LCHS Physics Mark Ewoldsen, Ph.D. Definitions Voltage: Electric Potential or Potential Difference...

LCHS PhysicsLCHS PhysicsMark Ewoldsen, Ph.D.Mark Ewoldsen, Ph.D.

DefinitionsDefinitions

• VoltageVoltage: Electric Potential or Potential : Electric Potential or Potential Difference (Energy added or used)Difference (Energy added or used)– V (volts) = Joules/Coulomb V (volts) = Joules/Coulomb

– One volt = one coulomb of charge gains One volt = one coulomb of charge gains or loses one joule of energyor loses one joule of energy

What is the difference between E and V?What is the difference between E and V?

• E is the voltage supplied by a batteryE is the voltage supplied by a battery

• V is the voltage measured across a resistorV is the voltage measured across a resistor

BatteriesBatteries• Positive and Negative terminals Positive and Negative terminals

– electrons leave negative terminalelectrons leave negative terminal

• Batteries use a chemical reaction to create Batteries use a chemical reaction to create voltagevoltage

• Construction: Two different metals and AcidConstruction: Two different metals and Acid– e.g. Copper, Zinc, and Citrus Acide.g. Copper, Zinc, and Citrus Acid

– e.g. Lead, Lead Oxide, Sulfuric Acide.g. Lead, Lead Oxide, Sulfuric Acid

– e.g. Nickel, Cadmium, Acid Pastee.g. Nickel, Cadmium, Acid Paste


• CurrentCurrent: Flow of electric charge: Flow of electric charge– I (amps) = Q/t = Coulombs/secondI (amps) = Q/t = Coulombs/second

– Higher current increases heat due to more Higher current increases heat due to more collisions of ‘free’ electrons with atomscollisions of ‘free’ electrons with atoms

Effect of Currents on the BodyEffect of Currents on the Body

• 0.001 A can be felt0.001 A can be felt

• 0.005 A is painful0.005 A is painful

• 0.010 A0.010 A causes involuntary muscle causes involuntary muscle contractionscontractions

• 0.015 A causes loss of muscle control0.015 A causes loss of muscle control

• 0.070 A can be fatal if the current last for more 0.070 A can be fatal if the current last for more than 1 second than 1 second

http://www.coolquiz.com/trivia/explain/docs/images/electric.gif

Speed of Electrons in CircuitSpeed of Electrons in Circuit

• Light goes on ‘instantly’ when switch turned onLight goes on ‘instantly’ when switch turned on– Electrons do not move at speed of lightElectrons do not move at speed of light

• c = 3 x 10c = 3 x 108 8

• Electrons – 6 x 10Electrons – 6 x 1055 m/s in random directions m/s in random directions

– Signal (energy) moves at speed of light due to electric Signal (energy) moves at speed of light due to electric energy fieldenergy field

• Wire acts as pipe to guide electric fieldWire acts as pipe to guide electric field

• Electrons in circuit do not come from battery but Electrons in circuit do not come from battery but are from the wireare from the wire

DefinitionsDefinitions• ResistanceResistance: : measure of a material’s ability measure of a material’s ability

to resist the flow of of electronsto resist the flow of of electrons– ΩΩ (ohms) = J-s/C (ohms) = J-s/C22

– Conductor – low resistanceConductor – low resistance• materials with free electronsmaterials with free electrons• e.g. copper, aluminum, gold, most metalse.g. copper, aluminum, gold, most metals

– Insulator – high resistanceInsulator – high resistance• materials with no free electronsmaterials with no free electrons• e.g. glass, plastics, ceramics, woode.g. glass, plastics, ceramics, wood

DefinitionsDefinitions• ResistanceResistance

– Increases with length Increases with length – L– L – Decreases with cross area Decreases with cross area – A – A – ResistivityResistivity – – ρρ

• material material dependentdependent • temperature dependenttemperature dependent

