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O HM’S L AW AND R ESISTANCE

Resistance is one of the basic principles of Ohm’s law, and can be found in virtually any device used to conduct electricity.

Georg Simon Ohm was a German physicist who conducted some very early experiments in electricity. His discovery of the relationship between current, voltage, and resistance is the basic law of current flow, and the formula that connects these three measurements is named in his honor.

Resistors are one of the basic building blocks of electrical circuits. Resistance occurs in all materials, but resistors are discrete components manufactured to create an exact amount of intended resistance in a circuit. Resistors are made of a mixture of clay and carbon, so they are part conductor part insulator. Because of this, they conduct electricity, but only with a set amount of resistance added.

The value of the resistance is carefully controlled. Most resistors have four color bands. The first band reveals the first digit of the value. The second band reveals the second digit of the value. The third band is used to multiply the value digits. The fourth band tells the tolerance of the accuracy of the total value. If no fourth band is present, it is assumed that the tolerance is plus or minus 20%.

Here are the digits represented by the colored bands found on a resistor:

Black 0

Brown 1

Red 2

Orange 3

Yellow 4

Green 5

Blue 6

Violet 7

Gray 8

White 9

Ohm’s law states this mathematical formula:

Voltage is equal to resistance multiplied by the current flow, or E=IR.

As with any algebraic formula, it is possible to rearrange the terms in order to solve the equation for a specific unit of measurement. Two algebraic equivalents of the formula would be:

I=E/R R=E/I

A very handy magic triangle is available that makes it easy to remember the different permutations of this formula.

Cover the value to be determined with your finger, and the relationship of the other two are already in the proper form.

(Example: you need to know the amount of current flowing in a circuit with 100Ω of resistance and 100 volts of pressure.

Cover I, the symbol for current, and the

remaining two symbols, E and R, appear

in their correct relationship E/R.)

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Ohms law and other formulae like it will yield an accurate result if and only if all of the units of measurement (such as Volts, Amps, and Ohms) use the same multiplier prefix within the same algebra problem.

Otherwise, your answer will be off by some order of magnitude, or power of ten.

Most often, it is easiest just to convert any readings you have into units, where no prefix is required. But this could leave you with a large number of 0s to keep track of.

On occasion, it may be more expedient to maintain a prefix such as Mega, if all of the measurements are given using that prefix. If the latter method is used, the answer to the problem will automatically come out in the same prefix used for the component parts.

For example:

#1

E (in volts) = I (in amps) x R (in ohms) E = 2A x 100Ω

E = 200v

___________________________

#2

E (in Megavolts) = I (in MegaAmps) x R (in MegaΩ)

E = 2MA x 100MΩ E = 200Mv

___________________________

In the first problem, units were used throughout, so the answer is simply given in volts. In the second problem the Mega prefix (M) is used for both amps and ohms, so the answer will also be given using the Mega prefix. Of course, these are very unusual values that are unlikely to occur in any sort of practical work.

It also possible to “mix and match”

prefixes to make the final answer come out as units.

For example:

E (in units) = I (in milliA) x R (in kiloΩ) In this problem the prefixes on the right hand side of the equation cancel each other out since “milli” means 1/1000 and Mega means 1000. 1/1000 x 1000 = 1.

These problems can also be worked with exponents using the form milli = 10

-3

and Mega = 10

3

. Again, 10

-3

x 10

3

= 1, so the end result would be an answer given in units.

Commonly used prefixes:

Mega Kilo Units milli micro

X,000,000 X,000 X .00X .000,00X

M millions K thousands units

m one thousandth μ one millionth

200,000Ω is equal to 200kΩ or 0.2MΩ.

0.002Ω is equal to 2m Ω.

3,300Ω can be written as either 3.3kΩ or 3k3Ω.

You should try to write any amount using

the form and prefix that requires the

fewest 0s written on the page.

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R ESISTANCE IN S ERIES:

A series of something generally means connected along a line, or in a row, or in an order of some sort. In electronics, series resistance means that the resistors are connected one after the other, and that there is only one path for current to flow through.

Here is an example of resistance in series:

R

1

=100 Ω R

2

=200Ω R

3

=300Ω E=24v Note that the resistors are labeled R

1

, R

2

, and R

3

. The numbers 1, 2, and 3 are given as subscripts. Subscripts are very common in electronics work. In this case, the resistors are given identifiers on the schematic, and the values are listed separately.

Resistances in series are seen by the circuit as only one resistance, so it is necessary to add the values together to get a total resistance. In this example:

R

1

+ R

2

+ R

3

= R

T

100Ω + 200Ω + 300Ω = 600Ω

Use the R

T

value to find the current draw on this circuit using ohm’s law:

I = E/R I = 24v/600Ω

or I = 0.04A

or 40mA

Current and resistance are inversely proportional, as one goes up the other goes down. A high resistance value will lead to a low current flow. A low resistance value will lead to a high current flow. This presupposes that the voltage remains constant. A higher voltage will pass more current at a static resistance value.

