Electrical Fundamentals
Module 1: Ohm's Law
PREPARED BY
IAT Curriculum Unit
August 2008
© Institute of Applied Technology, 2008
ATE310 – Electrical Fundamentals
Module 1: Ohm's Law
 Module Objectives
Upon completion of this module, students should be able to:
1. Use prefixes to convert electrical quantities.
2. State Ohm's Law and define the relationship between
current,voltage, and resistance.
3. Use Ohm's Law to solve unknown quantities of current,resistance,or
voltage.
4. Apply the power formula to calculate the power in a circuit.
 Module Contents
1. Introduction to Ohm's Law
2. Investigations of Ohm's Law using Edison Version 4 simulator
3. Electrical Units and Prefixes
4. Electrical Power
5. Practical Tasks
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
1.1 Introduction to Ohm's Law
Ohm's Law applies to electrical
circuits; it states that the current
passing through a conductor between
two points is directly proportional to
the potential difference (i.e. voltage
drop or voltage) across the two points,
and inversely proportional to the
resistance between them, at a constant
temperature.
The
mathematical
equation
describes this relationship is:
I
Simple electric circuit
that
V
R
Where I is the current in amperes, V is
the potential difference between two
points of interest in volts, and R is a
circuit resistance in ohms
The current (I) in a circuit is
directly proportional to the
applied voltage (V) and
inversely proportional to the
circuit resistance (R)
Ohm's Law can be expressed in the form
of three formulas as shown in the table
below and the triangular on the right hand
side. Using these three formulas, and
knowing any two of the values for voltage,
current, or resistance, it is possible to find
the third value.
Find Current
V
R
Current equals
voltage divided by
resistance
I
Ohm's Law Formulas
Find Voltage
V= I x R
Voltage equals
current multiplied
by resistance
Ohms law
Find Resistance
V
I
Resistance equals
voltage divided by
current
R
Module 1: Ohm's Law
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ATE310 – Electrical Fundamentals
1.2 Investigations of Ohm's Law
using Edison Version 4 simulator.
Edison 4 provides a unique new
environment for learning electricity and
electronics.
Teachers,
students
and
electronics enthusiasts can use digitally
scanned photorealistic components, a
solder
less
breadboard,
virtual
instruments, sound and animation to
create, test, and safely repair lifelike 3D
circuits and simultaneously see the
corresponding circuit schematic. Edison
also comes with over 100 experiments
and problems that teachers and students
can use immediately.
Edison 4 has a double panel screen layout
as shown in Fig 1.1.
1. The left main window shows a 3-D
perspective view of the working area
with "component shelves" on both
sides. You can pick up parts from the
shelves by clicking on them with the
left button of the mouse.
2. The right panel presents the same
parts
using
standard
schematic
symbols.
Fig 1.1: Edison Screen Layout
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Aim:
The goal of this lab is to complete our
understanding of how voltage, current and
resistance relate to each other in circuits
using a computer simulation instead of
using real wires, bulbs, resistors and
measuring instruments.
Creating a new circuit
1. Click the top 4.5V battery on the
shelf. The cursor will assume the
shape of a battery which you can
move any where on the left side of
the table as shown in Fig 1.2.
2. Let's add a bulb and switch to
complete the rest of the circuit as
shown in Fig 1.3
Fig 1.2: 4.5V battery
Fig 1.3: Main Components
3. The schematic symbol of the battery
will appear at the same time in the
right side window as shown in Fig
1.4.
4. Let's connect the parts with wire as
shown in Fig 1.5. When you position
the cursor over the terminal of a
part, the cursor changes to a small
circle. Click the mouse button and
use the mouse to trace the wire.
When the wiring cursor reaches the
target terminal, the cursor once
more changes to a small circle .Click
the left mouse button to end the
wire.
5. Move the cursor over the switch and
see it change into a hand symbol:
click the switch to complete the
current path as shown in Fig 1.6
Fig 1.4: Circuit Symbols
Fig 1.5: Circuit Connections
Fig 1.6 Circuit in Operation
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ATE310 – Electrical Fundamentals
Measuring the Current and Voltage
1. To measure current, you will need
to break your circuit. The current
needs to flow through the ammeter,
the ammeter needs to be placed in
series with the element (bulb,
battery) that we are measuring.
2. Switch off the circuit.
3. Drag and insert the ammeter
between the bulb and the battery.