R = R = ρLρL AA


• PowerPower is the rate energy is converted is the rate energy is converted into another forminto another form– Resistors transform electrical energy into Resistors transform electrical energy into

light, mechanical or heat energylight, mechanical or heat energy

• Equation for Power:Equation for Power:P = I V P = I V

(Watts) – Joules/second(Watts) – Joules/second


AAmmeterAmmeter: measures amp(ere)s: measures amp(ere)s

BatteryBattery: Source of energy: Source of energy

ResistorResistor: Removes energy: Removes energy

VVoltmeterVoltmeter: measures volt(age): measures volt(age)

WiresWires::

Reading a Voltmeter Reading a Voltmeter

Reading a Ammeter Reading a Ammeter

Making CircuitsMaking Circuits


A


A

V


A

V

V

V

A

How many batteries?How many batteries?

V

A

How many ammeters?How many ammeters?

V

A

How many voltmeters?How many voltmeters?

V

A

How many light bulbs?How many light bulbs?

Ohm’s Law:Ohm’s Law:V = I RV = I R

V = voltage, I = current, R = resistanceV = voltage, I = current, R = resistance

For a given voltage, as the resistance goes up, For a given voltage, as the resistance goes up, the current will go downthe current will go down

OR

If the resistance is less, the current is moreIf the resistance is less, the current is more

Ohm’s LawOhm’s Law

V

A

Constant VoltageConstant Voltage• For a given pressure on a car’s acceleratorFor a given pressure on a car’s accelerator

– Voltage or energy inputVoltage or energy input

• By increasing the steepness of a hillBy increasing the steepness of a hill– Changing resistanceChanging resistance

• The car’s speed will go downThe car’s speed will go down– Decrease the current or ampsDecrease the current or amps

Turned on

When light is first turned onWhen light is first turned on• Filament is coldFilament is cold

– The resistance is low andThe resistance is low and– since V = IR, current is highsince V = IR, current is high

• High current means filament is more likely High current means filament is more likely to burn outto burn out

http://www.glenbrook.k12.il.us/gbssci/phys/p163/ec/u9ec.html

Ohm’s Law:Ohm’s Law:V/I = RV/I = R

V = voltage, I = current, R = resistanceV = voltage, I = current, R = resistance

For a given resistance, as the voltage For a given resistance, as the voltage goes up, the current goes upgoes up, the current goes up

Ohm’s Law – Set ResistanceOhm’s Law – Set Resistance

A

V

A

V

A

V

A

V

http://www.physics.udel.edu/wwwusers/watson/scen103/98w/note0105.html

http://www.physics.udel.edu/wwwusers/watson/scen103/98w/note0105.html

In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Resistance if the Voltage is 12-V and the Resistance if the Voltage is 12-V and

Current is 3-A?Current is 3-A?

In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Resistance if the Voltage is 120-V the Resistance if the Voltage is 120-V

and Current is 5-A?and Current is 5-A?

In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Voltage if the Resistance is 12-the Voltage if the Resistance is 12- and and


In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Voltage if the Resistance is 3-the Voltage if the Resistance is 3- and and


In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Current if the Resistance is 3-the Current if the Resistance is 3- and and

Voltage is 17-V?Voltage is 17-V?

In a simple one resistor circuit, what is In a simple one resistor circuit, what is the Current if the Resistance is 3-the Current if the Resistance is 3- and and

Voltage is 120-V?Voltage is 120-V?

Series CircuitSeries Circuit

Finding V, I & R for a Finding V, I & R for a Series CircuitSeries Circuit

1.1. Determine total E (V from battery)Determine total E (V from battery)2.2. Find the total Resistance Find the total Resistance

RRT T = R= R11 + R + R2 2 + R+ R33 … …

3.3. Determine IDetermine ITT by by

I = VI = VTT/R/RTT

3.3. Since I is constant in a series circuit, find Since I is constant in a series circuit, find V (energy used) by each resistor using V (energy used) by each resistor using

V = IV = ITTRR


1.1. Determine E (V from battery)Determine E (V from battery)

12 V12 V

1 Ω2 Ω

3 Ω


1.1. Determine E (V from battery)Determine E (V from battery) - 12V- 12V2.2. Find the total ResistanceFind the total Resistance