The behavior of series circuits are governed by three specific laws that can be used to determine the relationship between volts, amps, and ohms within that circuit.

LAWS OF SERIES CIRCUITS

1) Individual resistances add up to the total circuit resistance.

2) Current through the circuit is the same at every point.

3) Individual voltages throughout the circuit add up to the total voltage.

Law # 1 was addressed on the previous page as R

1

+ R

2

+ R

3

= R

T

. Law # 2 should be somewhat intuitive, because it seems self evident that the same number of electrons should return to the power source as the number that left it.

According to Rutherford’s atomic theory, electrons are not being created or destroyed, they are just being pushed along through the circuit.

Law # 3 requires a bit more explanation.

As it turns out, the voltage pressure is

shared throughout the circuit,

proportional to the amount of resistance

at specific points in that circuit. In

keeping with that rule, voltage readings

taken at various points in a series circuit

will vary in accordance with the resistances

present at the particular point in the

circuit where the reading is taken.

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R

1

= 100Ω R

2

= 200Ω R

3

= 300Ω E = 24v

The junction between each of the components of this circuit is considered a node. If this circuit were built, it would be possible using a volt meter to take six different voltage readings for this circuit by measuring between these points:

A/B B/C C/D A/D A/C B/D It is also possible to determine the values mathematically, using what we know about Ohm’s law, and using the following procedure.

STEP ONE:

Use the first law of series circuits to determine R

T

for the circuit by adding together the individual resistances. In the earlier section, this was determined to be;

100Ω+200Ω+300Ω=600Ω

STEP TWO:

Use Ohm’s law to determine the current flow through the circuit. I = E/R

T

The current flow is the same at every point in a series circuit. This is the second law of series circuits. Again, we have already worked the current out to be:

The answer could be converted to milliamps, but this would just confuse the rest of the problem. It is best to wait until the end.

STEP THREE:

Remember that the current (I) remains constant throughout the entire circuit.

Ohm’s law in the configuration E = IR can be used to determine the voltage drop across any two nodes in the circuit.

Between A and B

E = IR or E = .04A x 100Ω or E = 4v Between B and C

E = .04A x 200Ω or E = 8v Between C and D

E = .04A x 300Ω or E = 12v

The voltage between any other points can be determined by adding together the appropriate legs of the circuit.

STEP FOUR:

Add all of the voltages together to check your work. The individual voltages should add up to the total voltage from the power source, which is the third law of series circuits.

4v + 8v + 12v = 24volts, the answers we

had were correct.

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Here is an example in which a meter is used to reference the same concept:

SOME PRACTICAL EXAMPLES

ACLs (aircraft landing lights) are sometimes used for special effects lighting on stage because they put out a tightly focused beam of light. These lamps were literally designed to be used on airplanes, which traditionally use a voltage rated at 28.5v. If they were simply plugged into a standard wall voltage of 120v, the filaments would burn out immediately because the voltage pressure would be far too much for them to carry. A special power supply could be used, but there is an easier way around this problem.

Ohm’s law can be used to determine a way to divide up the voltage in a series circuit created by four of the lamps in series. This will allow the lamps to be used with a standard 120 volt source, however it should be noted that if one of the lamps burns out, all of them will go dark because the completed circuit has been opened.

Electrons cannot move through an open circuit.

As a second example, if you were to use a very long power cord for tool with an electric motor, it might not operate properly. Quite often the motor might stall or have trouble starting. This problem comes about because the resistance of the electric motor is in series with the resistance present in the power cord itself, and a voltage divider is created.

The voltage pressure is shared between the cable and the tool. Any type of cable has a certain resistance to the flow of electrons, but a larger diameter conductor has less resistance per foot. Also, a shorter cable has less resistance than a longer one of the same diameter because there are fewer feet of it creating resistance. This problem with the motor can be avoided by using a larger gauge cable that is as short as possible.

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In electronics work, voltage dividers are often used to lower the voltage applied to a specific part of a particular circuit. This is especially true for circuits using transistors that we will look at later.

Another case might involve an indicator light that shows that the power is switched on to a particular device. Adding a resistor in series with the lamp will “share” the voltage between the resistor and the lamp so that the light runs on a smaller voltage.

Higher voltages make it much more likely that a person can receive electrical shock, so it is common to use a lower voltage for control devices.

RESISTANCES IN PARALLEL:

There is another way to place more than one resistance into a circuit rather than in series. Here is a standard type of parallel circuit.

In this example, each resistor has its own discrete path to the voltage source, and if one of the pathways is opened, the other will still operate. In a parallel circuit, the voltage in each part of the circuit remains constant, but the current varies in accordance with where a reading is taken.

This is the opposite of the way a series circuit operates.

There are many different ways to organize a parallel circuit. In the practical world, most wiring is done in parallel so that the voltage to any one part of the network is the same as the voltage supplied to any other part of it. Having a constant voltage is very important because electrical devices are designed to operate from a specific pressure. It would be impractical to change that voltage at will throughout the electrical service.

