Download R - Physics

ANNOUNCEMENT
REMINDER
There will be a review session Monday night in Forney
(FRNY) Room G124 (note change, many evening exams
on Monday). The building just South of Civil Engr. and
across the inside street on the West side of Physics.
Time: from 7pm to about 8:30 pm.
This will be an opportunity for you to ask questions
about problem solutions, either homework or practice
exam or other.
Week 11
Physics 214
Spring 2009
1
Electrical circuits
All circuits are basically the same.
There is an external source of energy
which produces a voltage.
In a charged battery there is chemical
separation of charge. When the
circuit is made positive charge flows
from high to low voltage or negative
charge flows from low to high
voltage releasing the stored energy.
Normally it is electrons which flow.
As they move they “collide” with the
atoms of the wire and lose some of
their energy in the form of heat.
There is resistance to the flow.
Week 11
Physics 214
Spring 2009
2
Current and resistance
The rate of flow of charge determines how
much charge is transferred per unit time
I = q/t Amperes (Coulombs/sec)
The direction of I is the flow of positive charge
or opposite to the flow of negative charge.
OHM’s Law
R = ΔV/I [Ohms, Ω ]
or:
V = IR where it is
understood that V is just the voltage
drop across the resistor, and we omit
the Delta symbol.
Every part of a circuit has resistance
including an internal resistance in the
battery. The higher the resistance the lower
is the current for a given voltage difference
Week 11
Physics 214
Spring 2009
3
Electromotive force
The electromotive force of a battery , ε is
the voltage difference between the two
terminals when no current is being drawn.
When it is connected to a simple circuit
I = ε/(Rcircuit + Rbattery)
A voltage difference is the energy stored
or released per Coulomb of charge. So if
charge +q goes from high to low voltage
then the energy released is qε or qΔV
Part of the energy is wasted in internal
resistance, the rest is delivered to the
external circuit.
Week 11
Physics 214
Spring 2009
(20 Ω+5 Ω)*60 mA = 1.5 V = ε
Internal voltage drop is
0.3 V
4
Series circuit
Rbattery
If we add more identical light bulbs in the circuit in series the total
resistance increases and the current will be reduced. The current is
the same in all parts of the circuit: It’s like water flowing in a pipe
ε = I(Rbattery + R + R + R)
The voltage difference across each bulb is ΔV = IR
QUESTON: what is the internal voltage drop in the battery now?
Week 11
Physics 214
Spring 2009
5
Series circuit
Rbattery
ε = I(Rbattery + R + R + R)
What is the internal voltage drop in the battery?
ε= 1.5
R total = 5+20+20+20 = 65 Ω .
I=
ε /R = 23 mA
Week 11
so
as before
ΔVinternal = 5*23 = 0.115 V
Physics 214
Spring 2009
6
Series circuit
Rbattery
ε = I(Rbattery + R1 + R2 + R3) [bulbs not the same]
The voltage difference across a bulb
is Vi = IRi. The internal resistance is ALSO
in series, and causes the voltage at the +
terminal to be LESS THAN ε
All the voltage drops add up to ε
Week 11
Physics 214
Spring 2009
7
Series circuit
Rbattery
ε = I(Rbattery + R1 + R2 + R3) [bulbs not the same]
ε - IRbattery = I(R1 + R2 + R3) = Vexternal , the available voltage
Here, ε is the chemical EMF of the battery, but if some
current flows, the available voltage at the battery terminals
is reduced by an amount IRbattery.
Week 11
Physics 214
Spring 2009
8
Series circuit
Rbattery
ε = I(Rbattery + R1 + R2 + R3) [bulbs not the same]
In any series circuit, if one element “burns out”
and causes an “open circuit”, then no currrent
can flow. Some strings of Christmas tree lights
have this design, which can be a big nuisance.
A burned out bulb has, in effect, infinite
resistance.