RRT T = R= R11 + R + R2 2 + R+ R33 … …

12 V

1 1 ΩΩ2 2 ΩΩ

3 3 ΩΩ

6 6 ΩΩ = = 3 3 ΩΩ + + 2 2 ΩΩ + + 1 1 ΩΩ


1.1. Determine E (V from battery)Determine E (V from battery) - 12V- 12V

2.2. Find the total Resistance - Find the total Resistance - 6 6 ΩΩ



2.2. Find the total Resistance Find the total Resistance - - 6 6 ΩΩ

3.3. Determine IDetermine ITT by I by ITT = V = VTT/R/RTT

2 A2 A = 12 V/ 6 = 12 V/ 6 ΩΩ



2.2. Find the total Resistance Find the total Resistance - - 6 6 ΩΩ

3.3. Determine that IDetermine that IT T = = 2 A2 A

4.4. Since I is constant in a series circuit, find Since I is constant in a series circuit, find V (energy used ) for each resistor using V (energy used ) for each resistor using V = IV = ITTRR

12 V

1 Ω2 Ω

3 Ω

2V2V = 2A * 1 = 2A * 1ΩΩ4V4V = 2A * 2 = 2A * 2ΩΩ

6V6V = 2A * 3 = 2A * 3ΩΩ

1212V = V = 6V6V + + 4V4V + + 2V2V

Series CircuitSeries Circuit• Current is the same at all points Current is the same at all points

I = II = I11 = I = I22 = I = I33 = I = I44

• Volt is divided among all the resistorVolt is divided among all the resistor

E = VE = V11 + V + V22 + V + V33

• Resistance accumulatesResistance accumulates

R = RR = R11 + R + R22 + R + R33

12 V12 V

1 Ω4 Ω

3 Ω

What is the total Resistance?What is the total Resistance?

12 V12 V

1 Ω4 Ω

3 Ω

What is the total Resistance? 8 What is the total Resistance? 8 ΩΩWhat is the current leaving the Battery?What is the current leaving the Battery?

12 V12 V

1 Ω4 Ω

3 Ω

What is the total Resistance? 8 ΩWhat is the total Resistance? 8 Ω

What is the current leaving the Battery? 1.5AWhat is the current leaving the Battery? 1.5A

What is the voltage used by the 1-What is the voltage used by the 1- resistor? resistor?

12 V12 V

1 Ω4 Ω

3 Ω




12 V12 V

1 Ω4 Ω

3 Ω

Parallel CircuitParallel Circuit

Finding V, I & R for a Finding V, I & R for a Parallel CircuitParallel Circuit

1.1. Determine EDetermine E2.2. Find the total Resistance Find the total Resistance

1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …

3.3. Determine IDetermine ITT leaving battery by leaving battery by

IITT = V = VTT/R/RTT

4.4. Since V is constant for each part of a parallel circuit, Since V is constant for each part of a parallel circuit, find I through each resistor using find I through each resistor using

I = V/RI = V/R

5.5. Check to verify that Kirchhoff’s Law is trueCheck to verify that Kirchhoff’s Law is true

1 Ω

2 Ω

12 V


1.1. Determine EDetermine E 12 V12 V


1.1. Determine EDetermine E

2.2. Find the total Resistance Find the total Resistance

1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …

1 Ω

2 Ω

1/R1/RT T = 1/1= 1/1ΩΩ + 1/2 + 1/2ΩΩ1/R1/RT T = 3/2= 3/2ΩΩ

12 V

RRT T = = 22ΩΩ/3/3




1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …

RRT T = 2= 2ΩΩ/3/3




1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …

3.3. Determine IDetermine IBB leaving battery by leaving battery by

IIBB = V = VBB/R/RTT

IIBB = 12V / 2 = 12V / 2ΩΩ/3 /3 oror 12V x 3/2 12V x 3/2ΩΩ

IIBB = 18A = 18A

1 Ω

2 Ω

12 V

IIBB = 18A = 18A




1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …



4.4. Since V is constant for each part of a parallel Since V is constant for each part of a parallel circuit, find I through each resistor using circuit, find I through each resistor using

