BASIC ELECTRICAL CIRCUITS
Course Code: EE 131
Aditta Chowdhury
Assistant Professor
Department of EEE, CUET
• Textbooks
• 1. Introductory Circuit Analysis (11th Edition) by Robert
L. Boylestad
2
Types of Circuital Elements
1. Active Element: is one that is capable of
continuously supplying energy to a circuit,
such as a battery, a generator, an operational
amplifier, etc.
2. Passive Element: physical elements such as
resistors, capacitors, inductors, etc, which
cannot generate electrical energy by
themselves but only consume it.
Types of Electrical Sources
1. Independent Voltage Source: also known as ideal
voltage source as its voltage does not depend on
either the value of the current flowing through the
source or its direction but is determined solely by the
value of the source alone.
2. Dependent Voltage Source: also known as
controlled voltage source, provides a voltage supply
whose magnitude depends on either the voltage across
or current flowing through some other circuit element.
A dependent voltage source is indicated with a
diamond shape and are used as equivalent electrical
sources for many electronic devices, such as
transistors and operational amplifiers.
Types of Electrical Sources
Fig. 1. Symbols of Electrical Sources
Law of
Equivalent resistance
Ohm‘sLaw KVL
VDR KCL
CDR
Basic Laws
Your logo
Law of equivalent resistance
Chapter 4,5,6,7
Series Circuit
Conditions to be series
1. One common terminal.
2. No current-carrying element is connected to that common point.
Total resistance for series resistances
2Ω One common point
R1 and R2 in series
4Ω
2Ω
current-carrying element
So, R1 and R2 are not in series
4Ω
Parallel Circuit
Condition to be parallel
They have two common point
Total resistance for parallel resistances
If all the resistance are same (i.e. R1=R2=…..=RN=R)
then, RT = R / N
common point 1
2Ω 4Ω
common point 2
2 common point
Here R1, R2, R3 are in parallel
Because they have 2 common point
Q/A
Quiz
4Ω 2Ω 2Ω
4Ω 4Ω 2Ω
4Ω
Ans: 10Ω
Ans: 25Ω
Ans: 20Ω
Ans: 110Ω
Ans: 1.05Ω
Ans: 0.5Ω
All the resistance are 9Ω
Ans: 12Ω
All the resistance are 8Ω
Ans: 12Ω
All the resistance are 4Ω
Ans: 1.60Ω
4Ω
2Ω
6Ω
Ans: 1.09Ω
Ans: 2.86Ω
Ans: 9.78KΩ
Ans: 10.55Ω
3Ω
6Ω
10Ω
2Ω 4Ω
Ans: 10Ω
Ans: 4Ω
Open Circuits & Short Circuits
Reference: Chapter 6 (11th Ed.)
Article 6.8
Also Follow Class Lecture
Open Circuits & Short Circuits
Q. What happens during Short Circuits?
Ans.
Ans: 10Ω
Ans: 1.88Ω
Ans: 12Ω
Ans: 1Ω
Ans: 20Ω
Ans: 174.12Ω
Ans: 3.43Ω
Ans: 5Ω
Ans: 5Ω
Ans: 5.6Ω
Statement: It states that the current I flowing
through a resistor is directly proportional to the
voltage V across the resistor.
Here, V = E
Plotting Ohm’s Law
Consider current (I) in Y-axis and voltage (V) in X-axis. If the resistance (R) is 5Ω, using
ohm’s law we get the current which is given below-
V (volt) 0 5 10 15 20
I (amp) 0 1 2 3 4
The linear (straight-line) graph reveals that
the resistance is not changing with current
or voltage level; rather, it is a fixed quantity
throughout.
We can calculate the resistance at any point of the graph using the formula given
below-
For example, at V= 20V and I = 4A,
1) R1 > R2
2) R1 < R2
3) None
Q. If an electric heater draws 9.5 A when connected
to a 120-V supply, what is the internal resistance of
the heater?
Chapter 4 (11th Ed.)
