CHAPTER 4
OHM'S LAW
CURRENTRESISTANCEVOLTAGE
You learned in chapter 1 that the flow of CURRENT is
influenced by both VOLTAGE and RESISTANCE. INCREASE
the VOLTAGE, and you also INCREASE the CURRENT. But
if you INCREASE the RESISTANCE, the CURRENT DECREASES.
In some respects, the flow of current in a circuit is like
the movement of water in a pipe. The rate of water
flow is influenced by the size or the pipe and by the
amount of pressure applied to the water. Naturally a
large pipe with few obstacles will allow more water to
move through it than a smaller one. And if a high pressure is applied a greater flow will take place than if the
pressure is lower.
Don't become confused by thinking that voltage and
water pressure are exactly the same thing. While water
pressure has an effect on the movement of water similar
to the effect of voltage on the movement of electrons, they
are not the same. Water pressure is a "push" from

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behind, while electrons are "pulled" toward the higher
voltage. The results of the two are much the same, but
you should keep the difference in mind.
The relationship of the CURRENT to both VOLTAGE and
RESISTANCE is given in Ohm's Law. It reads, "The current flowing in a circuit VARIES DIRECTLY as the VOLTAGE
and INVERSELY as the RESISTANCE." Since Ohm's Law
states the relationship as a PROPORTION, you may write
the law in a simple formula
Current (I) = Voltage (E) / Resistance (R) or just I = E / R
FINDING THE CURRENT BY USING OHM'S LAW
Ohm's Law provides a simple method for finding the
CURRENT flowing in a circuit if you know the VOLTAGE and
RESISTANCE.


Figure 28.Simple circuit.

In figure 28, a lamp is connected to a battery with an
emf of 1.5 volts. The resistance of the lamp is 15 ohms.
To find the current flowing, substitute the values of E
and R in the formula of Ohm's Law and solve

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I = E / R = (1.5 / 15) = 0.10 amperes
Thus whenever you know the applied VOLTAGE and the
RESISTANCE of a circuit, you can find the current by using
Ohm's Law.
FIND THE RESISTANCE BY USING OHM'S LAW
By performing an easy mathematical maneuver, Ohm's
Law can be changed
from I = E / R to R = E / I
In this new form you use Ohm's Law to find the RESISTANCE of a circuit if you know the CURRENT and the
APPLIED VOLTAGE.
The ammeter, in figure 29, indicates a current of 0.02
amperes flowing through the circuit. The applied E is


Figure 29.Simple circuit.

three volts. To find the resistance of the circuit, substitute in the values of E and I in the formula
R = E / I = 3/0.02 = 150 ohms

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FINDING THE VOLTAGE BY USING OHM'S LAW
When a current flows through a resistance, a DROP in
voltage occurs. The LARGER the current and the HIGHER
the resistance, the GREATER will be the DROP in voltage.
To find this voltage drop, Ohm's Law is again changed
from I = E / R to E = IR
The voltage in the new formula is equal to the PRODUCT
of the current and voltage. Hence if you know the resistance, and the current flowing through it, you can find
the VOLTAGE DROP ACROSS THE RESISTANCE. Voltage drops
across resistances are usually called "IR" drops.


Figure 30.Simple circuit.

The battery in figure 30 is forcing a current of 0.02
amperes to flow through a lamp whose resistance is 100
ohms. The IR drop across the lamp will be
E = IR = 0.02 X 100 = 2 volts
IR DROPS ABOUT A CIRCUIT
It is not unusual for a circuit to contain several resistances in series. Since the same current flows through
all the resistances, IR drop will be present across each
resistor.

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Figure 31.IR drops about a series circuit.

Figure 31 is a basic filament circuit used in some
small ACDC receivers. The filaments of the 5 tubes
T_{1} ..... T_{5} are connected in series with an outside resistor R. The resistance of each filament is given below
the tubes. An ammeter in the circuit indicates a current
of 0.15 amperes to be flowing. The IR drop across each
tube and resistor will be equal to
resistance X 0.15
So for the series circuit the IR drops will be
R  80 X .15 =  12.00 volts 
T_{1}  94 X .15 =  14.00 volts 
T_{2}  80 X .15 =  12.00 volts 
T_{3}  80 X .15 =  12.00 volts 
T_{4}  233 X .15 =  34.95 volts 
T_{5}  233 X .15 =  34.95 volts 
Total 119.90 volts 
All the IR drops added together equal 119.90 volts.
Now look at the applied voltage120 volts. It suggests
that the sum of the IR drops about a closed circuit are
EQUAL to the APPLIED E.
What happened to the other 0.10 of a volt
(120  119.90)? It's distributed over the connecting
wires and the meter.
Hence, in any circuit, "The sum of the individual IR
drops is EQUAL TO the APPLIED VOLTAGE." That statement
is known as Kirchhoff's First Law.

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Kirchhoff also discovered a law about current. It
states, "As much current flows AWAY from a point as
INTO it."


Figure 32.Current flowing through a divided circuit.

The battery in figure 32 causes a current of 2 amperes
to flow through the circuit. At point A, the circuit
divides and 1 ampere flows through each leg. At point
B, the circuit branches again with 1/2 ampere flowing
through each branch. At point C, the three branches
come together, and 2 amperes flow away.
In the case of each of the points, A, B, and C, AS MUCH
CURRENT FLOWS AWAY AS INTO THE POINT. While this
law may sound simple, it is one of the most useful in the
study of electricity.
POWER
POWER is the RATE OF DOING WORK. Suppose you are
assigned to help load ammunition. If you have a liberty
coming up when the loading is done, you will work hard
to get the job done in a hurry. But if you are not going
any place, chances are you will do the job in a more
leisurely manner.
In the first case, the RATE of doing work is high. You
will use up a lot of energy in a short period of time. In
other words, a lot of POWER will be exerted in doing the

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job. But when you go about your job leisurely, energy
is being expended at a slow rate and the power consumption is low.
In the same way, when a lot of electrons are forced
through a resistor in a short period of time, a lot of energy
is used and the power consumption is great. If fewer
electrons are forced through the same resistor, less power
is dissipated.
Actually, the amount of power consumed in forcing
electrons through a resistor is proportional to the SQUARE
of the CURRENT times the RESISTANCE, or
Power (in watts) = I^{2}R
Thus, if 0.02 amperes is flowing through 1,000 ohms of
resistance, the power being dissipated will be
Watts = .02^{2} X 1,000
Watts = 0.4
But if the current is increased to .04 amperes, the
power goes up to
Watts = .04^{2} X 1,000
Watts = 1.6
There are two other formulas that you can use to find
the power of a circuit. One is
Watts = E X I
and the other is
Watts = E^{2} / R
You may use any of these three formulas to find the
power. They all express rate of doing work.

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