I = V/RI = V/R

1 Ω

2 Ω

12 V

I = 12V / 1I = 12V / 1Ω = 12AΩ = 12A

I = 12V / 2I = 12V / 2Ω = 6AΩ = 6A




1/R1/RT T = 1/R= 1/R11 + 1/R + 1/R2 2 + 1/R+ 1/R33 … …



4.4. Since V is constant for each part of a parallel circuit, Since V is constant for each part of a parallel circuit, find I through each resistor using find I through each resistor using

I = V/RI = V/R

5.5. Check to verify that Kirchhoff’s Rule is trueCheck to verify that Kirchhoff’s Rule is true

1 Ω

2 Ω

12 V

18A18A

12A12A

6A6A

18A18A

IIBB = 18A = 18A

Parallel CircuitParallel Circuit

Volt is the same at all points Volt is the same at all points

E = VE = V11 = V = V22 = V = V33

ResistanceResistance 1 1 1 11 1 1 1--- = --- + --- + ------ = --- + --- + --- R RR R11 R R22 R R33

Current accumulatesCurrent accumulates

I = II = I11 + I + I22 + I + I33

2 2 ΩΩ

2 2 ΩΩ

20 V20 V

2 2 ΩΩ2 2 ΩΩ

20 V20 V

What is the voltage?What is the voltage?

2 2 ΩΩ2 2 ΩΩ

20 V20 V

What is the voltage? 20VWhat is the voltage? 20V


2 2 ΩΩ2 2 ΩΩ

20 V20 V


What is the total Resistance? 1.0 ΩWhat is the total Resistance? 1.0 Ω What is the total current?What is the total current?

2 2 ΩΩ2 2 ΩΩ

20 V20 V

What is the voltage? 20VWhat is the voltage? 20VWhat is the total Resistance? 1.0 ΩWhat is the total Resistance? 1.0 Ω

What is the total current? 20 AWhat is the total current? 20 A

What is the voltage through each What is the voltage through each resistor?resistor?

2 2 ΩΩ2 2 ΩΩ

20 V20 V


What is the total current? 20 AWhat is the total current? 20 AWhat is the voltage through each resistor? 20VWhat is the voltage through each resistor? 20V

What is the current through each What is the current through each resistor?resistor?

2 2 ΩΩ2 2 ΩΩ

20 V20 V


What is the total current? 20 AWhat is the total current? 20 AWhat is the voltage through each resistor? What is the voltage through each resistor?

20V20VWhat is the current through each resistor?What is the current through each resistor?

10A10A

2 2 ΩΩ

3 3 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ

24 V24 V


2 2 ΩΩ3 3 ΩΩ

24 V24 V




12A12A + 8A = 20A+ 8A = 20A

3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ




4A + 8A4A + 8A + 12A = 24A+ 12A = 24A

3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ

Complex CircuitsComplex Circuits

2 Ω

12 V

2 Ω

2 Ω

Finding V, I & R for a Finding V, I & R for a Complex CircuitComplex Circuit

1.1. Find the most complex portion of the Circuit – usually Find the most complex portion of the Circuit – usually part of the Parallelpart of the Parallel

2.2. Find the Total Resistance for that portionFind the Total Resistance for that portion3.3. Continue until the Total Resistance for the Circuit is Continue until the Total Resistance for the Circuit is

knownknown4.4. Starting with the least complex resistor, find the Starting with the least complex resistor, find the

voltage it usesvoltage it uses5.5. Continue until the parallel portion where the Continue until the parallel portion where the

remaining voltage will used on both sides will be remaining voltage will used on both sides will be identical, Find Iidentical, Find I

6.6. Make sure Kirchhoff's Rule is followedMake sure Kirchhoff's Rule is followed


1.1. Find the most complex portion of the Find the most complex portion of the Circuit – usually part of the ParallelCircuit – usually part of the Parallel

