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Ohm’s Law
George Simon Ohm
The law which governs most simple and many complex electrical phenomena is known as Ohm’s
Law. It is the most important law in electricity. In 1827, a German locksmith and mathematician
named Georg Simon Ohm, introduced the law which bears his name. His many years of
experimenting with electricity brought about the fact that the amount of E.M.F., or voltage, which
flowed in a circuit was directly proportional to the applied current, or amperage. In other words,
when current increases, voltage increases. If the voltage is held constant, the current will only
change when resistance changes, but in the opposite direction. Current, on the other hand, will
decrease as resistance increases and will increase as resistance decreases.
Ohm's Law States: The Current Which Flows In A Circuit Is Directly
Proportional To The Applied Voltage And Inversely Proportional To The Resistance.
Ohm’s Law Formula’s
There are three ways of fundamentally stating Ohm's Law. Translating these fundamentals into
mathematical formulas will allow us to put numerical values to quantities of volts, amps, watts,
and resistance. These numerical values will allow us to better understand and control electricity.
Here are the three fundamental laws...
Law #1
E = I x R
The pressure in volts (E) is equal to the current in amps (I) multiplied by the
resistance in ohms (R).
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Law #2
I=
E
R
The current in amps (I) is equal to the pressure in volts (E) divided by the
resistance in ohms (R)
Law #3
R=
E
I
The resistance in ohms (R) is equal to the pressure in volts (E) divided by the
current in amps (I)
The Ohm’s Law Wheel
From the three previous formulas we can derive all of the Ohm's Law Wheel formulas above.
The values for volts, amps, watts, and ohms can be found, when we know any two of the other
values.
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For example: We have a resistor in the above series circuit reading 2 Amps (I) with 1 Ohm (R)
of resistance. If we are trying to find how many Volts (E) there are we would check the volts
formulas knowing the values of amps (A) and ohms (R) in the Ohm's Law Wheel. The correct
formula for volts would be I x R, or amps times ohms (E = I x R).
E=IxR
E=2x1
E (Volts) = 2
Another example...
The resistor in the above series circuit is reading 6 Watts (W) and 1 Volt (E). If we are trying to
find how many Amps (I) there are we would check the current (amps) formulas. Knowing the
values of watts and volts in the Ohm's Law Wheel. The correct formula for Amps (I) would be
W/E , or Watts (W) divided by Volts.
I=W/E
I=6/1
I (Amps) = 6
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And one more...
The resistor in the above series circuit is reading 12 Volts and 4 Amps. If we are trying to find
how many ohms of Resistance (R), or (!) there are we would check the current (amps) formulas
knowing the values of watts and volts in the Ohm's Law Wheel. The correct formula for
Resistance (R) would be E/I , or Volts (E) divided by Amps (I).
R=E/I
R = 12 / 4
R (Resistance) = 3
There has been a long tradition in the electrical industry for using the basic ohm’s law equations
used in the previous examples, but, remembering twelve different formula’s out in the field, or
on an examination can be daunting.
Although deriving mathematic equations based on the relationship of amps, volts, resistance and
watts has importance in understanding the world around us, a quick way of finding ohm’s law
values can be, at times, very helpful.
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The Ohm’s Law Ladder
There’s a simple alternative to using the Ohm’s Law Wheel and trying to memorize all of those
crazy formulas. All you have to do is learn the elements of Ohm's Law; in the order above (top to
bottom)....
Watts (W)
Volts (E)
Amps (I)
Ohms (R)
Or, you can try to remember which is W.I.R.E. spelled wrong (W.E.I.R)! It can be more easily
thought of as steps in a ladder. In order to use the ladder just put the known values next to each
item. Then multiply the value below by the value above, stepping up the ladder, or divide the
value above by the value below stepping down the ladder. For example...
If Amps (I) equals 10, and Ohms (R) equals 5, then by multiplying the values stepping up the
ladder we would multiply 5 Ohms times 10 Amps, or 5 x 10 = 50.
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Now we can take the next step which is to multiply the Volts (50) times the Amps (10) or 50 x 10 =
500. Remember, we multiply going up the ladder.
We can double check our answers by stepping down the ladder. Remember we divide going
down each step; to find Amps we would divide Watts (500) by Volts (50) or 500/50 = 10.
We’re correct so far. How about finding Resistance; we can divide Volts (50) by Amps (10) or
50/10 = 5. How about that !
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Can you climb two rungs of a ladder without falling and breaking your neck? Slight problem !
What if you're missing two steps (Volts and Amps). There are two ways to climb two steps up the
ladder. You can find Amps by dividing Watts (40) by Ohms (10) but you must find the square root
of your answer. In other words √40/10 = √4.The square root of √4 equals 2.
The other way is to find volts by multiplying Watts (40) times Ohms (10) but again, you must find
the square root of your answer. In other words 40 x 10 = √400. The square root of √400 is of course
20.
Finding volts and amps while knowing watts is a bit academic. Fact is that volts, amps and
resistance are primary values, at least two of which are givens (easily measurable) values. Watts,
on the other hand, are derived. In other words, watts are just a product of volts and amps. For
instance, a wattmeter actually is a combination of a volt-meter and a amp-meter with the results
a value of both (volts x amps).
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Ohms Law Problems
1."A 10 ohm resistance carrying 10 amps of current uses ______ watts of power.
" (a)"100"
" (b)"200"
(c)" 500
(d)"1,000
2."The number of watts of heat given off by a resistor is expressed by the formula I2R. If
" 10 volts is applied to a 5 ohm resistor, the heat given off will be _______ watts.
" (a)"500"
" (b)"250"
(c)" 50
(d)"20
3."The computed load for a 15,000 watt heater at 240 volts is _______ amps.
" (a)"52"
" (b)"63"
(c)" 65
(d)"80
4."A 240 volt source has a 20 amp load, what is the wattage of this circuit ?
" (a)"240"
" (b)"2,400"
(c)" 480
(d)"4,800
5."A 1000 watt, 120 volt lamp uses electrical energy at the same rate as a 14.4 ohm resistor.
" (a)"True
" (b)"False
6."A 20 amp fuse will blow when a load of _______ watts is connected to it. (115 volt source)
" (a)"1,500"
" (b)"2,000"
(c)" 2,200
(d)"2,500
7."A 10 ohm resistor is connected to a 120 volt circuit, how many amps are flowing in the circuit ?
" (a)"8 "
" (b)"10"
(c)" 12
(d)"20
8."A 1500 watt heater is rated at 230 volts, what would the wattage be if it was connected to a
" 208 volt source ?
" (a)"1,500"
" (b)"1,350"
(c)" 1,240
(d)"1,226
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