NAS-Ch1-Part1-Circuit Elements

NETWORK ANALYSIS AND SYNTHESIS
B. Tech EE III Semester
Course Code: BTEEC302
Prepared By
Dr. K Hussain
Associate Professor & Head
Dept. of EE, SITCOE

Note: Some of the figures in this slide set are taken from the book (C.K.
Alexander and M.N.O Sadiku, Fundamentals of Electric Circuits,
Second Edition, 2004, McGraw Hill)
Overview
In this slide set, we will review basic concepts, electrical
quantities and their units, circuit elements, and basic circuit
laws.

What is an Electric Circuit?
• In electrical engineering, we are usually interested in transferring
energy or communicating signals from one point to another.
To do this, we often require an interconnection of
electrical components.
“An electric circuit is an interconnection of
electrical components.”
• Typical circuit or electrical components that we will see in this
course:
batteries or voltage sources, current sources, resistors,
switches, capacitors, inductors, diodes, transistors, operational
amplifiers,

What is an Electric Circuit?
• According to Merriam-Webster Dictionary:
“The complete path of an electric current including usually the source of
electric energy.”
• According to Encyclopedia Britannica: “Path that transmits electric
current.”
“A circuit includes a battery or a generator that gives energy to the
charged particles; devices that use current, such as lamps, motors, or
electronic computers; and connecting wires or transmission lines.
Circuits can be classified according to the type of current they carry (see
alternating current, direct current) or according to whether the current
remains whole (series) or divides to flow through several branches
simultaneously (parallel).
Two basic laws that describe the performance of electric circuits are Ohm's
law and Kirchhoff's circuit rules."

Electric Circuit
All electric circuits have three main parts:
1. A source of energy
2. A closed path
3. A device which uses the energy
If ANY part of the circuit is open, the device will notwork!
• According to Merriam-Webster Dictionary:
“The complete path of an electric current including usually the source of
electric energy”
An electric circuit is an interconnection of electrical components.
Dr. K Hussain, Associate Professor & Head, Dept. of EE, SITCOE
Fig. An Electric Circuit

System of Units
The International System of Units, or Système
International des Unités (SI), also known as metric
system uses 7 mutually independent base units. All other
units are derived units.
SI Base Units

Review of Basic Circuit Concepts
electrical• Electric Charge is the basis for describing all
phenomena .
• Charge is an electrical property of the atomic particles of which
matter consists and is measured in coulombs (Charles Augustin
de Coulomb (1736-1806) a French Scientist)
• Inside an atom, there is negative charge on electrons, positive
charge on protons and no charge on neutrons.
• The charge of an electron is equal to that of an proton and is:
e =1.602  10 -19 C

Charge
• Note that in 1C of charge there are:
1/ 1.602  10 -19 = 6.24  10 18 electrons
• Laboratory values of charges are more likely to be a fraction of a
Coulumb (e.g., pC, nC, C, or mC).
• Law of conservation of charge: charge can neither be created
nor destroyed, only transferred. (This is a law in classical
physics and may not be true in some odd cases!. We are not
dealing with those cases anyway.)
• Electrical effects are attributed to both separation of charges
and/or charges in motion!

A Material Classification
• Conductor: a material in which charges can move to
neighboring atoms with relative ease.
– One measure of this relative ease of charge movement is
the electric resistance of the material
– Example conductor material: metals and carbon
– In metals the only charged particles
that canmove are electrons
• Insulator: a material thatopposes the charge
movement (ideally infinite opposition, i.e., no
charge movement)
– Example insulators: Dry air and glass
• Semi-conductor: a material whose conductive properties are
somewhat in between those of conductor and insulator
– Example semi-conductor material: Silicon with some added
impurities

Electric Current (Charges in Motion!)
• Current: net flow of charge across any cross section of a
conductor, measured in Amperes (Andre-Marie Ampere (1775-
1836), a French mathematician and physicist)
• Current can be thought of as the rate of change of charge:
i 
dq
dt

Electric Current
• Originally scientists (in particular Benjamin Franklin (1706-1790)
an American scientist and inventor) thought that current is only
due to the movement of positive charges.
• Thus the direction of the current was considered the direction of
movement of positive charges.
current

Electric Current
• In reality in metallic conductors current is due to the movement
of electrons, however, we follow the universally accepted
convention that current is in the direction of positive charge
movement.
• Two ways of showing the same current:

Two Important Types of Currents
• Direct current (DC) is a current that remains constant with time.
• Alternating current (AC) is a current that varies sinusoidally with
time.

Voltage (Separation of Charge)
• Voltage (electromotive force, or potential) is the energy required
to move a unit charge through a circuit element, and is
measured in Volts (Alessandro Antonio Volta (1745-1827) an
Italian Physicist).
v 
dW
dq
•Similar to electric current, there are two important
types of
voltage: DC and AC

Voltage Polarity
• The plus (+) and minus (-) sign are used to define voltage
polarity.
• The assumption is that the potential of the terminal with (+)
polarity is higher than the potential of the terminal with (-)
polarity by the amount of voltage drop.