2 2 ΩΩ

12 V

2 2 ΩΩ

2 Ω



2.2. Find the Total Resistance for that portionFind the Total Resistance for that portion

12 V

2 2 ΩΩ

2 2 ΩΩ

2 Ω

1/R1/RT T = 1/2= 1/2ΩΩ + 1/2 + 1/2ΩΩ1/R1/RT T = 2/2= 2/2ΩΩ

RRT T = 1= 1ΩΩ

1Ω




3.3. Continue until the Total Resistance for the Continue until the Total Resistance for the Circuit is knownCircuit is known

12 V

2 2 ΩΩ

2 2 ΩΩ

2 Ω

RRT T = 2= 2ΩΩ + 1 + 1ΩΩRRT T = 3= 3ΩΩ

1Ω




3.3. Continue until the Total Resistance for the Circuit is Continue until the Total Resistance for the Circuit is knownknown

4.4. Find the Current that leaves the batteryFind the Current that leaves the battery

12 V

2 2 ΩΩ

2 2 ΩΩ

2 Ω

V = IR or I = V/RV = IR or I = V/RI = 12V / 3I = 12V / 3ΩΩ

1Ω

I = 4AI = 4A

, 4A, 4A




3.3. Continue until the Total Resistance for the Circuit is Continue until the Total Resistance for the Circuit is knownknown

4.4. Starting with the least complex resistor, Starting with the least complex resistor, find the voltage it usesfind the voltage it uses

2 2 ΩΩ

12 V, 4A

2 2 ΩΩ

V = IRV = IRSince it is series, I = 4ASince it is series, I = 4A

V = 4A x 2 V = 4A x 2 ΩΩ

V = 8V V = 8V 2 2 ΩΩ8V8V





voltage it usesvoltage it uses

5.5. Continue until the parallel portion where Continue until the parallel portion where the remaining voltage will used on both the remaining voltage will used on both sides will be identical, Find Isides will be identical, Find I

12 V, 4A

22Ω Ω

22Ω Ω

2 Ω8V

12V – 8V = 4V12V – 8V = 4V

I = 4V / 2I = 4V / 2Ω = 2AΩ = 2A

I = 4V/2I = 4V/2Ω = 2AΩ = 2A

4V 4V

4V4V





voltage it usesvoltage it uses5.5. Continue until the parallel portion where the Continue until the parallel portion where the

remaining voltage will used on both sides will be remaining voltage will used on both sides will be identical, Find Iidentical, Find I

6.6. Make sure Kirchhoff's Rule is followedMake sure Kirchhoff's Rule is followed

12 V, 4A

2 2 Ω, 4VΩ, 4V

2A + 2A = 4A2A + 2A = 4A

2A2A

2A2A

4A4A

4A4A

4A4A

2 Ω8V

2 2 Ω, 4VΩ, 4V

Try this!Try this!

‘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

R6=5

R1= 5

R2=10

R5=6

R3=6R4=12

Series-Parallel CircuitSeries-Parallel Circuit

R6=5

R1= 5

R2=10R5=6

R3=6R4=12

R7

EXAMPLE: What is the equivalent resistanceand current for the following circuit?


resistors 3 and 4 are in parallelresistors 3 and 4 are in parallel

1 1 1 1 1 1 1 1 1 1 --- = --- + --- = ----- + ---------- = --- + --- = ----- + -------

RR77 R R33 R R44 66 1212

= 0.25= 0.25 -1-1

RR77 = 4 = 4



R6=5

R1= 5

R2=10R5=6

R7=4


R6=5

R1= 5

R2=10R5=6

R7=4R8




resistor 5 and equivalent resistance 7 are in resistor 5 and equivalent resistance 7 are in seriesseries

RR88 = R = R77 + R + R55 = (4 + 6) = (4 + 6) = = 1010



R6=5

R1= 5

R2=10 R8=10



R6=5

R1= 5

R2=10 R8=10R9



resistor 2 and equivalent resistance 8 in resistor 2 and equivalent resistance 8 in parallelparallel

1 1 1 1 11 1 1 1 1--- = ----- + ----- = ------- + ---------- = ----- + ----- = ------- + ------- RR99 R R22 R R88 10 10 1010

= 0= 0.2 .2 -1 -1 ; R ; R9 9 = 5= 5



R6=5

R1= 5

R9=5V=7.5v


3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ

Try this!Try this!