4. To measure the voltage difference
across the bulb, a voltmeter is
needed. The voltmeter needs to be
placed in parallel with the bulb.
5. Drag and connect the voltmeter
across the bulb terminals as shown
in Figures 1.7 and 1.8.
6. Switch ON the circuit as shown in Fig
1.9 and record the readings of the
voltmeter and the ammeter in the
table below.
7. Switch OFF the circuit.
8. Use an ohmmeter to measure the
resistance of the bulb as shown in
Fig 1.10
9. Take the measured figures for
voltage and current; use Ohm's Law
equation
to
calculate
circuit
resistance. Compare this calculated
figure with the measured figure for
circuit resistance.
Fig 1.7 Current and Voltage
Measurements
Fig 1.8: Symbols of the
Measuring Instruments
Fig 1.9: Ammeter and
Voltmeter Readings
Bulb 1
(V)
I (A)
RC (Ω)
RM (Ω)
RC: Calculated resistance of the bulb
RM: Measured resistance of the bulb
Where :
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RC= V/I
Module 1: Ohm's Law
Fig 1.10: Using the
Ohmmeter to Measure the
Bulb Resistance
ATE310 – Electrical Fundamentals
10. Let's change the bulb by another
one
with
lower
resistance
(Higher power). You can do this
by double clicking on the bulb
and changing the value of the
power to a higher value.
11. Repeat steps 6, 7 and 8.
12. Record your results in the table
below.
13. Repeat step 9 to study the
voltage and current formulas of
Ohm's Law.
Bulb 2
(V)
I (A)
RC (Ω)
RM (Ω)
RC: Calculated resistance of the bulb
RM: Measured resistance of the bulb
Where :
RC= V/I
Conclusions :
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Module 1: Ohm's Law
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ATE310 – Electrical Fundamentals
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1.3 Electrical Units and Prefixes
There are three basic measurements
made on electrical circuits: 1- Voltage, 2current and 3- resistance. Figure 1.11a
lists these basic electrical quantities their
symbols and their functions.
In certain circuit applications, the basic
units, volt, ampere, and ohm, are either
too small or too big to work with. In such
cases, metric prefixes table, shown in
Fig 1.11 b, is used.
Quantity
Unit of Measure
Function
Name
Symbol Name
Symbol
Voltage
U
Volt
V
Voltage is the electromotive
V
force or pressure which
Emf
makes current flow in a
circuit
Current
I
Ampere
A
Current is the flow of
electrons through a circuit
Resistance
R
Ohm
Ω
Resistance is the opposition
to current flow offered by
electric devices in a circuit
Fig 1.11 a: Basic Electrical Quantities
One
One
One
One
One
One
One
One
Number
1,000,000,000
1,000,000
1,000
1
thousandth
0.001
million
0.000001
billionth
0.000000001
trillion
0.000000000001
billion
million
thousand
Power of Ten Prefix
109
giga
6
10
mega
3
10
kilo
0
10
--10-3
milli
10-6
micro
-9
10
nano
-12
10
pico
Fig 1.11 b: Metric Prefixes Table
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Module 1: Ohm's Law
Symbol
G
M
K
--M
μ
n
p
ATE310 – Electrical Fundamentals
Prefix Chart
Movement of decimal point to and from base units
3
3
Mega M
Kilo K
3
3
3
Base Units
milli m
3
3
micro μ
3
Example 1.1
To convert amperes (A) to milli amperes
(mA) , it is necessary to move the
decimal point three places to the right
(this is the same as multiplying the
number by 1000)
0.012 A= ? mA
0.012 A= 0.012
0.012 A = 12 mA
Example 1.2
To convert ohms (Ω) to kilohms (K Ω), it
is necessary to move the decimal point
three places to the left.
47000.0 Ω =? K Ω
47000.0 Ω= 47000.0
47000.0 Ω= 47.0 k Ω
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ATE310 – Electrical Fundamentals
Applied examples on Ohm's Law
Example 1.3
We measure the voltage across a resistor
and find it to be 17 V. We measure the
current through the resistor and find that
it is 0.4 A. What is the resistance of the
resistor?
Since we know voltage and current, we
use the following form of Ohm's Law
R = V/I.