Voltage Polarity
• Figures (a) and (b) are two equivalent representation of the
same voltage:
• Both show that the potential of terminal a is 9V higher than the
potential of terminal b.

Power
• The rate of change of (expending or absorbing) energy per unit
time, measured in Watts (James Watt (1736-1819) a Scottish
inventor and mechanical engineer)
p 
dW

dW

dq
 vi
dt dq dt

Other useful Power formulas
These formulas can also
be used! They are
simply derivations of
the POWER formula
with different versions
of Ohm's law
substituted in.

t>t0
 t
t t
t0 0
W  W (t ,t) 0 )i()dp()d  v(
Remainder
of Circuit
Circuit element
consuming/generating
power p(t)
+
-
•Energy absorbed or supplied by an element from time t0 to time
v(t)
• Instantaneous power: p(t)  v(t)i(t)
i(t)
Energy Calculation

Circuit Elements
• Circuit elements can be broadly classified as being either
active or passive.
• An active element is capable of generating energy.
– Example: current or voltage sources
• A passive element is an element that does not generate energy,
however, they can either consume or store energy.
– Example: resistors, capacitors, and inductors
• The most important active elements are voltage or current
sources that generally deliver power to the circuit connected to
them.

Independent Sources
• Two kinds of Sources: Independent and dependent sources
• Independent Sources: An (ideal) independent source is an active
element that provides a specified voltage or current that is
completely independent of other circuit elements.
• Symbols for independent voltage sources:
(a) Used for constant or time-varying voltage
(b) Used for constant voltage (dc)

Independent Sources
• Equivalent representation of ideal independent current sources
whose current i(t) is maintained under all voltage requirements
of the attached circuit:

Common Voltage and Current Source
Labeling

Dependent (Controlled) Sources
• An ideal dependent (controlled) source is an active element
whose source quantity is controlled by a voltage or current of
another circuit element.
• Dependent sources are usually presented by diamond-shaped
symbols:
Dependent sources are useful in modeling elements such as transistors,
operational amplifiers, and integrated circuits.

Dependent (Controlled) Sources
+
V(t)
-
+
- s
V (t)=V(t)
There are four types of dependent sources:
1. Voltage-Controlled Voltage Source (VCVS)
I(t)
+
V(t)
-
+
- s
V (t)=I(t)
2. Current-Controlled Voltage Source (CCVS)
I(t)

Dependent (Controlled) Source
+
V(t)
-
Is(t)=V(t)
3. Voltage-Controlled Current Source (VCCS)
I(t)
+
V(t)
-
Is(t)=I(t)
4. Current-Controlled Current Source (CCCS)
I(t)

Example: Dependent Source
• In the following circuits, identify the type of dependent sources:

Resistance
The resistance R of an element denotes its ability to resist the flow
of electric current; it is measured in ohms.
• The resistance of any material with a uniform cross-sectional
area A and length l is inversely proportional to A and directly
proportional to l.
of the proportionality is the resistivityof the• The constant
material, i.e., 
A
R 
l
A
lR  
(a) Resistor, (b) Circuit symbol for resistance

• In honor of George Simon Ohm (1787-1854), a German
physicist, the unit of resistance is named Ohm ().
• A conductor designed to have a specific resistance is called a
resistor.
Resistance

Ohm’s Law
• One can also write:
• Instantaneous power dissipated in a resistor
• Ohm’s law states that the voltage v across a resistor is directly
proportional to the current i flowing through the resistor.
• The proportionality constant is the resistance of the resistor,
i.e., v(t)  Ri(t)
R
i(t) 
1
v(t)  i(t)  Gv(t)
2
R
v2
(t)
p(t)  v(t)i(t)   Ri (t)

Linear and Nonlinear Resistors
• Linear resistor Nonlinear resistor
• In this course, we assume that all the elements that are
designated as resistors are linear (unless mentioned otherwise)

Conductance
• G=1/R is called the conductance of the element and is
measured in siemens (S) or mho ( ) .
German inventor
Ernst Werner von Siemens
(1816-1892)
• Conductance is the ability of an element to conduct current.
• A device with zero (no) resistance has infinite conductance and
a device with infinite resistance has zero conductance.

Short and Open Circuits
• A device with zero resistance is called short circuit and a device
with zero conductance (i.e., infinite resistance) is called open-
circuit.
Remaining
network
Remaining
network

Q. The essential component of a toaster is an electrical element (a resistor) that
converts electrical energy to heat energy. How much current is drawn by a toaster with
resistance 15 Ohms at 110 V?
Sol:
V =I*R
I=V/R
Answer: 7.333 A.
Example Problem

• Determine the power absorbed or supplied by the elements of
the following network:
Practice Problems

Practice Problems
• The power absorbed by the 10-kΩ resistor in the following circuit
is 3.6 mW. Determine the voltage and the current in the circuit.