3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ


What is the total current? 6.0 AWhat is the total current? 6.0 A


3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ



3 Ω: 18V and 6V3 Ω: 18V and 6V


3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ



3 Ω: 18V and 6V3 Ω: 18V and 6V What is the current through each resistor?What is the current through each resistor?

3 Ω: 6A and 2 Ω: 3A 3 Ω: 2A 6 Ω: 1A3 Ω: 6A and 2 Ω: 3A 3 Ω: 2A 6 Ω: 1A

3 3 ΩΩ

6 6 ΩΩ

24 V24 V

2 2 ΩΩ3 3 ΩΩ

Power• Power is the amount of energy that is transferred Power is the amount of energy that is transferred

every secondevery second

• Power is measured in wattsPower is measured in watts

P = I x VP = I x V

• PP = The power transferred by the component = The power transferred by the componentII = The current going through a component = The current going through a componentVV = The voltage across a component = The voltage across a component

• Remember that a kilowatt = 1,000 wattsRemember that a kilowatt = 1,000 watts

http://www.bbc.co.uk/schools/gcsebitesize/physics/electricity/mainselectricityrev5.shtml

An electric iron draws a current of An electric iron draws a current of 4A at 250V. What is its power 4A at 250V. What is its power usage?usage?

A.A. 0.0166W 0.0166W

B.B. 60W 60W

C.C. 1000W1000W

A common lightbulb reads 60W, A common lightbulb reads 60W, 120V. How much current in 120V. How much current in amperes will flow through the amperes will flow through the bulb?bulb?

A.A. 7200 amps 7200 amps

B.B. 0.5 amps0.5 amps

C.C. 2.0 amps2.0 amps

PracticePracticeDetermine the cost of using the following appliances for the Determine the cost of using the following appliances for the time indicated if the average cost is 9 cents/kWh.time indicated if the average cost is 9 cents/kWh.

(a) 1200W iron for 2 hours(a) 1200W iron for 2 hours

1.2 kW x (2h) x 9 cents = kWh

12.2 cents

PracticePracticeDetermine the cost of using the following appliances for the time indicated if the average cost is 9 Determine the cost of using the following appliances for the time indicated if the average cost is 9 cents/kWh.cents/kWh.


(b) 160W color TV for 3 hours and 30 minutes(b) 160W color TV for 3 hours and 30 minutes


12.2 cents

0.16 kW x (3.5h) x 9 cents = kWh

5.04 cents

PracticePracticeDetermine the cost of using the following appliances for the time indicated if the average cost is 9 cents/kWh.Determine the cost of using the following appliances for the time indicated if the average cost is 9 cents/kWh.


(b) 160W color TV for 3 hours and 30 minutes(b) 160W color TV for 3 hours and 30 minutes

(c) Six 60W bulbs for 7 hours.(c) Six 60W bulbs for 7 hours.


12.2 cents

0.16 kW x (3.5h) x 9 cents = kWh

5.04 cents

6 x .06 kW x (7h) x 9 cents = kWh

22.68 cents

Concept TestConcept Test

For resistors in For resistors in seriesseries, what is the same , what is the same for every resistor? R, V or I?for every resistor? R, V or I?

A.A.RR

B.B.VV

C.C.II

Concept TestConcept Test

For resistors in For resistors in parallelparallel, what is the same , what is the same for every resistor? for every resistor?

A.A.RR

B.B.VV

C.C.II


R6=5

R1= 5

R2=10R5=6

R3=6R4=12


Bibliography 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

LCHS Physics Mark Ewoldsen, Ph.D. Definitions Voltage: Electric Potential or Potential Difference...

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