Putting in the values for voltage and
current gives:
R = 17/0.4 = 42.5 ohms
Example 1.4
Suppose a 10KΩ carbon resistor is
connected to a 12V battery .Calculate the
current flow.
I=V/R
I= 12/10000= 1.2 mA
Example 1.5
Suppose a solar cell provides a current of
2.5 mA to a 500 ohm load. Calculate the
output voltage of the solar cell:
V= I X R
V= 2.5mA X 500 ohm
V= 1.25 V
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Example 1.6
Suppose an electric kettle draws a
current of 8A when connected to a 120 V
electric outlet. The resistance of the
kettle heating element is :
R = V/I
R= 120/8 =15 Ω
1.4 Electrical Power
The electrical power, used to operate an
electrical device, is defined as the
potential energy or voltage times the
current passing through the device. This
could also apply to a whole electrical
system, such as the power used in
running your household appliances as
shown in Fig 1.12. The electric company
uses the consumed power over a period
of time to calculate the consumed energy
and thus your electric bill.
Fig 1.12: Household
Appliances consuming
Power
Determining Electric Power (P)
The electrical power required to operate a
device is the input voltage times the
current required.
P = VI
Where:




P = electrical power in Watt
V = voltage used in Volt
I = current in Amperes
VI is V times I
Module 1: Ohm's Law 11
ATE310 – Electrical Fundamentals
Electrical power is measured in Watts. If
the amount of watts is large, kilowatts are
used. 1 kilowatt = 1000 watts, just as 1
kilometer = 1000 meters. The
abbreviation for kilowatt is usually kW.
From the above formula and by using
Ohm's Law V=I.R, it is possible to get two
other commonly used power formulas.
P=I2. R --------------------Watt
P= V2/R -------------------Watt
Example 1.7
Suppose an electric heater draws a current
of 8A when connected to its rated voltage of
235V. The power rating of the heater is :
P = VI=235x8=1880W=1.88 KW
Example 1.8
Suppose a current of 30A is being
supplied to an electric load. The total
resistance of the wire used to supply this
current is 0.1 Ω. The power that lost in
the wire is:
P = I2 x R = (30)2 X 0.1 = 90 W
Example 1.9
Suppose a 48 Ω resistor is to be
connected to a 6-V source. The wattage
that must be dissipated by the resistor
is:
P= V2/ R= (6)2/ 48
P= 0.75 W
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Summary of Ohm's Law Calculations
Module 1: Ohm's Law 13
ATE310 – Electrical Fundamentals
Practical Tasks
Investigations of Ohm's Law
Let us investigate the effect of the resistance on the current of the circuit.
This can be achieved by changing the resistance value to control lamp
brightness. Also, Ohm's Law will be applied to find a component
resistance in a circuit by measuring the current passes through the
resistance and the voltage across it.
Objectives
1. To explore the idea of the resistance of a component.
2. To measure the resistance in a simple circuit and investigate
Ohm's Law.
Equipment Required (Per group)
Equipment
Quantity
Electricity & Electronics Constructor EEC470.
1
Basic Electricity and Electronics Kit EEC471-2
1
Power supply unit 0 to 20 V variable dc regulated.
1
Multimeters
2
Task1: Effect of Resistance
Experimental Procedure
1. Construct the simple circuit shown in fig (1.13) as shown in the
patching diagram of Figure (1.14) (use a 10 Ω resistor for
component R).
Fig 1.13: Simple Electric Circuits
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Fig 1.14: Patching diagram
2. Set the variable voltage control to zero and switch it on.
3. Increase the voltage until the voltmeter reads 12 V and observe the
brightness of the lamp.
4. Switch off the power supply and replace the 10 Ω resistor with one
of 100 Ω.
5. Switch on the power supply and again observe the brightness of the
lamp.
6. Finally with the power supply switched off, replace the resistor with
one of1kΩ.
7. Switch on the power supply and note the state of the lamp. It
probably will not lit.
8. Increase the voltage to its maximum value and the lamp should
glow dimly.
Module 1: Ohm's Law 15
ATE310 – Electrical Fundamentals
Conclusions:

Explain the differences you observed in the lamp brightness when
using different values of resistances.
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Task2: Resistance Measurement and Ohm's Law
Let us investigate these simple circuits more thoroughly in order to
measure a resistance value using Ohm's Law by measuring the voltage
applied to it and the current which flows through the same resistance.
