Here are the steps to solve this circuit using the super node method: 1. Identify the voltage source and nodes connected to it as the super node. In this circuit, the super node contains nodes 1, 2 and the voltage source. 2. Write a KCL equation for the super node equating the sum of currents entering and leaving the super node to 0. I1 + I2 - 10/5 = 0 3. Replace the branch currents with expressions involving the nodal voltages using Ohm's law. (V1 - V2)/10 + (V2 - 0)/5 - 10/5 = 0 4. Solve the equation to get the nodal voltage V

1. ECE18R201 Network Theory
Unit1

2. Introduction
• Charge – fundamental concept
• Two types – positive - protons, negative –electrons
• Unit – columb
• Current - Charge in motion
• Charge conservation - neither create nor destroy electrons (or
protons) when running electric circuits
• Static charge - A quantity of charge that does not change with time is
typically represented by Q
• Time variant charge - The instantaneous amount of charge is
commonly represented by q(t) or q

3. Current
• “transfer of charge” or “charge in motion”
• It is a measure of the rate at which charge is moving
past a given reference point in a specified direction
• i = dq / dt
• Unit – Amphere
• The motion of positive charge

4. QUIZ 1
In the wire of Fig, electrons are moving
left to right to create a current of 1 mA.
Determine I1 and I2

5. Voltage
• Simple circuit element
• Ex - fuses, light bulbs, resistors, batteries, capacitors, generators,
and spark coils
• There are two paths by which current may enter or leave the
element
• An electrical voltage (or a potential difference) exists between the
two terminals, or that there is a voltage “across” the element
• Voltage = work done per charge
• Unit – volts,
• 1 volt = 1J/C

6. (a, b) These are inadequate
definitions of a voltage. (c) A correct
definition includes both a symbol for
the variable and a plus-minus symbol
pair.
(a, b) Terminal B is 5 V positive with respect to terminal A;
(c, d) terminal A is 5 V positive with respect to terminal B.

7. Quiz 2
For the element in Fig, v1 = 17 V.
Determine v2

8. Power
• If one joule of energy is expended in transferring
one coulomb of charge through the device in one
second, then the rate of energy transfer is one watt
• The absorbed power must be proportional both to
the number of coulombs transferred per second
(current) and to the energy needed to transfer one
coulomb through the element (voltage)
• P = VI or p = vi

9. Contd...
• The power absorbed by the element is given by the product p = vi.
Alternatively, we can say that the element generates or supplies a
power −vi.
• If the current arrow is directed into the “+” marked terminal of an
element, then p = vi yields the absorbed power. A negative value
indicates that power is actually being generated by the element

10. Contd...
• If the current arrow is directed out of the “+”
terminal of an element, then p = vi yields the
supplied power. A negative value in this case
indicates that power is being absorbed.
• Tutorial

11. Examples

12. Determine the power being absorbed by the
circuit element in Fig.
Quiz 3

13. Determine the power generated or absorbed
by the element in Fig. State whether the
power is generated or absorbed.
Quiz 4

14. Determine the power being generated by the circuit element in Fig.
Quiz 5

15. Determine the power absorbed or generated by the circuit
element A in the circuit below.
Contd...

16. ● Based on terminal voltage-current characteristics,
sources in a circuit –
Ideal Voltage source
Ideal Current Source
● Classified further as
Independent Source
Dependent Source
Energy sources

17. Independent voltage sources
• An independent voltage source is characterized by a terminal
voltage which is completely independent of the current through it
• b- absorbs power, c – delivers power
• (a) DC voltage source symbol;
• (b) battery symbol; (c) ac voltage source symbol

18. Voltage Source in Parallel
ideal voltage sources can be connected
in parallel provided they are of the same
voltage value.

19. Quiz 6
What is not allowed or is not best practice, is connecting together ideal
voltage sources that have_____________________

20. Voltage source in series

21. Independent current sources
• an independent current source as having zero
voltage across its terminals while providing a
fixed current

22. Current Source in Parallel
whose total current
output is given as the
algebraic addition of the
individual source
currents.

23. Current source in series

24. Quiz 7
Calculate the total current

25. Dependent sources
• In Independent sources the value of the source
quantity is not affected in any way by activities in
the remainder of the circuit
• the dependent, or controlled, source, in which the
source quantity is determined by a voltage or
current existing at some other location in the
system being analyzed

26. Cont...
• The four different types of dependent sources:
(a) current-controlled current source;
(b) voltage-controlled current source;
(c) voltage-controlled voltage source;
(d) current controlled voltage source

27. Cont...

28. Problem
Calculate the power supplied or absorbed by
each element in the following figure

29. Quiz 8
Compute the power absorbed or supplied by
each component of the circuit

30. Ideal & Practical voltage source
1. Two terminal element in
which the voltage is
completely independent
of current through its
terminals
2. Practical voltage sources
have internal resistance
in series => voltage
decreases as current
increases

31. Ideal & Practical current source
Two terminal element in which
current is independent of
voltage across its terminals
Practical current sources have
internal resistance in parallel
=> current falls as voltage
increases

32. Resistor
A (linear) resistor is an element for which
• v=iR
• where the constant R is a resistance.
• The equation is known as “Ohm’s Law.”
• The unit of resistance is ohm (Ω)
• Since R is never negative, a resistor always absorbs
power

33. Resistors absorb power: since v=iR
p=vi = v² /R = i²R
Positive power means the device is absorbing
energy.
Power is always positive for a resistor!

