ELECTRONICS
INDUSTRIAL MATH & SCIENCE
OHMS LAW
Recall earlier when we defined the “basic
electrical circuit.” The 3 things needed for a
complete circuit are a source, load and conductors.
In the circuit shown below these 3 elements are
connected so that electrical current can flow out of
the battery, through the wires to the load and from
the load back to the battery. This represents a
complete circuit.
The load.
Current flows
through the
wires.
It also says that if we increase the load the current
will go down (an inverse relationship).
I is the current in amps. (A)
V is the voltage in volts. (V)
Voltage in an electrical system is like pressure in a
water system. It is the thing that pushes
the electrons along.
I = V R is the load in ohms. The load in an
R electrical system resists the flow of
electrons much like a partially open
valve in a water system. The value of
the load is equivalent to the amount the valve is
open. A smaller value lets more current through
and a higher value lets less current through.
Quantity
Formula symbol Measured in
voltage
Battery.
FIG. 74
A simple circuit.
If we break this circuit by disconnecting a wire
or flipping a switch then we have an “open
circuit.” If we connect the positive and negative
posts of the battery directly to each other then we
have a “short circuit” NOTE this condition will
result in a very high current. This will cause the
battery to overheat, sparks or a flame will result
and the wire will overheat.
Electrical current is the flow of electrons. We
can measure electrical current with a flow meter
much in the same way that the flow of water
through a pipe is measured. Electrical current is
measured in Amperes (Amps) and so an electrical
flow meter is called an Ammeter.
In a water pipe a flow meter is attached by
cutting the pipe and inserting the meter so that all
the water must flow through the meter. An
ammeter is connected into a circuit by cutting a
wire (or breaking a connection) and inserting the
meter so that all the current flows through the
meter. One Ampere is the flow of
6.24 x 1018 electrons per second.
V
V (volts)
current
I
A (amps)
load
R
Ω (ohms)
The Ohms Law formula can be applied to any
circuit or part of a circuit and it can easily be
verified by setting up your own experiment.
Since this is a formula the unknown could be
any of V, I or R. As long as we know 2 of these
the third can be calculated.
I=V R=V V=IxR
R
I
These 3 formulas are really the same formula. The
unknown value is the value to the left of the equals
sign. An easy way to remember this formula is to
put V, I and R into a triangle like the one shown
here.
Really big number.
That is 6 240 000 000 000 000 000 electrons
every second. Fortunately this number is not
relevant in our work with Ohms Law.
Ohms law tells us that a voltage source of 1 volt
connected to a 1Ω load will result in 1 amp. of
current. It also says that if we increase the voltage
then the current will go up (a direct relationship).
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V
I
Use your thumb to cover up the
unknown quantity (the thing
you are trying to find). The
formula you should use
becomes apparent from the
letters that are left over.
R
ELECTRONICS
INDUSTRIAL MATH & SCIENCE
EXAMPLES
How much current will a 2Ω motorcycle
headlamp use? Motorcycles use a 6V electrical
system.
I = V = 6 = 3A
R
2
PROBLEMS (29)
Find the unknown values using Ohms Law.
Problem
1
How much load is needed to keep a 12V
camcorder battery at exactly 0.2A?
R = V = 12 = 60Ω
I 0.2
What is the load if a 60V power source delivers
20mA?
You must first change 20mA to base units.
20mA = 0.02A
R = V = 60 = 3000Ω = 3kΩ
I 0.02
Find the voltage needed to deliver 81mA to a
2.7kΩ load.
You must first change 81mA to base units.
81mA = 0.081A
You must first change 2.7kΩ to base units.
2.7kΩ = 2700Ω
V = I x R = 0.081 x 2700 = 218.7V
I
9V
2
What voltage will a 40Ω disk drive motor need to
run at 0.125A?
V = I x R = 40 x 0.125 = 5V
How much current will a 2kΩ load use if the
voltage is 120V?
You must first change 2kΩ to base units.
2kΩ = 2000Ω
I = V = 120 = 0.06A or 60mA
R
2000
V
#
56Ω
0.1A
3
120V
4
12V
5
R
250Ω
2.2kΩ
18A
5mA
1200
PROBLEMS (30)
1. How much current does an automobile
headlamp use if the resistance is 2.4Ω?
2. Find the resistance of a windshield wiper motor
if it uses 2.5A.
3. What is the voltage of a flashlight that uses a
160Ω lamp operating at 28mA?
4. Calculate the resistance of a toaster that draws
5A.
5. How much current will a fan use with a 50Ω
motor?
PROBLEMS (31)
Find the unknown values using Ohms Law.
V
I
1
12V
14mA
2
50V
Problem
#
3
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R
2.5kΩ
33mA
400Ω
4
1kV
600Ω
5
14V
12kΩ
ELECTRONICS
INDUSTRIAL MATH & SCIENCE
PROBLEMS (32)
P
Find the unknown values using Ohms Law.
Problem
1
#
V
I
180V
R
V
1.5MΩ
2
3mA
PROBLEMS (33)
89kΩ
3
2kV
4
4.5V
800Ω
5
120V
1.6kΩ
I
Find the unknown values using Ohms Law.
50mA
P
V
1 1000W
120V
Ohms Law can also tell us how much power a
load will use in Watts. Power is directly proportional to the voltage and current in a circuit. If you
increase the voltage the power will go up. If you
increase the current the power will go up.
2
120V
P=VxI
5
60W
3
4
1.2kV
5kW
I
4A
18A
6V
50mA
EXAMPLES
Our 6V motorcycle lamp uses 3A.
P = V x I = 6 x 3 = 18W
PROBLEMS (34)
The 12V camcorder battery giving a steady output
of 0.2A is supplying
P = V x I = 12 x 0.2 = 2.4W
Find the unknown values using Ohms Law.
The 5V disk drive motor that draws 125mA is
using (125mA = 0.125A)
P = V x I = 5 x .125 = 0.625W
1
2
The formula P = V x I can also be rearranged to
find V or I
I=P
V
V=P
P
P=VxI
3
4
5
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100W
V
I
12V
6A
24V
3.1V
1.8kW
18mA
15A
1kV
35mA
ELECTRONICS
INDUSTRIAL MATH & SCIENCE
PROBLEMS (35)
Find the unknown values using Ohms Law.
P
1
2
V
I
120V
150W
R
10Ω
3A
3
72mA
5kΩ
PROBLEMS (36)
Find the unknown values using Ohms Law.
1
P
V
50mW
5V
2
12V
3
I
R
65mA
14mA
18kΩ
PROBLEMS (37)
Find the unknown values using Ohms Law.
1
P
V
5kW
220V
2
3 180mW
I
R
12.5V 135mA
15mA
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