage applied.
That is, I ! V
And, V = IR (Ohm’s Law)
R = resistance, in ohms (!)
If I = 0.4 A when V = 3 V, then
“Conventional” flow of
current
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R = V/I = (3 V)/(0.4 A) = 7.5 !
6
Conduction of heat
Rate of flow of heat:
Q=
∆T
kA
∆T
L
k = thermal conductivity
Conduction of charge
Rate of flow of charge:
I=
A
∆V
ρL
ρ = electrical resistivity
Electrical conductivity = 1/"
∆V
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Conduction of charge
Rate of flow of charge:
I=
A
∆V
ρL
∆V
∆V = I ×
ρL
A
Compare with Ohm’s law: ∆V = I × R
→ R=
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ρL
A
8
Resistivity
The resistance of a piece of wire of length, L, and crosssectional area, A, is
R=
!L
A
and V = IR for a resistive material
" is the “resistivity” of the material of the wire
!=
RA ".m2
=
= ".m
L
m
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Table of Resistivities
The resistivity varies with temperature
- basis of temperature gauge
Superconductors - zero resistivity
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10
Prob. 20.12/11: 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...
9.93x10-3 m
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11
Test for deep vein thrombosis – how fast does
leg recover?
Measure the resistance of
part of the calf:
!L
!L
=
A
Vcalf/L
!L2
R=
Vcalf
R=
Volume V,
cross sectional
area A
Vcalf = volume of calf of length L
=LA
Inflate cuff to cut off blood flow
from the leg, but not to it – volume
of calf increases, R drops.
Release cuff, volume and resistance return to normal, but how
quickly? Should be fast.
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12
Impedance Plethysmography
http://www.medis-de.com/en/ipg.html
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Variation of resistivity with temperature
! = !0[1 + "(T − T0)]
Resistivity = "0 when temperature = T0.
α = temperature coefficient of resistivity (or resistance)
Same relation holds for resistance:
R = R0[1 + !(T − T0)]
as R = "L/A
Similar to variation of length or volume with temperature.
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Prob. 20.13/105: A platinum resistance thermometer has a resistance
of 125 ! at 20º C. When immersed in boiling chlorine, its resistance
drops to 99.6 !.
The temperature coefficient of resistance of Pt is α = 0.00372 ºC-1.
What is the temperature of boiling chlorine?
• How does resistance change with temperature?
-34.6oC
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15
Prob. 20.121/19: 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 total resistance does not change with temperature. What is
the ratio of the lengths?
W: ρ01 = 5.6 ! 10-8 ΩΩ.m,
α1 = 0.0045 ºC-1
Cu: ρ02 = 3.5 ! 10-8 ΩΩ.m,
α2 = –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...
LW/LCu = 0.069
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Electrical Power
A charge Δq in falling through a potential difference V
loses potential energy VΔq.
This energy is supplied by the battery or power supply.
The current flow is:
I=
!q
!t
The rate at which energy is delivered by the supply is:
P =
energy
V ∆q
=
=VI
time
∆t
which is the power supplied.
Ohm’s law: V = IR, so P = VI = V2/R = I2R
Wednesday, January 20, 2010
17
A car battery is being charged at a voltage of 12 V and a current of
19 A.
How much power is being produced by the charger?
Power, P = VI = (12 V) ! (19 A) = 228 W.
Prob. 20.28/26: 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.
Estimate the length of wire.
" = 10-6 !.m
• From P and V, what is the resistance of the wire?
48 m
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Chapter 20 so far...
Ohm’s law
V = IR
Resistance, resistivity
R=
!L
A
Temperature dependence
! = !0[1 + "(T − T0)]
R = R0[1 + !(T − T0)]
Power
P = VI = V2/R = I2R
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19
Clicker Question
Two materials have different resistivities. Two wires of the
same length are made, one from each of the materials.
Is it possible for the two wires to have the same resistance?
A) Yes, if the material with the greater resistivity is used for
the thinner wire.
B) Yes, if the material with the greater resistivity is used for
the thicker wire.
C) No, it is not possible.
Answer: B) as R = "L/A
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