Experimental Procedure
1. Construct the simple circuit of Figure (1.15) as shown in the patching
diagram of Figure (1.16) (use a 100 Ω resistor for component R).
Fig 1.15: Resistance Measurements
Fig 1.16: Patching Diagram
Module 1: Ohm's Law 17
ATE310 – Electrical Fundamentals
2. Set the meter to monitor the variable dc voltage.
3. Make sure that the variable dc control knob is fully counter clockwise,
and then switch on the power supply.
4. Increase the applied voltage in 2 V steps from 0 V up to 10 V.
5. At each step measure the current flowing in the resistor, as shown on
the ammeter. Record your reading in table 1.
Table 1
Resistance
Voltage
Current
Calculated value
R
V
I
( R=V / I)
2V
100 Ω
4V
6V
8V
10 V
2V
1K Ω
4V
6V
8V
10 V
6. Turn the voltage back to zero, and replace the 100Ω resistor with one
of 1kΩ, as shown in Figure (1.17).
Fig 1.17: Circuit Diagram
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
7. Now repeat the procedure shown in steps 4 and 5.
8. From table 1, determine the value of resistance used in the circuit for
each
step. Tabulate your results in the same table
9. Measure the actual resistances of the resistors (100 Ω, and 1K Ω) by
using the ohmmeter. Record the measured values in table 2.
Table 2
Resistance
Value
Calculated
value
(From table 1)
(R = V/I)
Measured
value using
Ohmmeter
Resistance
value using
color code
100 Ω
1K Ω
10. Now use the color code chart to determine the resistances of the
resistors used (100 Ω, and 1K Ω). Write down the values obtained in
table 2.
Conclusions

Comment on the value of the resistors obtained in each case. Are the
values the same? Write down your conclusions based on this
observation.
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Module 1: Ohm's Law 19
ATE310 – Electrical Fundamentals
 Compare the calculated, measured and color coded results of the
resistors used in the experiment.
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Exercise
For the table shown below, calculate and record the missing values:
Current
I
100 mA
Resistance
R
Voltage
V
250V
4.7 kΩ
24V
3A
Power
P
40 Ω
Calculations:
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Task 3: Effect of changes voltage on the current
1- This experiment demonstrates the effect of voltage changes on the
current. Use Fig. 11-18 to complete this experiment
a) Breadboard the circuit as shown in Fig. 11-18
Figure 1.18: Circuit Diagram
b) Vary the voltage (in 1V steps) from 0 to 10 V. Measure the current
at each voltage value. Then, record your results in the following
table:
Voltage (V)
Current (mA)
0
1
2
3
4
5
6
7
8
9
10
c) From your results, explain the effect on current due to voltage
change.
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Module 1: Ohm's Law 21
ATE310 – Electrical Fundamentals
Critical thinking
In an electric circuit that consists of a battery and a fixed resistor, if
the resistor value is doubled, what will happen to the current?
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
Review Questions:
1- The symbol used to represent electric current is:
(a) P
(b) V
(c) R
(d) I
2- The base unit used to measure resistance is:
(a) The ohm.
(c) The volt
(b) The ampere
(d) The watt
3- The symbol for the prefix kilo (metric measure) is:
(a) k
(b) M
(c) m
(d) µ
4- The symbol that can be used to represent voltage is
(a) V
(b) emf
(c)U
(d) all of these
5- The multiplier represented by the metric prefix kilo is:
(a) 106.
(b) 103
(c) 10-3
(d) 10-6.
6- The milli ampere is:
(a) one thousand times smaller than an ampere
(b) one thousand times larger than an ampere
(c) one million times smaller than an ampere
(d) one million times larger than a microampere
7- The symbol for the prefix milli (metric measure) is:
(a) m.
(b) M.
(c) 
(d) k.
8-The symbol for the prefix micro (metric measure) is:
(a)m.
(b) M.
(c) 
(d) mic.
9- A 47,000-  -resistor may be designated as :
(b) 470k  (b) 470k  (c) 47k 
(d) 4.7 M 
10- Converting 100mA to its base unit produces:
(a) 0.1 A (b) 0.001 A (c) 10.000 A
(d) 100.000 A.
11- A 2.2- M  resistor expressed in its base unit is :
(a) 2200  (b) 0.220  (c) 22.000 
(d) 2.200,000 
Module 1: Ohm's Law 23
ATE310 – Electrical Fundamentals
12- A voltage of 0.048 V can be expressed as :
(a) 4800mv.
