DFV10113-CHPTR 3

WELCOME TO
ELECTRICAL
CIRCUIT CLASS…

1



1 UNIT 3 - LEARNING OUTCOME
2 3.1 DEFINITION OF OHM’s LAW
3 3.2 CALCULATING CURRENT, VOLTAGE

& RESISTANCE

4 3.3 ANALYZING OHM’s LAW
5 3.4 INTERNAL RESISTANCE & EMF
6

3

1 LEARNING
OUTCOME

After completing the unit, students
should be able to:

1  Explain Ohm’s law.
2  Apply Ohm’s law to determine voltage,

current and resistance of a circuit.

3  Interpret the relationship voltage, current
and resistance in circuit using Ohm’s
Law.

4  Calculate internal resistance of measuring
equipment .

REVIEW OF V, I AND R

Voltage is
 the amount of potential energy available to move electrons

from one point to another in a circuit.

Current is
 the rate of charge flow and is measured in amperes.

Resistance is
 the opposition that a substance offers to the flow of electric

current and is measured in ohms.

Quantity Symbol Unit of Unit

Current I Measurement Abbreviation
Voltage V or emf Ampere (Amp) A
Resistance R
Volts V

Ohm Ω

21 DEFINILTEIAORNNOINF G
OHMO’sULATCWOME
7

TOPIC LEARNING OUTCOME “

After completing the topic, students
should be able to:

“Explain OHM’s Law

OHM’s LAW

The CURRENT in a circuit between two points is
directly proportional to the VOLTAGE across the two
points, and inversely proportional to the RESISTANCE

between them

Georg Simon Ohm (16 March 1789 – 6 July 1854)
was born in Erlangen, Bavaria. He worked as a
mathematics lecturer, and as a physics teacher.
Being the son of a locksmith he was able to build
equipment to research the recently invented
electrochemical cell. These experiments resulted in
the discovery of Ohm's Law.

OHM’s LAW

 Ohm’s Law defines that in an electrical circuit,
current is proportional to the voltage.
The circuit current is inversely proportional to
the resistance R.

 Current and voltage have a linear relationship
with resistance remain constant.

V ∝I Where,
I ∝1 V= voltage in volts(V)
I = current in ampere(A)
R R= resistance in ohms(Ω)

Equation:

THE LINEAR RELATIONSHIP OF
CURRENT AND VOLTAGE

Voltage Current current Voltage vs current Current
(V) (A)
10 1 10 I = V
20 2 9 R
30 3 8
40 4 7
50 5 6
60 6 5
70 7 4
80 8 3
90 9 2
1
0 10 20 30 40 50 60 70 80 90

Voltage

Note: Resistance value is constant(10Ω)

THE INVERSE RELATIONSHIP OF CURRENT
AND RESISTANCE

Resistance Current Current
(Ω) (A)
10 1 1.2 Current
20 0.5 1
30 0.8 I =V
40 0.333 0.6 R
50 0.25 0.4
60 0.2 0.2
70 0.167 0
80 0.143
90 0.125
0.111

Note: Voltage value is constant(10V)

SUMMARY

1. Current is proportional to the voltage.
2. Current is inversely proportional to

the resistance R.

1 CALCULATING
3 CURRENT, VOLTAGE

AND RESISTANCE

14

TOPIC LEARNING OUTCOME “

After completing the topic, students
should be able to:

“Apply Ohm’s law to
determine voltage,
current and resistance
of a circuit.

CALCULATING I,V AND R

 Formulated with three variables: V, I, and R
 Relationship called Ohm’s Law
 Three forms exist:

I =V V = IR R =V
R I

EXAMPLE 1

Let's see how these equations might
work to help us analyze simple circuits.

Direction of current flow : There is only one source of
voltage (the battery, on the
+ left) and only one source of
- resistance to current (the
lamp, on the right).
Simple Circuit 1 This makes it very easy to
apply Ohm's Law

EXAMPLE 1

In this example, we will calculate the
amount of current (I) in a circuit, given
values of voltage and resistance.

What is the amount of
current (I) in this circuit?

+ V=IR
- I=V/ R
I=12V/3Ω
Simple Circuit 1 I= 4 A

EXAMPLE 2

• In this example, we will calculate the
amount of resistance (R) in a circuit,
given values of voltage (E) and current
(I):
What is the amount of
resistance (R) in the
circuit?
V=IR
R=V/I
R=36V/4A
R=9Ω

EXAMPLE 3

• In the last example, we will calculate
the amount of voltage (V) supplied by a
battery, given values of current (I) and
resistance (R):
What is the amount of
voltage provided by the
battery?
V=IR
V=(2A)(7Ω)
V=14Ω

Ohm’s Law Chart (MEMORY CHART)

Solve for V Cover the quantity that is
unknown.
V=IR

V

I xR

Ohm’s Law Chart (MEMORY CHART)

Solve for I Cover the quantity that is
unknown.
I=V/R

V
IR

Ohm’s Law Chart (MEMORY CHART)

Solve for R Cover the quantity that is
unknown.
R=V/I

V
IR

Lets do the exercise
for better

understanding

If you need to solve for current, Ohm’s
law is:
I =V
R

What is the current in
from a 12 V source if the
resistance is 10 Ω?

Answer: 1.2 A

Determine the value of voltage across a
680 Ω resistor if the current is 26.5 mA.
If you need to solve for resistance, Ohm’s law is:

V = IR

Answer: 18 V

Determine the value of (hot) resistance in
. the bulb.

115 V OFF V
Hz

V

mV

A Ra ng e 1s
Auto ra ng e 1s
10 A
Touc h/Hold

V

40 mA COM

Fused

Answer: 132 Ω

A circuit has a voltage of 12 V and a resistance
of 100 ohms. Calculate the current going
. around the circuit.