• Given the following network, find R and VS.
Practice Problems

• Given the following circuit, find the value of the voltage source
and the power absorbed by the resistance.
Practice Problems

B.Tech EE III Semester
Prepared By
Dr. K Hussain
Dept. of EE, SITCOE

Terminology (Nodes and Branches)
• A branch represents a single element such as a voltage source or a
resistor.
• A node is usually indicated by a dot in a circuit.
• A node is the point of connection between two or more circuit branches.
(The node includes the interconnection wires.)
• A loop is any closed path in a circuit.
• A loop is a closed path formed by starting at a node, passing through a
set of nodes, and returning to the starting node without passing through
any node more than once.
• A loop is said to be independent if it contains at least one branch which is
not a part of any other independent loop. Independent loops or paths result
in independent sets of equations.

• A network with b branches, n nodes, and l independent loops will
satisfy the fundamental theorem of network topology:
b=l+n-1
Note:
• Two or more elements are in series if they exclusively share a
single node and consequently carry the same current.
• Two or more elements are in parallel if they are connected to the
same two nodes and consequently have the same voltage across
them.

Example
• Are the following two circuits different? Identify the nodes in
each circuit.

Example Problem
• In the following circuit, find the number of branches and nodes.
Ans: Five branches and 3 nodes

Series and Parallel Circuits
• Two or more elements are connected “in series” when they
belong to the same branch.(even if they are separated by other
elements).
• In general, circuit elements are in series when they are
sequentially connected end-to-end.
• Elements that are in series carry the same current.

Series and Parallel Circuits
• Two or more circuit elements are “in parallel” if they are
connected between the same two “ nodes”.
• Consequently, parallel elements have the same voltage
The equivalent conductance of resistors connected in parallel is the sum of their
individual conductances.

Potential Difference: Series…
1) the cell has a potential difference
of 3.0 V and the resistor has a
resistance of 8.0 . The ammeter
reading is 0.2 A.
a) Calculate the potential
difference across the resistor?
b) Calculate the potential
difference across the lamp?
PD Rule
 In a series circuit the
energy is shared
between components
 V = IR
Answers
a) V = IR so 0.2A x 8  = 1.6V
b) 3V – 1.6V = 1.4V

Potential Difference: Series...
2) A battery, an ammeter, a 10 
resistor and a 15  resistor are
connected in series. The ammeter
reading is 0.36 A. Calculate the
potential difference across:
a) the 10  resistor.
b) the 15  resistor.
c) the battery.
d) What would be the resistance of
a single resistor that would have
the same current if it was
connected on its own to the
same battery?
Answers
a) V = IR so 0.36A x 10  = 3.6V
b) V = IR so 0.36A x 15  = 5.4V
c) Vs = 3.6V + 5.4V = 9V
d) V=IR or V/I = R 9V/0.36A = 25 

Potential Difference: Series…
3) A 6.0 V battery, a 10  resistor and
a 20  resistor are connected in
series with each other.
a) Draw out a circuit diagram
b) Calculate the total resistance of
the two resistors in series.
c) Calculate the current in the
circuit.
d) Calculate the potential
difference across the 20 
resistor.
Answers
a) All connected in series (see figure)
b) RT = R1 + R2 = 10 +20  = 30 
c) V=IR or V/R = I; I = 6V / 30  = 0.2A
d) V = IR = 0.2A x 20  = 4V

Parallel Connections
4) A 12 V battery is connected to a 10 
resistor in parallel with a 15  resistor, as
shown. Calculate the current through:
a) the 10  resistor.
b) the 15  resistor.
c) the battery.
d) What would be the resistance of a
single resistor that would have the
same current if it was connected on
its own to the same battery?
Answers:
a) V = IR; V/R = I ;
So 12V /10  = 1.2A
b) V = IR; V/R = I
So 12V /15  = 0.8A
c) 0.8A + 1.2A = 2 A
d) V = IR; V/I = R; 12V / 2A = 6 
Or 1/R T=1/10+1/15 = 0.167 RT = 6 
21
111
RRRT


Summary
1. An electric circuit consists of electrical elements connected together.
2. The International System of Units (SI) is the international measurement
language, which enables engineers to communicate their results.
From the seven principal units, the units of other physical quantities can
be derived.
3. Current is the rate of charge flow past a given point in a given
direction.
4. Voltage is the energy required to move 1C of charge through an
element.
5. Power is the energy supplied or absorbed per unit time. It is also the
product of voltage and current.
6. An ideal voltage source produces a specific potential difference across
its terminals regardless of what is connected to it. An ideal current source
produces a specific current through its terminals regardless of what is
connected to it.
7. Voltage and current sources can be dependent or independent.
A dependent source is one whose value depends on some other circuit
variable.