Week 11
Physics 214
Spring 2009
9
Parallel circuit
Rbattery
The current q at A
divides into the three
light bulbs and then
recombines at B
In a parallel circuit, the voltage difference across each
light is the SAME. The total current is the SUM of the
three currents
I = I1 + I2 + I3 and since a current = ΔV /R
ΔV /Rcircuit = ΔV /R1 + ΔV /R2 + ΔV /R3
and
1/Rcircuit = 1/R1 + 1/R2 + 1/R3
The total resistance of the circuit is smaller than any of the single
resistances. It is also true that if one bulb fails the other two will
stay lit at the same brightness.
Week 11
Physics 214
Spring 2009
10
QUIZ
For circuits that obey OHM’s law, ΔV = IR
A 1.5 volt flashlight battery delivers 1 Amp of current
to a lightbulb. There is a voltage drop of 1.25 V
across the terminals of the lightbulb. What is the
internal resistance of the battery? Here, the
chemical EMF, ε, is 1.5 V and all of this will
appear across the batterey terminals ONLY at
zero current.
A. 1.5 Ω B. 1.0 Ω C. 0.75 Ω D. 0.5 Ω E. 0.25Ω
.
.
30 Mar 09
Answer: 0.25= ΔV across the internal resistance, hence since the current
is 1 A, the internal resistance must be 0.25 Ω because
V = IR
Note: we use V and ΔV somewhat interchangeably for Voltage drops.
Week 11
Physics 214
Spring 2009
11
Review for Test 2, Ch. 7 - 12
FORMULAE AND CONSTANTS
Conversion Factors
1 inch = 2.54 cm
1 ft = 30.5 cm
1 m = 3.281 ft
1 km = 0.621 miles
1 mile = 5280 ft
1 nautical mile = 1.1508 miles
1 kg = 2.205 lbs (where g = 9.8 m/s2)
Equations
s = d/t speed
s = d/Δt
“
v = d/Δt
(vector velocity)
a = Δv/Δt
(vector acceleration)
v = v0 + at
d = v0t + ½at2
v2 = v02 + 2ad
d = ½(v + v0)t
F = ma (vector) “N2“
F12 = - F21
“N3“
W = mg
Week 11
P = 1015 Peta
T = 1012 Tera
G = 109 Giga
M = 106 Mega
k = 103 kilo
m = 10-3 milli
μ = 10-6 micro
n = 10-9 nano
p = 10-12 pico
f = 10-15 femto
Pressure = Force/Area
[N/m2 = Pa = Pascal]
1 Atmosphere = 101.3 kPa = 760mmHg ≈ 33 ft H2O
P = ρgd where d is depth below surface (or
height of a liquid column
Buoyant force = weight of displaced liquid
Density H20 is 1 Tonne/m3 or 1 gram/cm3
Physics 214
Spring 2009
12
Review for Test 2, Ch. 7 - 12
MOMENTUM AND IMPULSE -- ELASTIC AND INELASTIC
COLLISIONS, CONSERVATION OF MOMENTUM
ROTATIONAL MOTION OF SOLID OBJECTS -- TORQUE, MOMENT
OF INERTIA, CONSERVATION OF ANGULAR MOMENTUM,
ANGULAR ACCELERATION, KE of ROTATION
FLUIDS – PRESSURE, DENSITY, BUOYANCY
TEMPERATURE AND HEAT -- TEMPERATURE SCALES, HEAT
CAPACITY, PHASE CHANGES AND LATENT HEATS
HEAT ENGINES AND 2d LAW OF THERMODYNAMICS -- CARNOT
EFFICIENCY
ELECTRICITY AND MAGNETISM: ELECTROSTATICS -ELECTRIC FIELDS AND FORCES, VOLTAGE (ELECTRIC
POTENTIAL), ELECTRIC CHARGE
Week 11
Physics 214
Spring 2009
13
Review for Test 2, Ch. 7 - 12
Circumference = 2πR
g = 9.8 m/s2
G = 6.67 × 10-11N·m2/kg2
π = 3.14159
ω = θ/t = Δθ/Δt angular velocity
e = 1.6x10-19 C the size of the charge on an electron or a proton
me = 9.1x10-31kg
the mass of the electron
mp = 1836 me
the mass of the proton
cH2O= 1 Cal/gm
heat capacity of water
Lf = 80 Cal/gm
latent heat of fusion of water
Lv = 540 Cal/gm
latent heat of vaporization of water
Week 11
Physics 214
Spring 2009
14
Review for Test 2, Ch. 7 - 12
DEFINITIONS
a = v2/R
F = mv2/R