Although the wiring running between the

lights is arranged differently, these lamps

have the same electrical connection as the

lamps depicted in the previous schematic

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wiring in a lighting system may be, all of the circuits involved are still in parallel, and all of the outlets have the same 120v service.

LAWS OF PARALLEL CIRCUITS

1) The reciprocals of all the individual resistances add up to the reciprocal of the total circuit resistance. 1/R

T

= 1/R

1

+ 1/R

2

+ 1/R

3

… 2) Voltage through the circuit is the same at every point.

3) Individual current draws throughout the circuit add up to the total current draw. I

T

= I

1

+ I

2

+ I

3

Remember that the voltage in every part of a parallel circuit is the same. As a result, Ohm’s Law is most often used in a parallel circuit to determine what the total current draw for the entire network will be.

Theatre lighting systems (like all others) are protected by either fuses or circuit breakers which will disconnect the flow of electricity if too much demand is placed on that system. The purpose is to protect the component parts from damage from overloading. If too many electrons pass through the wiring inside the wall, or through a jumper, the wire will overheat just like the filament inside a light bulb.

However, the filament is housed inside a vacuum, and the entire lamp is constructed to withstand such overheating.

Wiring is not intended to withstand that sort of extreme use, and will likely cause a fire if too great a load is place on it.

Circuit breakers have a current limit stamped on the switch so that users know in advance how much current can be drawn through them. This maximum allowable value is determined by engineers to limit the current to an amount safe for all the components in the system. It is often necessary to add up all of the current

draws on a system to see if that total amount exceeds the limitation set by the circuit breaker.

HOW TO DETERMINE CURRENT FLOW IN A PARALLEL CIRCUIT:

1) Use Ohm’s law to determine the current flow in each branch.

2) Add the currents together to find I

T

for the entire circuit.

In this procedure, we will assume that both the resistance and the voltage are givens for any particular problem, since it is I, the current, that is being determined.

Here is a drawing of a parallel circuit:

Use the magic triangle to find the correct formula for finding the current value of the entire circuit. The formula is I = E/R, the voltage divided by the resistance. The voltage is 120v. But what should the total resistance be? In series circuits R

T

was discovered by adding together all of the individual resistances because the current must flow through each one of them to complete the circuit. But in examining the drawing it is apparent that the current will be split between two branches of the circuit. Splitting the current has the effect of lessening the resistance against it because the pathway has in effect become

“wider”.

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Referring back to the first law of parallel circuits we see that it states:

The reciprocals of all the individual resistances add up to the reciprocal of the total circuit resistance. 1/R

T

= 1/R

1

+ 1/R

2

+ 1/R

3

So in determining R

T

for the parallel circuit shown, it is apparent that:

1/R

T

= 1/6 + 1/8 or 1/ R

T

= .167 + .125 or 1/ R

T

= .292

R

T

= 3.42Ω

Continuing on, the formula I = E/R is used to determine the total current used in the circuit:

I = 120v/3.42Ω I = 35A

There is an alternative method to using the reciprocals, that determines the value of I one branch at a time. If instead of two lamps in this circuit, there were only the first one with its 6Ω resistance, the current would be determined as follows:

I = 120v/6Ω I = 20A

Solving for the 8Ω resistance we see:

I = 120v/8Ω I = 15A

The 20A and the 15A can be added together for a sum of 35A, the same answer as with the first method.

Determining the individual currents in a system, and adding them together to get a figure for the entire system is another way to solve for I

T

.

I = I + I + I …

Notice that adding lamps to the system causes the current draw to go up. It seems intuitive that the more lights in a circuit the more power will be consumed. Also, the lower the resistance of each lamp, the more power will be consumed.

Most of the time, we do not know the resistance of a particular stage lamp, although it could easily be determined with a VOM. Instead, the power rating is given. Power ratings are stated in Watts.

P is the symbol used for mathematical computations using power.

W is the symbol used to express an amount of power. Example: P=100W.

H OW TO D ETERMINE P OWER U SAGE IN W ATTS:

Another of Ohm’s formula’s is used to determine wattage, and is often used in conjunction with the formula E = IR that we have been using so far. This formula is called the Pie Formula for obvious reasons.

P = IE

The same sort of magic triangle that was used previously will work for this formula as well:

The following derivations are possible:

P = IE

I = P/E

E = P/I

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Suppose that the following schematic is given, and that the problem is to find the total current draw:

We are seeking the value of I, and covering that symbol on the magic triangle gives the formula:

I = P/E

First the total power consumed by the circuit must be determined.

P

T

= P

1

+ P

2

+ P

3

… P

T

= 500W + 1000W P

T

= 1500W

Solving:

I = P/E

I = 1500W/120v I = 12.5A

TERMS USED IN THIS CHAPTER

Aircraft Landing Lights Color Band

Laws of Parallel Circuits Laws of Series Circuits Magic Triangle Ohm’s Law Open Circuit Order of Magnitude Parallel

Pie Formula

Power

Resistance

Resistor

Series

Subscript

References

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