Download Electric Circuits

Electric Circuits
LCHS Physics
Mark Ewoldsen, Ph.D.
Definitions
• Voltage: Electric Potential or Potential
Difference (Energy added or used)
– V (volts) = Joules/Coulomb
– One volt = one coulomb of charge gains
or loses one joule of energy
What is the difference between E and V?
• E is the voltage supplied by a battery
• V is the voltage measured across a resistor
Batteries
• Positive and Negative terminals
– electrons leave negative terminal
• Batteries use a chemical reaction to create
voltage
• Construction: Two different metals and Acid
– e.g. Copper, Zinc, and Citrus Acid
– e.g. Lead, Lead Oxide, Sulfuric Acid
– e.g. Nickel, Cadmium, Acid Paste
Capacitors
• Both batteries and capacitors store electrical
energy
• The difference between a capacitor and a battery
is that a capacitor can dump its entire charge in a
tiny fraction of a second, where a battery would
take minutes to completely discharge itself
– Sometimes, capacitors are used to store charge for
high-speed use.
• a camera flash or a TV
capacitor
Definitions
• Current: Flow of electric charge
– I (amps) = Q/t = Coulombs/second
– Higher current increases heat due to more
collisions of ‘free’ electrons with atoms
Effect of Currents on the Body
•
•
•
•
•
0.001 A
0.005 A
0.010 A
0.015 A
0.070 A
can be felt
is painful
causes involuntary muscle contractions
causes loss of muscle control
can be fatal if the current last for more
than 1 second
Speed of Electrons in Circuit
• Light goes on ‘instantly’ when switch turned on
– Electrons do not move at speed of light
• c = 3 x 108
• Electrons – 6 x 105 m/s in random directions
– Signal (energy) moves at speed of light due to electric
energy field
• Wire acts as pipe to guide electric field
• Electrons in circuit do not come from battery but
are from the wire
Definitions
• Resistance: measure of a material’s ability
to resist the flow of of electrons
– Ω (ohms) = J-s/C2
– Conductor – low resistance
• materials with free electrons
• e.g. copper, aluminum, gold, most metals
– Insulator – high resistance
• materials with no free electrons
• e.g. glass, plastics, ceramics, wood
Definitions
• Resistance
– Increases with length – L
– Decreases with cross area – A
– Resistivity – ρ
• material dependent
• temperature dependent
R = ρL
A
Definitions
• Power is the rate energy is converted
into another form
– Resistors transform electrical energy into
light, mechanical or heat energy
• Equation for Power:
P=IV
(Watts) – Joules/second
Definitions
Ammeter: measures amp(ere)s
A
Battery: Source of energy
Resistor: Removes energy
Voltmeter: measures volt(age)
Wires:
V
Ohm’s Law:
V=IR
V = voltage, I = current, R = resistance
For a given voltage, as the resistance goes up,
the current will go down
OR
If the resistance is less, the current is more
Ohm’s Law
A
V
Ohm’s Law
A
V
Constant Voltage
• For a given pressure on a car’s accelerator
– Voltage or energy input
• By increasing the steepness of a hill
– Changing resistance
• The car’s speed will go down
– Decrease the current or amps
Turned
on
When light is first turned on
• Filament is cold
– The resistance is low and
– since V = IR, current is high
• High current means filament is more likely
to burn out
Ohm’s Law:
V/I = R
V = voltage, I = current, R = resistance
For a given resistance, as the voltage
goes up, the current goes up
Ohm’s Law – Set Resistance
A
V
http://www.physics.udel.edu/wwwusers/watson/scen103/98w/note0105.html
Series Circuit
Finding V, I & R for a Series Circuit
1. Determine total E (V from battery)
2. Find the total Resistance
RT = R1 + R2 + R3 …
3. Determine IT by
I = VT/RT
3. Since I is constant in a series circuit, find
V (energy used) by each resistor using
V = ITR
Finding V, I & R for a Series Circuit
1. Determine E (V from battery)
12 V
3Ω
2Ω
1Ω
Finding V, I & R for a Series Circuit
1. Determine E (V from battery) -
2. Find the total Resistance
RT = R1 + R2 + R3 …
12V
12 V
3Ω
2Ω
1Ω
6 Ω= 3 Ω + 2 Ω+ 1 Ω
Finding V, I & R for a Series Circuit
1. Determine E (V from battery) - 12V
2. Find the total Resistance - 6 Ω
Finding V, I & R for a Series Circuit
1.