1. The unit of current is:
(a) coulomb (b) ampere (c) volt (d) joule
2. Which of these is not an electrical quantity?
(a) charge (b) time (c) voltage (d) current (e) power
3. The voltage 2,000,000 V can be expressed in powers of 10 as:
(a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV
4. A charge of 2C flowing past a given point each second is a current of 2A.
(a) True (b) False
5. The voltage across a 1.1-kW toaster that produces a current of 10 A is:
(a) 11 kV (b) 1100 V (c) 110 V (d) 11 V
6. Voltage is measured in:
(a) watts (b) amperes (c) volts (d) joules per second
7. A 4-A current charging a dielectric material will accumulate a charge of
24 C after 6 s.
(a)True (b) False
Review Questions

Mr Powell 2012
Index
Circuit Symbols
Candidates will be
required to interpret and
draw circuit
diagrams.
Knowledge and
understanding of the use
of thermistors in circuits
e.g. thermostats is
required.
Knowledge and
understanding of the
applications of light-
dependent resistors
(LDRs) is required, eg
switching lights on when
it gets dark

ammeter
voltmeter
cell
Indicator / light source
diode
Light emitting diode
resistor
Variable resistor
thermistor
Light dependent resistor
heater
Electric motor
Many cells = a battery
No more semi circles!
of a specific value
Resistance falls as temp rises
Low resistance, connect in series
High resistance, connect in parallel
Emits light when forward biased
Resistance falls as light level rises
Conducts when forward biased

B. Tech EE III Semester
By
Dr. K Hussain
Dept. of EE, SITCOE

Series Resistors
• The equivalent resistance of any number of resistors connected
in series is the sum of the resistors (Why?)
GnGeq
1
G1 G2
1

1

1
Req  R1 R2 Rn or

Voltage Division
• In a series combination of n resistors, the voltage drop across
the resistor Rj for j=1,2, …, n is:
• What is the formula for two series resistors?!
v (t)in
Rj
jv (t) 
R1 R2 Rn

Parallel Resistors
• The equivalent conductance of resistors connected in parallel is
the sum of their individual conductances:
nRReq R1 R2
1

1

1

1
Geq  G1  G2  Gn or
• Why?

Current Division
• In a parallel combination of n resistors, the current through the
resistor Rj for j=1,2, …, n is:
• Why?
i (t)in
Gj
ji (t)
G1 G2  Gn

Parallel Resistors and Current Division
• For the special case of two parallel resistors
21 i(t)
R1R2R1R2
eq
R1 R2R1 R2
, i (t) 
R1 R2
R  i(t), andi (t) 
Example:
The equivalent resistance of two parallel resistors is equal to the product of their
resistances divided by their sum.
In general,

Example
• In the following circuit, find Req:
Answer: 6 Ohms

Example
• In the following circuit find I1, I2, I3, Va, and Vb.
Sol: 3K, 3K resistors are in series, so, total is 6K
then 6K ||6K -> 3K
then, 9K and 3K are in series, total : 12K
I1= 12/12K=1mA
I2=1mA*(6K/12K)=0.5mA and
I3=1mA*(6K/12K)=0.5mA

Example
• In the following circuit, find the resistance seen between the two
terminals A and B, i.e., RAB

Example
• In the following circuit, find the current i.
20
12V
30
30
30
10
i

B.Tech EE III Semester
By
Dr. K Hussain
Dept. of EE, SITCOE

Kirchhoff’s Current Law (KCL)
• Gustav Robert Kirchhoff (1824-1887), a German physicist,
stated two basic laws (1847) concerning the relationship
between the currents and voltages in an electrical circuit.
• KCL: The algebraic sum of the currents entering a node (or a
closed boundary) is zero.
• The current entering a node may be regarded as positive while
the currents leaving the node may be taken as negative or vice
versa.

• KCL is based on the law of conservation of charge.
• Example: Write the KCL for the node in this circuit:
Circuit

• Alternative statement of KCL: For lumped circuits, the
algebraic sum of the currents leaving a node (or a closed
boundary) is zero.
•
Σiin=Σiout
Alternate statement for KCL:
The sum of the currents entering a node is equal to the sum of
the currents leaving that node.

Example
• The following network is represented by its topological diagram.
Find the unknown currents in the network.

Example
• In the following circuit, use a closed surface to find I4.

Kirchhoff’s Voltage Law (KVL)
• KVL: The algebraic sum of the voltage drops around any closed path
(or loop) is zero at any instance of time.
• Write KVL for the above circuit.
Sum of voltage drops=Sum of voltage rises

Example
• In the following circuit, assume VR1=26V and VR2=14V. Find VR3.
Solution:
By using KVL, Vin-VR1+5-VR2+15-VR3=0
30-26+5-14+15-VR3=0
VR3 =10 V
Vin=30 V

Example
• In the following circuit use KVL to determine Vae and Vec.
Note that we use the convention Vae to indicate the voltage of point
a with respect to point e or Vae=Va-Ve

Problem 1: For the circuit shown in Figure 1, determine
(a) the battery voltage V,
(b) the total resistance of the circuit, and
(c) the values of resistance of resistors R1, R2 and R3.
Given that the P.D.s across R1, R2 and R3 are 5V, 2V and 6V respectively.
(a) Battery voltage V =V1 + V2 + V3 =5 + 2 + 6=13V
(b)Total circuit resistance R= V/ I
= 13/4=3.25Ω
(c) Resistance R1 = V1/ I = 5/4 =1.25 Ω
Resistance R2 = V2/ I = 2/4 =0.5 Ω
Resistance R3 = V3/ I = 6/4 =1.5 Ω
Solution:
Dr. K Hussain