(b) 480mv.
(c) 48mv.
13- A current of 60 µA is equal to :
(a) 60.000 A.
(b) 60 × 106 a.
(d) 4.8 mv.
(c) 60 × 10-6 A. (d) 600 A.
14- The multiplier represented by the prefix milli(metric measure)
is:
(a) 106
(b) 103
(c) 10-3
(d)10-6
15-Ohm’s law states that :
(a) current is directly proportional to the resistance and inversely
proportional to the voltage.
(b) voltage is directly proportional to the current and inversely
proportional to the resistance .
(c) current is directly proportional to the voltage and inversely
proportional to the resistance .
(d) voltage is directly proportional to the resistance and inversely
proportional to the current .
16- Technician A says that in an electric circuit, voltage can exist
without current . Technician B says that current cannot exist
without voltage . who is correct ?
(a) Technician A only
(c) both technician A and Technician B.
(b) Technician B only
(d) neither Technicians a nor Technician B
17- If The voltage applied to a circuit is doubled and the
resistance remains the same . The current
(a) remains the same
(b) decreases by half
(c) is doubled
(d) decreases in proportion
18- If resistor value is doubled in an electric circuit consisting of a
battery and a fixed resistor the current :
(a) doubles
(b) does not change
(c) decreases by half
(d) increases
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Module 1: Ohm's Law
ATE310 – Electrical Fundamentals
19- Ohm’s law may be interpreted for purposes of calculation as :
(a) I = V × R
(b) R =V/I
(c) V =I/R
(d) R = VXI
20- How much current will a 0.75- car rear window defogger draw
when connected to a 12-v battery source?
(a) 9 amps
(b) 12 amps (c) 16 amps
(d) 18 amps
21- What is the internal resistance of an electronic – control
module that conducts a current of 6 mA when connected to a
12-V source?
(a) 0.5 
(b) 2 
(c) 500 
(d)200 
22- The current flowing through a 180-  resistor is measured and
found to be 0.5 A. How much voltage is being applied across it?
(a) 90 V
(b) 9 V
(c) 360 V
(d) 36 V
23- What is the resistance of a soldering iron element that
conducts a current of 3 A when connected to a 120- V electric
outlet?
(a) 480 
(b) 360 
(c) 160 
(d) 40 
24- If a 4.7- K  resistor is connected to a 12- V battery, how
much current flows through the resistor ?
(a) 2.55mA.
(b) 2553 A
(c) 0.39mA
(d)391A.
52- In an electric circuit , if the total current supplied to the circuit
is 25mA and the total resistance of the circuit is 5000  the
source voltage is:
(a) 125V.
(b) 200 v
(c) 1250V
(d) 5V
Module 1: Ohm's Law 25
ATE310 – Electrical Fundamentals
26- The current flowing through a 2.2- K  resistor is measured
and found to be 5mA. How much voltage is being applied
across the resistor?
(a) 11V
(b) 44V
(c) 44mV
(d) 11MV
27- A thermocouple provides a current of 0.750mA to an electric
gas- safety solenoid that has a resistance of 2.5 k  . The value
of the voltage being produced by the thermocouple is :
(a) 300mv. b) 300 v.
(c) 1.875mv.
(d) 1.875v.
28- What is the internal resistance of an electronic control module
that conducts a current of 45mA when connected to a 9-V
battery ?
(a) 500 
(b) 200 
(c) 180 
(d) 18 
29- The base unit used to measure electric power is:
(a) the ohm
(b) the watt.
(c) the ampere
(d) the volt.
30- Which is the general formula for electric power?
(a) P = V × I
(b) 300 V.
(c) 1.875mv
(d) 1.875V.
31- If a 3 V is applied across a 9  resistor, how many watts of
power does the resistor dissipate?
(a) 27 W
(b) 9 W
(c) 3 W
(d) 1 W
32- A 25  , 2-W resistor is to be connected to a 12-V source and
draw a current of 480mA.
Technician A says the resistor will overheat and burn up if
connected as planned. Technician B says the resistor can easily
dissipate this amount of power. Who is correct?
(a) Technician A only
(b) Technician B only
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Module 1: Ohm's Law
(b) both Technician A and Technician B
(d) neither Technician A not Technician B