##
__ __* 1 Law *

*1 Law*

**"The current in a circuit is directly proportional to the applied voltage and inversely proportional to the**

**circuit resistance."**Ohm's Law is expressed by the equation

**I = E / R**

Spoken as, "

**Amps**is equal to volts divided by

**resistance**."

**Directly proportional,**when one increases▲ the other increases▲,

*when one decreases ▼the other decreases ▼*

**and**

**inversely proportional,**when one increases▲ the other decreases▼

*when one decreases ▼the other increases ▲*

**and**The

**current**

**(called**

**amps**

**)**represented by the letter

**(**

**I )**in a circuit,

is directly proportional ▲▲to the applied

**voltage**represented by the letter

**(**

**E )**

**...**

*and*the

**current**is also inversely proportional ▲▼to the circuit

**resistance**represented by the letter

**(**

**R )**

again

**I = E / R**

*Another way to express Ohm's Law is with triangle charts >*

*4 triangle charts*

##
__ 4 Quantities __

__4 Quantities__**E**represents electromotive force, EMF and is measured in voltage or volts or V.

**I**represents the intensity of electrical current and is measured in amperage or amps or A.

**R**represents resistance and is measured in ohms or the omega symbol; Ω

**P**represents power or electrical energy and is measured in wattage or watts or W.

Multiply the first 2 quantities E x I and find the 4th quantity "watts"

Watts are commonly referenced in Ohm's Law calculations and originated in Watt's law,

**Electrical energy, expressed as "P" measured in watts**

**P = I x E**

*Watts is equal to amps times volts.*##
* 12 formulas *

This equation; *12 formulas*

**√**

*E = I x R = P / I =**(PxR)*represents the first 3 formulas;

Formula 1)

**Volts (represented by the letter E) is equal to amps (letter I ) times resistance (R).**

*E = I x R*Formula 2)

**Volts is equal to watts (P) divided by amps.**

*E = P / I*Formula 3)

**√**

*E =**(PxR)*

**Volts is equal to the square root of the answer to watts times resistance.**

This equation;

**√**

*I = E / R = P / E =**(P/R)*

**represents;**

Formula 4)

**Amps is equal to volts divided by resistance.**

*I = E / R*Formula 5)

**Amps is equal to watts divided by volts.**

*I = P / E*Formula 6)

**√**

*I =**(P/R)*

**Amps is equal to the square root of the answer to; watts divided by resistance.**

This equation;

**P = I x E = E**^{2}

**/ R = I**^{2}

*represents;*

**x R**Formula 7)

*Watts is equal to amps times volts.*

**P = I x E**Formula 8)

**P =**

**E**^{2}

*Watts is equal to volts squared divided by resistance.*

**/ R**Formula 9)

**P =**

**I**^{2}

*Watts is equal to amps squared times resistance.*

**x R**This equation;

*R = E / I = P /*

**I**^{2 }

*=*

**E**^{2}

**represents;**

*/ P*Formula 10)

**Resistance is equal to volts divided by amps.**

*R = E / I*Formula 11)

*R = P /*

**I**^{2}

**Resistance is equal to watts divided by the answer to; amps squared.**

Formula 12)

*R =*

**E**^{2}

**Resistance is equal to volts squared divided by watts.**

*/ P*If you know the value of any 2 quantities, you can find the third by using one of the above 12 formulas.

Example;

You want to find the watts (P) and you already know the amperage (I) is 5 amps and the voltage (E) is 10 volts. Look through the "P" formulas above for the one that contains I and E. Use formula 7: P = I x E.

Now replace the letters with your values; P (watts) = 5 (amps) x 10 (volts) or P = 5 x 10.

Now solve for P; 5 x 10 = 50, P = 50 watts.

##
__ 1 Flaw __

Missing, in the formulas, is temperature and it's effect on resistance.

__1 Flaw__As the temperature increases the resistance also increases. Because temperature is not constant but changing with time it also changes the values calculated with Ohms law.

It is as though you solved this equation; E = I x R finding 10 (E) volts = 5 (I) amps x 2 (R) ohms.

But then a change in temperature creates more resistance and when you are not looking it erases your 2 ohms and changes it to 3 ohms leaving you with an faulty calculation.

To electricians, the National Electrical Code addresses this flaw by adding temperature "correction factors" to the allowable conductor ampacity tables beginning with table 310-16.

Changing air temperature also has an effect on resistance as seen during a summer heatwave.

As the summer air temperature rises it increases the resistance in the cross country, overhead power lines.

This added resistance is like a "heat monster" consuming electricity and wasting it with no purpose.

The additional electrical demand overloads the utility and creates a blackout.

During an electrical blackout there is mention of the added electrical demand of air conditioners but many fail to note the added electrical demand of the "heat monster" .. that invisible electrical consumer that may have also been unseen by Professor Ohm.

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###
** Links **

Using facial muscles for resistance to electrical current;

**Links**Khan Academy Free online math lessons

http://www.khanacademy.org/math/arithmetic

Tony R. Kuphaldt A free series of textbooks on the subjects of electricity and electronics

http://www.ibiblio.org/kuphaldt/electricCircuits

http://www.grc.nasa.gov/WWW/K-12/airplane/ohms.html

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://phet.colorado.edu/en/simulation/ohms-law

http://www.kpsec.freeuk.com/ohmslaw.htm

http://www.ndt-ed.org/EducationResources/HighSchool/Electricity/ohmslaw.htm

http://www.wisc-online.com/Objects/ViewObject.aspx?ID=dce11904

http://www.teachersdomain.org/resource/hew06.sci.phys.maf.ohmslaw/

More from this author; http://120v.blogspot.com/

http://www.grc.nasa.gov/WWW/K-12/airplane/ohms.html

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://phet.colorado.edu/en/simulation/ohms-law

http://www.kpsec.freeuk.com/ohmslaw.htm

http://www.ndt-ed.org/EducationResources/HighSchool/Electricity/ohmslaw.htm

http://www.teachersdomain.org/resource/hew06.sci.phys.maf.ohmslaw/

More from this author; http://120v.blogspot.com/

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