= 12
= 120Ω
=?
=
12
= = 120Ω

Answer: 0.1A

Determine the current in the circuit.

.

Answer: 30mA

Determine the value of resistor in the circuit.

150 mA

R

Answer: 30mA

Determine the value of voltage source in the
circuit.

100 mA
Answer: 50V

The resistor is green-blue brown-gold.
Determine the DC ammeter reading.

+ DC Ammeter -

Power Supply

VA

+15 V

Gnd 5 V 2A - + - +

Answer :26.8 mA

SUMMARY

1. Voltage and current are linearly
proportional.

2. Ohm’s law gives the relationship of
voltage, current, and resistance.

3. Use: V = IR, when calculating voltage.
4. Use: I = V/R, when calculating

current.
5. Use: R = V/I, when calculating

resistance.

41 ANALYZING
OHM’S LAW
34

TOPIC LEARNING OUTCOME “

After completing the topic, students
should be able to:

“ Interpret the

relationship voltage,
current and

resistance in circuit
using Ohm’s Law

ANALYZING OHM’S LAW

CASE 1 : For R constant

Graph of voltage vs current CASE 1: If we can increase V and
keep R a constant, then from
V Ohm’s Law, we can see that
I will increase. Similarly, if we
V2 I decrease V and keep R a
V1 constant, from Ohm’s Law, I will
decrease.
I1 I2
V2 > V1
I2 > I1

27

ANALYZING OHM’S LAW

We can interpret the relationship between current,
voltage and resistance using the following circuit where
the current flow in a closed circuit through a variable
resistor R and where the voltage source is a variable
voltage V.

I

If V=5V and R=10Ω,then
using Ohm’s Law,
+ R current is
V

- ……………………..

V=IR, thus I=0.5A

Simple Circuit 1

EXAMPLE :CASE 1

Let V be increased to 10V, R remain at
10Ω, then using Ohm’s Law we have
current value

V=IR (Ohm’s Law)

I =V/R =10/10

+ R
V
I =1A
-
(I increased from 0.5A to 1A)

Simple Circuit 1

EXAMPLE :CASE 1

Now, let's decrease V to 2.5V, R remain
at 10Ω, then using Ohm’s Law we have
current value

V=IR (Ohm’s Law)

+ I =V/R =2.5V/10
V
R
-
I =0.25A
Simple Circuit 1
(I decreased from 0.5A to 0.25A)

If V=5V and R=100Ω,then using Ohm’s Law,
current is 0.05 A. Interpret the current value
if we decrease V to 1.5V and R remain at
100Ω.

V=IR (Ohm’s Law)
I =V/R =1.5V/100
I =0.015A

From the calculation , I decreased from 0.05A to 0.015A. So it
conclude that from Ohm’s Law, if we decrease V and keep R a
constant, I will be decrease.

If V=5V and R=100Ω,then using Ohm’s Law,
current is 0.05 A. Interpret the current value
if we change the value of V to 7.5V and R
remain at 100Ω.

V=IR (Ohm’s Law)
I =V/R =7.5V/100
I =0.075A

From the calculation , I increase from 0.05A to 0.075A. So it
conclude that from Ohm’s Law, if we increase V and keep R a
constant, I will be increase.

ANALYZING OHM’S LAW

CASE 2 : For V constant

Graph of Voltage vs Resistance CASE 2: If resistance is varied for
a constant voltage, the current
10 verses resistance curve plots a
hyperbola.
8.0
Current (mA)
(R1, I1 )

6.0

4.0

(R2, I2 )

2.0

0 R2 > R1
I2 < I1
0 1.0 2.0 3.0

Resistance (kΩ)

27

ANALYZING OHM’S LAW

We can interpret the relationship between current,
voltage and resistance using the following circuit where
the current flow in a closed circuit through a variable
resistor R and where the voltage source is a fixed voltage
V.
I Let’s say R is 15Ω, and

keep the value V at 5V,
then using Ohm’s Law we
+ R will get current

V= 5V
- ……………………..

Simple Circuit 2 V=IR
I =V/R I =5/15
I =0.33A

EXAMPLE :CASE 2

Let’s choose a larger R which is 50Ω, and
keep the value V at 5V, then using Ohm’s
Law we will get

V=IR

+ I =V/R

V= 5V R

- I =5V/50Ω =0.1A

Simple Circuit 2 (I decreased from 0.33A to 0.1A)

EXAMPLE :CASE 2

Let’s choose a smaller R which is 5Ω, and
keep the value V at 5V, then using Ohm’s
Law we will get

V=IR

+ I =V/R

V= 5V R

- I =5V/5Ω =1A
(I increased from 0.33A to 1A)
Simple Circuit 2

SIMULATION

Let’s try simulate the circuit.

https://phet.colorado.edu/en/simulation/ohms-law

SUMMARY

From the above analysis, we can
conclude that according to Ohm’s law:
1. The current in a circuit where the

resistance is a constant, is directly
proportional to the voltage.
2. The current in the circuit where the
voltage is a constant, is inversely
proportional to the resistance.

51 INTERNAL
RESISTANCE
AND EMF

48

TOPIC LEARNING OUTCOME “

After completing the topic, students
should be able to:

“ Calculate internal
resistance of
measuring
equipment .

INTERNAL RESISTANCE & EMF

01 All voltage sources contain INTERNAL
02 RESISTANCE, that is resistance that is

03 part of the voltage producing
device itself which cannot be eliminated.

The voltage that the device (battery for
example) could produce if no internal
resistance was present
is called its EMF.

EMF stands for Electromotive Force – the
force that moves the electrons.

The useable voltage which is available to
the circuit after the internal resistance
consumes its share of the EMF is called

The terminal voltage, VL.