F = GM1M2/R2
F = kq1q2/R2
W = Fd
P = W/t = Fv
KE = ½ mv2
PE = mgh
PE = ½ kx2
P = mv
FΔt = Pf – Pi = Δp
Fexternal = 0, Δp = 0
W = qΔV
P = I ΔV
E = ΔV/d
Week 11
centripetal acceleration
centripetal force
gravitational force G = 6.61x10-11Nm2/kg2
electrostatic (Coulomb) force k = 9x109 Nm2/C2
work
power
Kinetic energy
potential energy
potential energy
momentum
impulse
momentum conservation
work or energy, electrical
power, electrical
for uniform electric field, where d is distance along
the voltage gradient
Physics 214
Spring 2009
15
Review for Test 2, Ch. 7 - 12
DEFINITIONS
ε = (Thot – Tcold)/Thot
PV = NkT
Carnot efficiency of an ideal heat engine
Ideal gas law where P,V,T are pressure, volume,
and Kelvin temperature and k is Boltzmann’s constant
2
I = mR
Moment of rotational inertia of a point mass a
distance R from the axis of rotation
Torque = r x F
Torque is product of Force x “Lever arm” distance
to the axis of rotation (the component of distance at right angle to F)
Torque = I α
α
angular acceleration, in radians/s2
KE = 0.5 I ω2
Kinetic energy of rotation of a solid body
I = MR2
for hoop
I = 0.5 MR2
for disk
I = 0.4 MR2
for sphere
I = (1/12)ML2
for a rod of length L
KEPLER’s Laws
Elliptical orbit sweeps out equal areas in equal time intervals
T2 = r3 for all satellites of a common attracting body, where T is the
period and r is the average radius of the orbit
Week 11
Physics 214
Spring 2009
16
What is voltage?
We have seen that the definition of voltage is
ΔV = ΔPE/q when a charge q is moved in an electrical force field.
So energy is stored as potential energy when a positive
charge is moved against the direction to E (or a negative charge is
moved in the same direction as E.)
If we move a positive charge toward a positive charge, the
potential energy and ΔV increase. Also if we move a negative
charge away from a positive charge.
Just as in the gravitational field, what matters is DIFFERENCES
in PE. So normally we use the term ΔV. But very often for circuits
we choose one point, usually the negative terminal, to be zero and
then instead of ΔV we just use V.
When charge is free to move, the PE will transform into KE just
like dropping something. Positive charge will move to lower
voltage, negative charge to higher voltage, tending to neutralize
the system.
In a simple circuit with resistance the KE is turned
into heat and light and there is a voltage drop across every
element in the circuit.
Week 11
Physics 214
Spring 2009
17
Voltage drop
( here we assume negligible Rbattery )
If we have a circuit with many different
resistors then there is a voltage drop
across each resistor [in the direction of
the current] and there is also a voltage
drop for the whole circuit.
Current only flows if there is a voltage
difference. In a time t , charge q
passes through the resistor.
I = q/t and ΔV = IR
Case1
I = 6/60 = 1/10
ΔV15 = 15/10 ΔV20 = 20/10 ΔV25 = 25/10
Case 2
I = 12/8 I24 = 12/24
Case 3
I = VAB/5.5; ΔV33 = 1.5I; ΔV3 = 3I; ΔV333 =1I
Check: (1.5+3+1) I = 5.5 I = VAB
Week 11
Physics 214
Spring 2009
18
Voltage drop
( here we assume negligible Rbattery )
The voltage jumps from 0 to ε as we
go from the – to the + plate of the
battery.