2.
Determine E (V from battery) - 12V
Find the total Resistance - 6 Ω
3. Determine IT by IT = VT/RT
2 A = 12 V/ 6 Ω
Finding V, I & R for a Series Circuit
1.
2.
3.
Determine E (V from battery) - 12V
Find the total Resistance - 6 Ω
Determine that IT = 2 A
4. Since I is constant in a series circuit, find
V (energy used ) for each resistor using
V = ITR
12 V
12V = 6V + 4V + 2V
3 Ω 6V = 2A * 3Ω
2Ω
4V = 2A * 2Ω
1Ω
2V = 2A * 1Ω
Series Circuit
• Current is the same at all points
I = I1 = I2 = I3 = I4
• Volt is divided among all the resistor
E = V1 + V2 + V3
• Resistance accumulates
R = R1 + R2 + R3
Parallel Circuit
Finding V, I & R for a Parallel Circuit
1.
2.
3.
4.
5.
Determine E
Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
Determine IT leaving battery by
IT = VT/RT
Since V is constant for each part of a parallel circuit,
find I through each resistor using
I = V/R
Check to verify that Kirchhoff’s Law is true
12 V
1Ω
2Ω
Finding V, I & R for a Parallel Circuit
1. Determine E
12 V
Finding V, I & R for a Parallel Circuit
1. Determine E
2. Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
12 V
1/RT = 1/1Ω + 1/2Ω
1/RT = 3/2Ω
RT = 2Ω/3
1Ω
2Ω
Finding V, I & R for a Parallel Circuit
1. Determine E
2. Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
RT = 2Ω/3
Finding V, I & R for a Parallel Circuit
1.
2.
Determine E
Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
3. Determine IB leaving battery by
IB = VB/RT
IB = 12V / 2Ω/3 or 12V x 3/2Ω
IB = 18A
IB = 18A
12 V
1Ω
2Ω
Finding V, I & R for a Parallel Circuit
1. Determine E
2. Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
3. Determine IT leaving battery by
IT = VT/RT
4. Since V is constant for each part of a parallel
circuit, find I through each resistor using
I = V/R
12 V
I = 12V / 1Ω = 12A
1Ω
2Ω
I = 12V / 2Ω = 6A
Finding V, I & R for a Parallel Circuit
1.
2.
3.
4.
5.
Determine E
Find the total Resistance
1/RT = 1/R1 + 1/R2 + 1/R3 …
Determine IT leaving battery by
IT = VT/RT
Since V is constant for each part of a parallel circuit,
find I through each resistor using
I = V/R
Check to verify that Kirchhoff’s Rule is true
IB = 18A
12 V
12A
1Ω
18A
2Ω
6A
18A
Parallel Circuit
Current accumulates
I = I1 + I 2 + I 3
Volt is the same at all points
E = V 1 = V2 = V3
Resistance
1
1
1
1
--- = --- + --- + --R R1 R2 R3
Complex Circuits
12 V
2Ω
2Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
2.
3.
4.
5.
6.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
Continue until the Total Resistance for the Circuit is
known
Starting with the least complex resistor, find the
voltage it uses
Continue until the parallel portion where the
remaining voltage will used on both sides will be
identical, Find I
Make sure Kirchhoff's Rule is followed
Finding V, I & R for a Complex Circuit
1. Find the most complex portion of the
Circuit – usually part of the Parallel
12 V
2Ω
2Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
Find the most complex portion of the Circuit – usually
part of the Parallel
2. Find the Total Resistance for that portion
12 V
1/RT = 1/2Ω + 1/2Ω
1/RT = 2/2Ω
2Ω
RT = 1Ω
2Ω
1Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
2.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
3. Continue until the Total Resistance for the
Circuit is known
12 V
2Ω
RT = 2Ω + 1Ω
RT = 3Ω
2Ω
1Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
2.
3.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
Continue until the Total Resistance for the Circuit is
known
4. Find the Current that leaves the battery
12 V, 4A
V = IR or I = V/R
I = 12V / 3Ω
2Ω
I = 4A
2Ω
1Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
2.
