Transcript phys1442-lec8x

PHYS 1442 – Section 004
Lecture #8
Monday February 10, 2014
Dr. Andrew Brandt
CH 18
•
Electric Power
CH 19
•
Resistors
•
Series and parallel
•
“Simple” circuits
***Defer AC to Monday 17th along with short review
Mondady Feb. 10, 2014
PHYS 1442-004, Dr. Andrew Brandt
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Announcements
a) HW4 on ch 18 due Tues at 11:30 pm
c) HW5 on part of ch 19 due Monday 17th at 5:25 pm (can get a 24 hour
extension if in class)
d) Test on Weds. Feb. 19th ch 16-18 plus part of ch 19th
Mostly multiple guess, bring a scantron, I don’t know how many problems
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•
Electric
Energy
Why is electric energy useful?
– It can be transformed easily into different forms of energy:
• Motors, pumps, etc, transform electric energy to mechanical energy
• Heaters, dryers, cook-tops, etc., transform electricity to thermal energy
• Light bulb filaments transform electric energy to light energy
– Only about 10% of the energy turns to light with 90% lost via heat
– Typical household light bulb and heating elements have resistance of order few
ohms to few hundred of ohms
• How does electric energy transform to thermal energy?
– Flowing electrons collide with the vibrating atoms of the wire.
– In each collision, part of electron’s kinetic energy is transferred to the atom it
collides with.
– The kinetic energy of wire’s atoms increases, and thus the temperature of the wire
increases.
– The increased thermal energy can be transferred as heat through conduction and
convection to the air in a heater or to food in a pan; it can also be radiated as light.
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Electric Power
• How do we find out the power of an electric device?
– What is definition of the power?
• The rate at which work is done or the energy is transferred
• What energy is transferred when charge q moves through a
potential difference V?
– U=qV; So to move a small amount of charge q in a small amount of
time t through a potential difference V, the power P is
What is this?
– P  U t  V q t
2
– Thus, we obtain P  IV .
V
2
In
terms
of
resistance
P

I
R
– What is the unit? Watts = J/s
R
– What kind of quantity is the electrical power?
• Scalar
– P=IV can apply to any device, while the formulae involving resistance
only applies to Ohmic resistors.
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Example
Headlights: Calculate the resistance
of a 40-W automobile headlight
designed for a 12V battery.
Since the power is 40W and the voltage is 12V, we use the
formula with V and R.
V2
P
R
Solve for R
12V 
V

 3.6
R
P
40W
2
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Power in Household Circuits
• Household devices usually have small resistance
– But since they draw current, if it become large enough,
wires can heat up (overload) and cause a fire
• Why is using thicker wires safer?
– Thicker wires have less resistance, lower heat
• How do we prevent this?
– Put in a switch that disconnects the circuit
when overloaded
• Fuse or circuit breakers
• They open up the circuit when
the current exceeds a certain
value
Mondady Feb. 10, 2014
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Overload
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Example
Will a 30A fuse blow?
Determine the total current drawn
by all the devices in the circuit in
the figure.
The total current is the sum of current
drawn by the individual devices.
P  IV
Bulb
Solve for I
I PV
I B  100W 120 V  0.8 A
Stereo I S  135W 120 V  2.9 A
Heater I H  1800W 120 V  15.0 A
Dryer I D  1200W 120 V  10.0 A
Total current
I T  I B  I H  I S  I D  0.8 A  15.0 A  2.9 A  10.0 A  28.7 A
What
is theFeb.
total
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2014
Max power?
PB  PH Dr.
 PAndrew
 100W  1800W  350W  1200W 7 3450W
PTPHYS
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S  PD Brandt
EMF and Terminal Voltage
• What do we need to have current in an electric circuit?
– A device that provides a potential difference, such as battery or
generator
• typically it converts some type of energy into electric energy
• These devices are called sources of electromotive force (emf)
– This does NOT refer to a real “force”.
• The potential difference between terminals of the source,
when no current flows to an external circuit, is called the emf
( ) of the source.
• A battery itself has some internal resistance (r ) due to the
flow of charges in the electrolyte
– Why do headlights dim when you start the car?
• The starter needs a large amount of current but the battery cannot provide
charge fast enough to supply current to both the starter and the headlights
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EMF and Terminal Voltage
• Since the internal resistance is inside the
battery, we cannot separate the two.
• So the terminal voltage difference is Vab=Va-Vb.
• When no current is drawn from the battery, the
terminal voltage equals the emf which is determined
by the chemical reaction; Vab= .
• However when the current I flows from the battery,
there is an internal drop in voltage which is equal to
Ir. Thus the actual delivered terminal voltage is
Vab    Ir
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Resistors in Series
• Resistors are in series when two or
more of them are connected end to end
– These resistors represent simple electrical
devices in a circuit, such as light bulbs,
heaters, dryers, etc.
• What is common in a circuit connected in series?
– the current is the same through all the elements in series
• Potential difference across each element in the circuit is:
V1=IR1, V2=IR2 and V3=IR3
• Since the total potential difference is V, we obtain
V=IReq=V1+V2+V3=I(R1+R2+R3)
Thus, Req=R1+R2+R3
Req 