From there on, there are a
succession of voltage drops, in the
direction of the current flow through
the series of resistors.
These bring us back to V=0 at the plate
Case1
I = 6V/60Ω = 1/10 A
ΔV15 = 15/10; ΔV20 = 20/10; ΔV25 =
25/10. 1.5+2+2.5 = 6V checks OK.
Week 11
Physics 214
Spring 2009
19
Voltage drop
Case 3
I = VAB/5.5;
ΔV3 = 3I;
( here we assume negligible Rbattery )
ΔV33 = 1.5I;
ΔV333 =1I
Check: (1.5+3+1) I = 5.5 I = VAB
NOTE: Here we “lump together”
the two (or three) resistors in
parallel with each other, before
putting the “lumps” in series with
each other and with the single
3Ω
Week 11
1.5 Ω
3Ω
1Ω
resistor.
Physics 214
Spring 2009
20
Measuring current and Voltage
It is often very important to know
the current in a circuit or the
voltage difference between two
points.
A hand held meter is very useful to
test batteries or a circuit.
An ammeter is a device inserted
into a circuit. The resistance of an
ammeter is very small so as to
minimize the effect on the circuit.
A voltmeter is attached in parallel
and V is found by measuring the
current and V = ImeterRmeter. The
resistance has to be much larger
than the circuit resistance so that
the current is very small and does
not disturb the main circuit.
Week 11
Physics 214
Spring 2009
Both meters actually
measure current
21
Measuring current and Voltage
An old fashioned voltmeter is
a very sensitive coil in a
magnetic field, with a pointer
that swings when tiny
currents flow in the coil.
We put this in series with a
large resistor, so that it won’t
disturb the current flowing in
the circuit very much.
The combination is what’s
represented by the circle-V
Week 11
Physics 214
Spring 2009
Both meters actually
measure current
22
Measuring current and Voltage
Now we can use this
voltmeter to make an
Ammeter: The circle-A
contains a very low-R “shunt
resistor” which is put in
series with the circuit, so that
the current I flows though the
shunt. Low RA means tiny
Voltage drop, so little effect
on I.
Put (V) meter across the
shunt to measure the tiny
Voltage drop. Knowing RA,
we know the current I
Week 11
Physics 214
Spring 2009
Both meters actually
measure current
23
Measuring current and Voltage
Better modern meters use
power amplification, so that
they drain an even tinier
fraction of the power from the
circuit being measured.
Affordable DVMs (digital
voltmeters) have 10 MΩ “input
impedance” and can measure
tiny voltages (sub-mV), and
tiny currents (sub-μA).
Multi-meters can also measure
resistance, using an known V
internal battery to drive current
through a R and measuring the
current through the resistor.
Week 11
Physics 214
Spring 2009
Both meters actually
measure current
24
Power
Electromotive force or voltage difference between two points is the
difference in potential energy/unit charge.
So the energy delivered if charge q is transferred is
energy = Vq
power = Vq/t = VI watts
For any voltage difference ΔV and current I.
power = ΔVI but we will generally just use V for the voltage
difference across the R
For circuits that obey OHM’s law V = IR
P = VI = I2R = V2/R watts
The power used appears as heat or light
Choosing the most appropriate form for the electric Power
equation can simplify a solution.
Week 11
Physics 214
Spring 2009
25
Power
P = VI = I2R = V2/R watts
Just choose the most appropriate form.
Example: 100 W lightbulb designed for 115 V. What is R (when
filament is hot)?
What do we know?: P, V
What is the question? R=?
Use the last form, since we aren’t told OR asked what is I.
R = V2/P = 1152 V2/100 W = 132.25 Ω
Week 11
Physics 214
Spring 2009
26
Power
P = VI = I2R = V2/R watts
Just choose the most appropriate form.
Example: 100 W lightbulb designed for 115 V.
Now we ask what is the Power (compared to the 115V power) if the
voltage is cut to 100 V?