Problem 2. For the circuit shown in Figure, determine the P.D.
across resistor R3. If the total resistance of the circuit is 100 Ω,
determine the current flowing through resistor R1. Find also
the value of resistor R2.
P.D. across R3= V3 = 25 −10 −4
= 11V
Current = I = V/ R= 25/100
= 0.25A,
Resistance R2 = V2/ I
= 4/0.25 =16 Ω
Solution:
Dr. K Hussain

Problem 3: A 12V battery is connected in a circuit having
three series-connected resistors having resistances of 4 Ω, 9 Ω
and 11 Ω. Determine the current flowing through, and the P.D.
across the 9 Ω resistor. Find also the power dissipated in the
11 Ω resistor.
Total resistance R=4 + 9 + 11=24 Ω
Current I = V/ R = 12/24
=0.5A,
which is the current in the 9 Ω resistor.
P.D. across the 9 Ω resistor = V1 = I × 9 = 0.5 × 9 = 4.5V
Power dissipated in the 11 Ω resistor
= P = I2R=(0.5)2*(11)
= 0.25*(11)
= 2.75 Watts
Solution:
Dr. K Hussain

Problem 4: For the circuit shown in Figure , determine
(a) the reading on the ammeter, and
(b) the value of resistor R2.
P.D. across R1 is the same as the supply voltage V.
Hence supply voltage, V =8 × 5=40V
(a) Reading on ammeter = I = V/R3= 40/20=2A
(b) Current flowing through R2 =11−8−2=1A
Hence, R2 = V/I2= 40/1=40 Ω
Solution:
Dr. K Hussain

Problem 5: For the circuit shown in Figure, find (a) the value of the
supply voltage V and (b) the value of current I.
(a) P.D. across 20 Ω resistor = I2R2 = 3× 20 V,
Hence supply voltage V =60V
(since the circuit is connected in parallel)
(b) Current I1 = V/R1= 60/10= 6A;
I2 = 3A
I3 = V/R3= 60/60= 1A
Current I =I1+I2+I3
I =6+3+1=10A
Alternatively,
(1/R)= (1/60)+ (1/20)+ (1/10)= (1 + 3 + 6)/60 = 10/60
Hence total resistance = R = 60/10=6 Ω
Current I = V/R= 60/6=10A
Solution:
Dr. K Hussain

Problem 1: Find the current through each resistor of the
circuit shown in fig, using nodal analysis.
Solution:
At node1, I1+I2+I3 = 0
[V1-15/1]+[V1/1]+[V1-V2/0.5] = 0
V1-15+V1+2V1-2V2 = 0
4V1-2V2 = 15 --------------------- (1)
At node2, I3=I4+I5
[(V1-V2)/0.5] =(V2/2) +[(V2-20)/1]
2V1-2V2-0.5V2-V2 + 20 = 0
2V1-3.5V2 = -20 ------------------ (2)
Multiplying (2) by 2 & subtracting from (1)
5V2 = 55
V2 = 11V
V1 = 9.25V
I1 = (V1-5)/1 = (9.25-1)/5 = -5.75A
I2 = V1/1 = 9.25A
I3 = V1-V2/0.5 = -3.5A = 3.5A
I4 = V2/2 = 5.5A
I5 = (V2-20)/I = (11-20)/1
= -9A
Dr. K Hussain

Problem 6: For the circuit shown in Figure, calculate (a) the
value of resistor Rx such that the total power dissipated in the
circuit is 2.5kW, and (b) the current flowing in each of the four
resistors.
(a) Power dissipated P=VI watts,
Hence 2500=(250)(I)
i.e., I = 2500/250 = 10A
From Ohm’s Law RT=V/I = 250/10 =25 Ω,
where RT is the equivalent circuit resistance.
The equivalent resistance of R1 and R2 in parallel is
= (15 × 10)/(15 + 10)
= 150/25
= 6 Ω
The equivalent resistance of resistors R3 and Rx in parallel = 25 Ω−6 Ω,
i.e. , 19 Ω.
(38*Rx)/(38+Rx)=25-6=19 Ω
38* Rx=(38+Rx)*19
38* Rx = (38*19 +19*Rx)
Rx=38*19/38
= 19 Ω
Solution:
Dr. K Hussain

Problem 4: Two resistors are connected in series across a 24V
supply and a current of 3A flows in the circuit. Find value of Rx
resistance and (b) the p.d. across the 2 Ω resistor. If the circuit is
connected for 50 hours, how much energy is used?
(a) Total circuit resistance R= V/ I
= 24/3 = 8 Ω
Value of unknown resistance, Rx =8 −2=6 Ω
(b) P.D. across 2 Ω resistor, V1 =IR1 =3 × 2=6V
Alternatively, from above,
V1 = (R1/R1 + Rx))
V = (2/2 + 6) (24) = 6V
Energy used = Power × Time
= V × I × t = 24 V * 3 A * 50 h
= 3600 Wh = 3.6 kWh Dr. K Hussain

Problem 1: Use Thevenin’s theorem to find the current flowing
in the 10 Ω resistor for the circuit shown in Figure.
Following the above procedure:
The 10 Ω resistance is removed from the circuit as
shown in Figure.
(i) Since there is no current flowing in the 5 Ω
resistor and
Current I1 is given by:
I1 = 10/(R1 + R2)
= 10/(2 + 8)
= 1A
Solution:
Dr. K Hussain