Example
4.5
Exercise
13, 16
Voltage
Rise and Drop
20V
+
5V
- 15V
+
3V
-
=20V 12V
+
4V
-
8V
+
8V
- 0V
0V
0V
20V
+
5V
- 15V
+
+ Potential drop
3V
- -
=20V 12V
+
4V
- +
8V
+ Potential Rise
8V
- -
0V
0V
+ -
V Current flow
I R High potential +
to
- +
V Low potential -
R I
Potential Rise
Potential Drop
Statement: It states that the algebraic
sum of the potential rises and drops around
a closed loop (or path) is zero.
+ 2.8V -
V1 V2
+ +
- -
Applying KVL (clock wise),
+ E1 - V1 - V2 - E2 = 0
+ E1 - E2 = V1 + V2
+ 16V – 9V = 2.8V + 4.2V
7V= 7V
I + -
I
+
-
CW / CCW
+ E - V1 - V2 = 0
E = V1 + V2
Using KVL find V1
Using KVL find V1 and V2
Using KVL find V1 and V2
According to VDR the voltage across
the resistive elements will divide as
the magnitude of the resistance levels.
Find voltage across the
resistance R4
Find voltage VR4
Proof
Chapter 5 (11th Ed.)
Example
5.10, 5.12, 5.13, 5.15,
5.16, 5.18, 5.20, 5.26
Exercise
20-34
KIRCHHOFF’S CURRENT LAW (KCL)
Kirchhoff’s current law (KCL) states that the algebraic sum of
the currents entering and leaving an area, system, or junction
is zero.
Consider the unknown is leaving
Then apply KCL
If the sign of unknown current is (+) then the current is leaving
If the sign of unknown current is (-) then the current is entering
I3 = ? I3 = ?
I1 = 4A I1 = 2A
I2 = 5A
I2 = 1A
I4 = 2A I4 = 5A
CURRENT DIVIDER RULE (CDR)
For two parallel elements of equal value, the current will divide
equally.
For parallel elements with different values, the smaller the
resistance, the greater the share of input current.
For parallel elements of different values, the current will split with
a ratio equal to the inverse of their resistor values.
Chapter 6 (11th Ed.)
Example
6.14, 6.15, 6.17, 6.19, 6.21,
6.23, 6.26, 6.27
Voltage Sources in Series
Opposite polarity adds.
Similar polarity subtracts.
In case of similar polarity, direction of net Voltage
& Current will be along the higher potential.
Voltage Sources in Parallel
Voltage Sources in Parallel
• It is not a good practice to connect voltage sources
in parallel unless they are of exact same values.
• If unequal voltage sources are connected in
parallel, the smaller one may get shorted.
• If equal voltage sources are connected in parallel,
the net voltage will still be same but current will
increase.
• So, to increase net voltage connect voltage
sources of opposite polarity in series and to
increase current connect voltage sources of
same value in parallel.
Mathematical Problems
Example 5.20 Determine I & Voltage across the 7Ω
resistor for the network in Fig. 5.43.
Mathematical Problems
Example 5.26 Using the voltage divider rule,
determine the voltages V1 and V2 of Fig. 5.62.
Mathematical Problems
Example 6.26 Determine voltages Vab and Vcd for the
network in Fig. 6.55.
Mathematical Problems
Example 6.27 Determine the unknown voltage and
current for each network in Fig. 6.57.
Have we gone through Books?
What we have studied so far?
• Study only the presented topics from the reference books.
• And must solve all maths as given in the class.
Chapter 7 (Boylestad - 11th Edition)
Example: 7.1, 7.2, 7.3, 7.4, 7.7, 7.8, 7.10, 7.11
Exercise: 8, 9, 11, 17, 25, 28.
Thank
You
Q/A
Class Test 1: 23-09-24, Time: 8.30 AM
Syllabus
Some Definitions
Chapter 4: Example 4.5; Exercise: 13, 16
Chapter 5 Example 5.10, 5.12, 5.13, 5.15, 5.16, 5.18, 5.20, 5.26; Exercise: 28-34
Chapter 6 Example 6.14, 6.15, 6.17, 6.19, 6.21, 6.23, 6.26, 6.27
Chapter 7 Example: 7.1, 7.2, 7.3, 7.4, 7.7, 7.8, 7.10, 7.11; Exercise: 8, 9, 11, 17, 25, 28.