(We’ll assume, not quite correctly, that the cooler filament has the same
R. In fact, it will have a lower resistance than the hotter filament at 115V)
Here, we still use the last form, and ALSO use ratios so that the allegedly
constant R divides out:
P’/P = V’2/V2 = (100/115)2 = 0.872 = 0.756
Numerical observations: divide 100 by 100 + 15%. Result is 1.00 - 13%.
Squaring it roughly doubles the percentage difference: (1 - 24.4%)
(1+x)2 = 1 + 2x + x2 and if x is small, you can more or less neglect x2
(1-x)2 = 1 - 2x + x2 ;
(1+x)-2 = 1 - 2x + [(-2)(-3)/2]x2 + ...x3 + ….x4 ……
Using this last form, 1/(1+15%)2 ≈ 1-30% and you crudely get P’/P ≈ 0.700
Week 11
Physics 214
Spring 2009
27
Digression
Numerology, but may be useful in everyday life.
Let’s say you lose 50% of your investment in stock in Bank of
America. Then the market picks up and your BoA stock value
increases by 50%. Are you happy?
(1 - 0.5) x (1 + 0.5) = 0.5 x 1.5 = 0.75
Your BoA stock is worth 75% of your original investment. You would
have needed a 100% increase to get back to where you started.
Put simply, if your stock value drops to half, it needs to DOUBLE to
get back where it started. So percentage decreases and increases
are not exactly additive!!
For SMALL percentages, increases and decreases are ALMOST
additive: 1.01 x .99 = 0.9999
but you’re still a tiny bit (.012) low
Week 11
Physics 214
Spring 2009
28
Power
P = VI = I2R = V2/R
watts
Let’s rig three 100 Watt 115 V bulbs into a circuit where two
bulbs are in parallel, and the third bulb is in series with the
pair.
|------O------|
o--------O--------|
|---------o
|------O------|
All bulbs will run dim, if this entire arrangement is put across
the same Voltage of 115V
The total RT is R + R/2 where R is for one bulb and R/2 is for
the parallel pair. So now I will be 2/3 as big as before, and the
left bulb will have I2R power = (2/3)2 = 4/9 as big as a single
bulb across 115 V.
Week 11
Physics 214
Spring 2009
29
Power
P = VI = I2R = V2/R
watts
Let’s rig three 100 Watt bulbs into a circuit where two bulbs are in
parallel, and the third bulb is in series with the pair.
|------O------|
o--------O--------|
|---------o
|------O------|
More extremely: the current divides in the parallel portion, so each
of those bulbs gets only 1/3 the current of a single bulb across
115V, hence only 1/9 the power.
Does power add up? 4/9 + 1/9 + 1/9 = 6/9 = 2/3 for all three bulbs.
Same voltage, and remember 3/2 as much resistance, hence 2/3 as
much current. Now use the FIRST form of the Power:
P’ = VI = (same x 2/3) as much as for one single bulb. Checks OK
Week 11
Physics 214
Spring 2009
30
Power
P = VI = I2R = V2/R watts
Do things add up? 4/9 + 1/9 + 1/9 = 6/9 = 2/3 for all three bulbs.
Same voltage, and remember 3/2 as much resistance, hence
2/3 as much current. Now use the FIRST form of the Power:
P = VI = same x 2/3 as much as for one single bulb. Checks OK
***
But I put in an unnecessary extra step here, I could just have
used the THIRD Power form, and not bothered to calculate the
current, and found:
New Power = (V)2/(3/2R) = 2/3 {(V)2/R} = 2/3 of single bulb Pwr.
Everything checks out consistently. There is no wrong
approach, just one approach is somewhat simpler and quicker.
Week 11
Physics 214
Spring 2009
31
QUIZ
March/April 09
P = VI = I2R = V2/R [watts];
V = IR Ohm’s Law
Two 100 Watt bulbs are in series, across 115 Volts
0V o--------O-----------O---------o +115V
QUESTION: Compared to a single 100 Watt bulb across
115V, how much power does ONE of these bulbs
consume? Let R be the resistance of a single bulb and
work out the series resistance, etc.