3.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
Continue until the Total Resistance for the Circuit is
known
4. Starting with the least complex resistor,
find the voltage it uses
12 V, 4A
2Ω
8V
V = IR
Since it is series, I = 4A
V = 4A x 2 Ω
V = 8V
2Ω
2Ω
Finding V, I & R for a Complex Circuit
1.
2.
3.
4.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
Continue until the Total Resistance for the Circuit is
known
Starting with the least complex resistor, find the
voltage it uses
5. Continue until the parallel portion where
the remaining voltage will used on both
sides will be identical, Find I
12 V, 4A
12V – 8V = 4V
2Ω
8V
I = 4V / 2Ω = 2A
2Ω 4V
2Ω 4V
I = 4V/2Ω = 2A
Finding V, I & R for a Complex Circuit
1.
2.
3.
4.
5.
Find the most complex portion of the Circuit – usually
part of the Parallel
Find the Total Resistance for that portion
Continue until the Total Resistance for the Circuit is
known
Starting with the least complex resistor, find the
voltage it uses
Continue until the parallel portion where the
remaining voltage will used on both sides will be
identical, Find I
6. Make sure Kirchhoff's Rule is followed
4A
12 V, 4A
2Ω
8V
2A + 2A = 4A
2A
4A
2 Ω, 4V
2 Ω, 4V
2A
4A
Power
• Power is the amount of energy that is transferred
every second
• Power is measured in watts
P=IxV
• P = The power transferred by the component
I = The current going through a component
V = The voltage across a component
• Remember that a kilowatt = 1,000 watts
Practice
An electric iron draws a current of 4A at
250V. What is its power usage?
A. 0.0166W
B. 60W
C. 1000W
Practice
A common lightbulb reads 60W, 120V. How
much current in amperes will flow through
the bulb?
A. 7200 amps
B. 0.5 amps
C. 2.0 amps
Practice
Determine the cost of using the following appliances for the
time indicated if the average cost is 9 cents/kWh.
(a) 1200W iron for 2 hours
1.2 kW x (2h) x 9 cents = 12.2 cents
kWh
(b) 160W color TV for 3 hours and 30 minutes
0.16 kW x (3.5h) x 9 cents = 5.04 cents
kWh
(c) Six 60W bulbs for 7 hours.
6 x .06 kW x (7h) x 9 cents = 22.68 cents
kWh
Concept Test
• For resistors in series, what is the same
for every resistor? R, V or I?
• Answer: I
• For resistors in parallel, what is the
same for every resistor? R, V or I?
• Answer: V
Try this!
R6=5W
R5=6W
R2=10W
R4=12W
R1= 5W
R3=6W
‘Some Electrical Circuit Components and Circuits’ by Michael Condren, Professor of Chemistry, Christian Brothers University, Memphis, TN @ http://www.cbu.edu/~mcondren/chem415/c415_ele.ppt, 4/17/04
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R5=6W
R2=10W
R1= 5W
R3=6W
R4=12W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R5=6W
R2=10W
R1= 5W
R3=6W
R4=12W
R7
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
resistors 3 and 4 are in parallel
1
1
1
1
1
--- = --- + --- = ----- + ------R7 R3
R4
6W
12W
= 0.25 W-1
R7 = 4 W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R5=6W
R2=10W
R1= 5W
R7=4W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R5=6W
R2=10W
R1= 5W
R8
R7=4W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
resistor 5 and equivalent resistance 7 are in
series
R8 = R7 + R5 = (4 + 6) W = 10W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R2=10W
R1= 5W
R8=10W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
R2=10W
R1= 5W
R9
R8=10W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
resistor 2 and equivalent resistance 8 in
parallel
1
1
1
1
1
--- = ----- + ----- = ------- + ------R9 R2
R8
10 W
10 W
= 0.2 W-1 ; R9 = 5 W
Series-Parallel Circuit
EXAMPLE: What is the equivalent resistance
and current for the following circuit?
R6=5W
V=7.5v
R1= 5W
R9=5W
Bibliography
1. ‘Some Electrical Circuit Components and Circuits’ by Michael Condren,
Professor of Chemistry, Christian Brothers University, Memphis, TN @
http://www.cbu.edu/~mcondren/chem415/c415_ele.ppt, 4/17/04
2. ‘Fundamentals of Electronics,’ Dan Bruton, Professor of Astrophysics Stephen
F. Austin State University @ observe.phy.sfasu.edu/courses/phy262/
lectures262, 4/17/04