(ii) P.D. across R2 =I1R2 =1×8=8V
Hence P.D. across AB, i.e. the open-circuit voltage across the
break, E =8V
(iii) Removing the source of e.m.f. gives the circuit of Fig (c).
Resistance, r = R3 + (R1R2)/(R1 + R2)
=5+ (2×8/2+8)
= 5 + 1.6 = 6.6 Ω
(iv) The equivalent Thevenin’s circuit is shown in Fig (d).
Current I = E/(R+r)= 8/(10+6.6)= 8/16.6=0.482A
Dr. K Hussain

Problem 2: Use Norton’s theorem to determine the current I flowing in
the 4 Ω resistance shown in Figure (a).
The 4 Ω branch is short-circuited as shown in Figure (b).
From Figure (b), ISC =I1 +I2 =4A
If the sources of e.m.f. are removed, the resistance
‘looking-in’ at a break made be given by:
r = (2 × 1)/(2 + 1) = (2/3) Ω
From the Norton equivalent network shown in
Figure (c)
The current in the 4 Ω resistance is given by:
I =(4)*[(2/3)/((2/3) + 4)] = 0.571A,
Solution:
Dr. K Hussain

Problem 3: The terminal voltage of a voltage source is 12 V when
connected to a 2-W load. When the load is disconnected, the
terminal voltage rises to 12.4 V.
(a) Calculate the source voltage and internal resistance.
(b) Determine the voltage when an load is connected to the
source.
Solution:
(a) We replace the source by its Thevenin’s equivalent.
The terminal voltage when the load is disconnected is
the open-circuit voltage
When the load is connected, as shown in Fig (a),
vL =12 V and PL= 2W
Hence,
The load current is
Dr. K Hussain

(b) Now that we have the Thevenin’s equivalent of the source, we connect
the 8  load across the Thevenin’s equivalent as shown in Fig (b).
Using voltage division, we obtain
Dr. K Hussain
The voltage across Rs is the difference between the
source voltage vs and the load voltage vL or

Capacitors and Inductors
• Introduction
• Capacitors
• Series and Parallel Capacitors
• Inductors
• Series and Parallel Inductors
Dr. K Hussain

Introduction
• Resistor: a passive element which dissipates
energy only
• Two important passive linear circuit
elements:
1) Capacitor
2) Inductor
• Capacitor and inductor can store energy only
and they can neither generate nor dissipate
energy.
Dr. K Hussain

Michael Faraday (1971-1867)
Dr. K Hussain

Capacitors
• A capacitor consists of two conducting plates
separated by an insulator (or dielectric).
(F/m)10854.8
ε
12
0
0





 r
d
A
C
Dr. K Hussain

• Three factors affecting the value of
capacitance:
1. Area: the larger the area, the greater the
capacitance.
2. Spacing between the plates: the smaller the
spacing, the greater the capacitance.
3. Material permittivity: the higher the permittivity,
the greater the capacitance.
d
A
C
ε

Dr. K Hussain

(a) Polyester capacitor, (b) Ceramic capacitor, (c) Electrolytic capacitor
Dr. K Hussain

Variable capacitors
Dr. K Hussain

Charge in Capacitors
• The relation between the charge in plates and
the voltage across a capacitor is given below.
Cvq 
C/V1F1 
v
q Linear
Nonlinear
Dr. K Hussain

Voltage Limit on a Capacitor
• Since q=Cv, the plate charge increases as the
voltage increases. The electric field intensity
between two plates increases.
• If the voltage across the capacitor is so large
that the field intensity is large enough to
break down the insulation of the dielectric,
the capacitor is out of work.
• Hence, every practical capacitor has a
maximum limit on its operating voltage.
Dr. K Hussain

I-V Relation of Capacitor
dt
dv
C
dt
dq
iCvq  ,+
-
v
i
C
Dr. K Hussain

Physical Meaning
dt
dv
Ci 
• when v is a constant voltage, then i=0; a constant
voltage across a capacitor creates no current through
the capacitor, the capacitor in this case is the same as
an open circuit.
• If v is abruptly changed, then the current will have an
infinite value that is practically impossible. Hence, a
capacitor is impossible to have an abrupt change in
its voltage except an infinite current is applied.
+
-
v
i
C
Dr. K Hussain

• A capacitor is an open circuit to dc.
• The voltage on a capacitor cannot change
abruptly.
Abrupt change
Dr. K Hussain

• The charge on a capacitor is an integration of
current through the capacitor. Hence, the
memory effect counts.
dt
dv
Ci   

t
idt
C
tv
1
)(
 
t
t
o
o
tvidt
C
tv )(
1
)(
 0)( v
 Ctqtv oo /)()( 
+
-
v
i
C
Dr. K Hussain

Energy Storing in Capacitor
dv
p vi vC
dt
 
21
2
t t t
tdv
w pdt C v dt C vdv Cv
dt
  
     
)(
2
1
)( 2
tCvtw 
C
tq
tw
2
)(
)(
2

)0)(( v +
-
v
i
C
Dr. K Hussain

Model of Practical Capacitor
Dr. K Hussain

Example 1
(a) Calculate the charge stored on a 3-pF
capacitor with 20V across it.
(b) Find the energy stored in the capacitor.
Dr. K Hussain