i
Ri
Resistors
in series
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When Mondady
resistors
are
in series,
totalDr.resistance
increases and the current through
the
circuit decreases compared to a single resistor.
Energy Losses in Resistors
• Why is it true that V=V1+V2+V3?
• What is the potential energy loss when charge q passes
through the resistor R1, R2 and R3
U1=qV1, U2=qV2, U3=qV3
• Since the total energy loss should be the same as the
energy provided to the system by the battery , we obtain
U=qV=U1+U2+U3=q(V1+V2+V3)
Thus, V=V1+V2+V3
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Example
Battery with internal resistance. A 65.0- resistor is
connected to the terminals of a battery whose emf is
12.0V and whose internal resistance is 0.5-. Calculate
(a) the current in the circuit, (b) the terminal voltage of
the battery, Vab, and (c) the power dissipated in the
resistor R and in the battery’s internal resistor.
(a) Since Vab    Ir
Solve for I
We obtain Vab  IR   Ir

12.0V
I
 0.183 A

R  r 65.0  0.5
What is this?
A battery or a
source of emf.
(b) The terminal voltage Vab is Vab    Ir 12.0V  0.183 A  0.5  11.9V
(c) The power dissipated
in R and r are
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P  I R   0.183A  65.0  2.18W
2
2
P  I r   0.183A  0.5  0.02W
2
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Resistors in Parallel
• Resistors are in parallel when two or
more resistors are connected in
separate branches
– Most house and building wirings are
arranged this way.
• What is common in a circuit connected in parallel?
– The voltage is the same across all the resistors.
– The total current that leaves the battery, is however, split.
• The current that passes through every element is
I1=V/R1, I2=V/R2, I3=V/R3
• Since the total current is I, we obtain
I=V/Req=I1+I2+I3=V(1/R1+1/R2+1/R3)
Thus, 1/Req=1/R1+1/R2+1/R3
1

Req

i
1
Ri
Resistors
in parallel
Feb.
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PHYS 1442-004,
Dr. resistance
Andrew Brandtdecreases and the current through
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When Mondady
resistors
are
in parallel,
the total
the circuit increases compared to a single resistor.
Resistor and Capacitor Arrangements
C
• Parallel Capacitor arrangements
Ceq 
• Series Resistor arrangements
Req 
• Series Capacitor arrangements
1

Ceq

1

Req

• Parallel Resistor arrangements
Mondady Feb. 10, 2014
PHYS 1442-004, Dr. Andrew Brandt
i
i
R
i
i
i
i
1
Ci
1
Ri
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Example
Series or parallel? (a) The light bulbs in the figure
are identical and have identical resistance R. Which
configuration produces more light? (b) Which way do
you think the headlights of a car are wired?
(a) What are the equivalent resistances for the two cases?
1
R
Series
Parallel
So
Req  2R
 2
Req 
Req
R
2
The bulbs get brighter when the total power transformed is larger.
2
2
V
2V
V2 V2

 4 PS
series PS  IV 
parallel PP  IV 

Req
R
Req 2 R
So parallel circuit provides brighter lighting.
(b) Car’s headlights are in parallel to provide brighter light and also to
prevent both lights going out at the same time when one burns out.
MondadyisFeb.
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So what
bad
parallelPHYS
circuits?
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.
Example
Current in one branch. What is the current flowing through
the 500- resistor in the figure?
What do we need to find first? We need to find the total
current.
To do that we need to compute the equivalent resistance.
1
1
12
1



Req of the small parallel branch is:
RP 500 700 3500
Req of the circuit is: Req  400  3500  400  292  692
12
V
12

 17mA
Thus the total current in the circuit is I 
Req 692
RP 
3500
12
The voltage drop across the parallel branch is Vbc  IRP  17  103  292  4.96 V
The current flowing across 500- resistor is therefore
Vbc 4.96
3

9.92

10
 9.92 mA

I 500  R 500
I 700  I  I 500  17  9.92  7.08 mA
What is the current flowing in the 700- resister?
Mondady Feb. 10, 2014
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What is the current flowing in the 400- resister?
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Resistors in Series and in Parallel
Conceptual Example: An illuminating surprise.
A 100-W, 120-V light bulb and a 60-W, 120-V light bulb are connected in two
different ways as shown. In each case, which bulb glows more brightly? Ignore
change of filament resistance with current (and temperature).
Solution: a.) Each bulb sees the full 120V drop, as they are designed to do, so the 100W bulb is brighter. b.) P = V2/R, so at constant voltage the bulb dissipating more power
will have lower resistance. In series, then, the 60-W bulb – whose resistance is higher –
will beMondady
brighter.
(More of the voltage will drop across it than across the 100-W bulb).
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Resistors in Series and in Parallel
Conceptual Example: Bulb brightness in a
circuit. The circuit shown has three
identical light bulbs,
each of resistance R.
(a) When switch S is
closed, how will the brightness of bulbs
A and B compare with
that of bulb C? (b) What happens when
switch S is opened? Use a minimum of
mathematics in your answers.
Solution: a. When S is closed, the bulbs in parallel have half the
resistance of the series bulb. Therefore, the voltage drop across them is
smaller. Bulbs A and B will be equally bright, but much dimmer than C.
b. With switch S open, no current flows through A, so it is dark. B and C
are now equally bright, and each has half the voltage across it, so C is
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somewhat
dimmer than it was with the switch closed, and B is brighter.