A. same B. 2x C. 4x D. ½ x E. ¼ x
. For a multi-step solution, note that the resistance is 2R, the current is
0.5 I, and EACH BULB sees HALF the115 Volts drop – compared to
a single 100 W bulb. At that point, you can use form 1, P’=V’I’ = 0.5 V
x 0.5 I = 0.25 VI
Or you can save time by going directly to form 3 with ½ 115 V across the
one bulb, and you are squaring that, which gives a factor of ¼ if R is
unchanged
E. Is the correct answer.
Week 11
Physics 214
Spring 2009
32
Summary Chapter 13
Common I, voltage drops add
I = q/t Amperes (Coulombs/sec)
OHM’s Law: R = ΔV/I [Ohms, Ω]
I = ε/(Rcircuit + Rbattery)
P = ΔVI = I2R = (ΔV)2/R
watts
Common voltage drop, currents add
Week 11
Physics 214
Spring 2009
33
Current and power
I = q/t Amperes (Coulombs/sec)
Transferring charge q across a
voltage drop ΔV involved an
energy q ΔV in Joules
Then when current flows across
a voltage drop, there is a Power
P = I ΔV in Watts or Joules/sec
Week 11
Physics 214
Spring 2009
34
Transmission
In the distribution of electric power the
goal is to deliver to the user as large a
fraction as possible of the generated
power.
Practical cables have a specific
resistance so the power losses
will be I2Rcable and we need I to be as
small as possible.
But we also need the delivered power
P = VsourceIsource
to be as high as possible, therefore,
the electrical power is distributed at
very high voltage and low current.
The voltage is reduced from
375,000Volts to 230Volts and/or 115 V
for households by using a transformer.
The current increases by the same
factor since for an ideal transformer no
power is lost. Transformers are the
dominant reason electrical
transmission is alternating current
Week 11
i
Rcable
Vuser
Vsource
Physics 214
Spring 2009
Ruser
i
Vuser = Vsource – IRcable
Puser = iVuser = iVsource – i2Rcable
35
Transmission
AC allows voltage step-up
and step-down via “magnetic
induction” – the electric
fields that arise from
changing magnetic fields.
With modern solid state
electronics, we can step
i
steady (DC) voltages up and
Rcable
down, and convert DC to AC
Vuser Ruser
Vsource
for use in the household.
DC transmission lines are
i
better: more average voltage
Vuser = Vsource – IRcable
for a given peak voltage, and
Puser = iVuser = iVsource – i2Rcable
no phase-matching problems
over the nation-wide grid.
Week 11
Physics 214
Spring 2009
36
Transmission
Above about 375 kV AC,
power lines tend to “leak” via
ionization of the surrounding
air, and arcing at the
insulating supports of the
wires. This limits how far we
can push the HV strategy.
i
Rcable
Vuser Ruser
Vsource
Aluminum is lighter and
cheaper than copper, but has
i
poorer conductance (higher
Vuser = Vsource – IRcable
resistance.) Choice of metal
Puser = iVuser = iVsource – i2Rcable
is an economic tradeoff.
Week 11
Physics 214
Spring 2009
37
Household appliances
Household circuits are wired in parallel so that when more than
one appliance is plugged in each sees the same voltage and can
get the required current.
As we plug in more and more appliances the current in the circuit
increases and the I2R losses in wires could cause a fire. This is
why we have fuses and why major appliances use 230 volts and
many parts of the world use 230 volts for all household use. Half
the voltage means double the current and four times the
dangerous waste heat. But 115 V is less likely to electrocute.
As many people turn on appliances (air conditioners) the grid has
to supply more power by increasing the current.
Puser = iVuser = iVsource – i2Rcable
This results in a higher fraction of the power being lost in the
cable
In cases of very heavy load the power station reduces the
transmission voltage resulting in a “brown out” and in extreme
cases there are rolling blackouts.
Week 11
Physics 214
Spring 2009
38