Example 1
Solution:
(a) Since
(b) The energy stored is
pC6020103 12
 
q
pJ600400103
2
1
2
1 122
 
Cvw
,Cvq
Dr. K Hussain

Example 2
• The voltage across a 5- F capacitor is
Calculate the current through it.
Solution:
• By definition, the current is
V6000cos10)( ttv 
6
105 

dt
dv
Ci
 
6000105 6
)6000cos10( t
dt
d
A6000sin3.06000sin10 tt 
Dr. K Hussain

Example 3
• Determine the voltage across a 2-F capacitor
if the current through it is
Assume that the initial capacitor voltage is zero.
Solution:
• Since
mA6)( 3000t
eti 






t t
ev 0
3000
6
6
102
1
0
3000
3
3000
103 tt
e



 
t
vidt
C
v 0
)0(
1
,0)0(and v
3
10
dt
V)1( 3000t
e

Dr. K Hussain

Parallel connection
Neq CCCCC  ....321
Dr. K Hussain

Parallel Capacitors
• The equivalent capacitance of N parallel-
connected capacitors is the sum of the
individual capacitance.
Niiiii  ...321
dt
dv
C
dt
dv
C
dt
dv
C
dt
dv
Ci N ...321
dt
dv
C
dt
dv
C eq
N
k
K 




 
1
Neq CCCCC  ....321
Dr. K Hussain

Series Capacitors
Neq CCCCC
1
...
1111
321

Dr. K Hussain

Series Capacitors
• The equivalent capacitance of series-connected
capacitors is the reciprocal of the sum of the
reciprocals of the individual capacitances.
Neq
t
N
t
eq
C
tq
C
tq
C
tq
C
tq
id
CCCC
id
C
)()()()(
)
1
...
111
(
1
21
321

  


)(...)()()( 21 tvtvtvtv N
21
111
CCCeq

21
21
CC
CC
Ceq


Dr. K Hussain

Summary
• These results enable us to look the capacitor
in this way:
1/C has the equivalent effect as the resistance.
Dr. K Hussain

Example 4
• Find the equivalent capacitance seen between
terminals a and b of the circuit in Fig.
Dr. K Hussain

Example 4
Solution:
F4
520
520




F302064 
F20F
6030
6030
 


eqC
:seriesinarecapacitorsF5andF20  
F6with theparalleliniscapacitorF4  
:capacitorsF20and 
withseriesiniscapacitorF30 
capacitor.F60the 
Dr. K Hussain

Example 5
• For the circuit in Fig, find the voltage across
each capacitor.
Dr. K Hussain

Example 5
Solution:
• Two parallel capacitors:
• Total charge:
• This is the charge on the 20-mF and 30-mF capacitors,
because they are in series with the 30-v source.
• ( A crude way to see this is to imagine that charge acts
like current, since i = dq/dt)
mF10mF
1
20
1
30
1
60
1 

eqC
C3.0301010 3
 
vCq eq
Dr. K Hussain

Example 5
• Therefore,
• Having determined v1 and v2, we now use
KVL to determine v3 by
• Alternatively, since the 40-mF and 20-mF capacitors are in
parallel, they have the same voltage v3 and their
combined capacitance is 40+20=60mF.
,V15
1020
3.0
3
1
1 

 
C
q
v
V10
1030
3.0
3
2
2 

 
C
q
v
V530 213  vvv
V5
1060
3.0
mF60 33 

 
q
v
Dr. K Hussain

Joseph Henry (1979-1878)
Dr. K Hussain

Inductors
• An inductor is made of a coil of conducting
wire
l
AN
L
2

Dr. K Hussain

(H/m)104 7
0
0
2







r
l
AN
L
turns.ofnumber:N
length.:l
area.sectionalcross: A
coretheoftypermeabili:
Dr. K Hussain

(a) air-core
(b) iron-core
(c) variable iron-core
Dr. K Hussain

Flux in Inductors
• The relation between the flux in inductor and
the current through the inductor is given
below. Li
Weber/A1H1 
i
ψ Linear
Nonlinear
Dr. K Hussain

Energy Storage Form
• An inductor is a passive element designed to
store energy in the magnetic field while
• a capacitor stores energy in the electric field.
Dr. K Hussain

I-V Relation of Inductors
• An inductor consists of
a coil of conducting
wire.
dt
di
L
dt
d
v 

+
-
v
i
L
Dr. K Hussain

Physical Meaning
• When the current through an inductor is a
constant, then the voltage across the inductor is
zero, same as a short circuit.
• No abrupt change of the current through an
inductor is possible except an infinite voltage
across the inductor is applied.
• The inductor can be used to generate a high
voltage, for example, used as an igniting element.
dt
di
L
dt
d
v 

Dr. K Hussain

• An inductor are like a short circuit to dc.
• The current through an inductor cannot
change instantaneously.
Dr. K Hussain

 
t
t
o
o
tidttv
L
i )()(
1
 

t
dttv
L
i )(
1
memory.hasinductorThe
vdt
L
di
1

+
-
v
L
Dr. K Hussain

Energy Stored in an Inductor
• The energy stored in an inductor
i
dt
di
LviP 






   





t t
idt
dt
di
Lpdtw
 

)(
)(
22
)(
2
1
)(
2
1ti
i
LitLidiiL ,0)( i
)(
2
1
)( 2
tLitw 
+
-
v
L
Dr. K Hussain

Model of a Practical Inductor
Dr. K Hussain

Example 6
• The current through a 0.1-H inductor is i(t) =
10te-5t A. Find the voltage across the inductor
and the energy stored in it.
Solution:
V)51()5()10(1.0 5555
teetete
dt
d
v tttt
 
J5100)1.0(
2
1
2
1 1021022 tt
etetLiw 

,H1.0andSince  L
dt
di
Lv
isstoredenergyThe
Dr. K Hussain

Example 7
• Find the current through a 5-H inductor if the
voltage across it is
Also find the energy stored within 0 < t < 5s.
Assume i(0)=0.
Solution:






0,0
0,30
)(
2
t
tt
tv
.H5and L)()(
1
Since
0
0 
t
t
tidttv
L
i
A2
3
6 3
3
t
t
 
t
dtti 0
2
030
5
1
Dr. K Hussain

Example 7
  
5
0
6
5
kJ25.156
0
5
6
6060
t
dttpdtw
thenisstoredenergytheand,60powerThe 5
tvip 
before.obtainedas
usingstoredenergytheobtaincanweely,Alternativ
)0(
2
1
)5(
2
1
)0()5( 2
LiLiww 
kJ25.1560)52)(5(
2
1 23

Dr. K Hussain

Example 8
• Consider the circuit in
Fig (a). Under dc
conditions, find:
(a) i, vC, and iL.
(b) the energy stored
in the capacitor and
inductor.
Dr. K Hussain

Example 8
Solution:
,2
51
12
Aii L 


,J50)10)(1(
2
1
2
1 22
 cc Cvw
J4)2)(2(
2
1
2
1 22
 iL Lw
)(a :conditiondcUnder capacitor
inductor
circuitopen
circuitshort
)(b
V105  ivc
Dr. K Hussain

Inductors in Series
Neq LLLLL  ...321
Dr. K Hussain

Series Inductors
• Applying KVL to the loop,
• Substituting vk = Lk (di/dt) results in
Nvvvvv  ...321
dt
di
L
dt
di
L
dt
di
L
dt
di
Lv N ...321
dt
di
LLLL N )...( 321 
dt
di
L
dt
di
L eq
N
K
K 




 
1
Neq LLLLL  ...321
Dr. K Hussain

Inductors in Parallel
Neq LLLL
1111
21
 
Dr. K Hussain

Parallel Inductors
• Using KCL,
• But
Niiiii  ...321
 
t
t k
k
k o
tivdt
L
i )(
1
0
  
t
t
t
t s
k
tivdt
L
tivdt
L
i
0 0
)(
1
)(
1
0
2
01  
t
t N
N
tivdt
L 0
)(
1
... 0
)(...)()(
1
...
11
00201
21
0
tititivdt
LLL
N
t
t
N






 
 







t
t
eq
N
k
k
t
t
N
k k
tivdt
L
tivdt
L 00
)(
1
)(
1
0
1
0
1
1
1 1N
k k eqL L
 
  
 

Dr. K Hussain

• The inductor in various connection has the
same effect as the resistor.
Dr. K Hussain

Example 9
• Find the equivalent inductance of the circuit
shown in Fig.
Dr. K Hussain

Example 9
• Solution:
10H12H,,H20:Series
H6
427
427



:Parallel
H18864  eqL
42H
Dr. K Hussain

Practice Problem 1
Dr. K Hussain

Example 10
• Find the circuit in Fig,
If find :
.mA)2(4)( 10t
eti 

,mA1)0(2 i )0((a) 1
i
);(and),(),((b) 21 tvtvtv )(and)((c) 21 titi
Dr. K Hussain

Example 10
Solution:
.mA4)12(4)0(mA)2(4)()(a 10
 
ieti t
mA5)1(4)0()0()0( 21  iii
H53212||42 eqL
mV200mV)10)(1)(4(5)( 1010 tt
eq ee
dt
di
Ltv 

mV120)()()( 10
12
t
etvtvtv 

mV80mV)10)(4(22)( 1010
1
tt
ee
dt
di
tv 

isinductanceequivalentThe)(b
Dr. K Hussain

Example 10
   t t t
dteidtvti 0 0
10
121 mA5
4
120
)0(
4
1
)(
mA38533mA5
0
3 101010 ttt
ee
t
e 

  t tt
dteidtvti 0
10
20 22 mA1
12
120
)0(
12
1
)(
mA11mA1
0
101010 ttt
ee
t
e 

)()()(thatNote 21 tititi 
 
t
idttv
L
i 0
)0()(
1
)(c
Dr. K Hussain

NAS-Ch1-Part1-Circuit Elements

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