34. Ohm’s law
I = V/R
Current is directly proportional to the voltage and inversely proportional
to the total resistance of the circuit

35. Problem
A 560 Ω resistor is connected to a circuit which
causes a current of 42.4 mA to flow through it.
Calculate the voltage across the resistor and the
power it is dissipating.
Sol.
v = iR = (0.0424)(560) = 23.7 V
p = i²R = (0.0424)2 (560) = 1.007 W

36. Quiz
An electric heater draws 8A from 250 V
supply.What is its power rating also find
resistance of heater element.
https://www.menti.com/vus6ujtsee

37. A dc power link is to be made between two islands
seperated by a distance of 24 miles.The operating voltage is
500 kv and the system capacity is 500MW.Calculate
maximum dc current flow and estimate resistivity of the
cable.
Sol...
Dc power=600x 10^6 W,
Operating voltage = 500 x 10^3V
I=P/V= 600x 10^6 W/500 x 10^3V=1200 A
Cable resistence= V/I=500 x 10^3V/1200= 417 Ω

38. Conductance
• Reciprocal of resistance (1/R), which is called
conductance (symbol G, unit siemens (S))
• A resistor R has conductance G=1/R
• The i-v equation (i.e. Ohm’s law) can be
written as i=Gv

39. Inductance
P = vi = Ldi/dt i
W = ∫p dt = ∫ L i di/dt = L i
2 /2
Inductor store energy even if voltage across is zero
Never dissipates energy. Only stores
Voltage across inductor is zero, if current through it is
constant
Small change in current in zero time gives infinite voltage
across the inductor

40. Capacitance
• Ability of a body to store an electrical charge
• Two conducting surfaces separated by an
insulating medium
• Conducting surfaces – Electrodes
• Insulating medium – dielectric
• Amount of charge per unit voltage - Capacitance

41. Cont….
Unit of capacitance – Farad
C= Q/V; C = q/v
i = C dv/dt => dv = 1/C x i dt =>
∫dv = 1/C ∫i dt => v(t) – v(0) =
1/C ∫i dt
v(t) = v(0) + 1/C ∫i dt
P = vi = v C dv/dt
W = ∫p dt = ½ Cv2
• Current in a capacitor is zero
if V is constant
• Open circuit to dc
• Small change in voltage
within zero time gives infinite
current
• Even if current is zero, store
finite amount of energy
• Pure capacitor never dissipate
energy

42. Open circuit
An open circuit between A and B means i=0.
• Voltage across an open circuit: any value.
• An open circuit is equivalent to R = ∞ Ω.
• The open circuit voltage is the voltage difference
measured between two terminals when no current is
drawn or supplied.

43. Quiz
What is current value in open circuit?

44. Short circuit
A short circuit between A and B means v=0.
• Current through a short circuit: any value.
• A short circuit is equivalent to R = 0 Ω.
The short circuit current is the current that flows when the
terminals are forced to have zero voltage difference

45. Quiz
What is current value in short circuit?

46. Node
• A point at which two or more elements have a common
connection is called a node.
• If no node was encountered more than once, then the set of
nodes and elements that we have passed through is defined
as a path.
• Mesh or Loop is a set of branches forming a closed path in a
network.
• A Branch as a single path in a network, composed of one
simple element and the node at each end of that element.

47. • (a) A circuit containing three
nodes and five branches.
• (b) Node 1 is redrawn to look
like two nodes; it is still one
node.
• If three is no element
between two or more
connected adjacent nodes,
these nodes can be
recombined as a single node

49. Quiz
A network which connects various points of the
network with one another is ____

50. RESISTORS IN SERIES
• All the current, voltage, and power relationships
in the remainder of the circuit will be unchanged

52. Equivalent resistance

53. Resistors in parallel
• Determine i in the circuit of Figure

54. • The special case of only two parallel resistors
is encountered fairly often, and is given by

55. Problems
• Use resistance and source combinations to determine
the current I in Figure and the power delivered by the
80 V source

56. • Determine v in the circuit of Figure by first
combining the three current sources, and then the
two 10 resistors.

57. VOLTAGE DIVISION
• If we know the total voltage over a series of resistors,
we can easily find the individual voltages over the
individual resistors.
• Since the resistors in series have the same current, the
voltage divides up among the resistors in proportion
to each individual

58. Cont...
To find the voltage over an individual resistance
in series, take the total series voltage and
multiply by the individual resistance over the
total resistance

59. Problem on voltage division
• Determine vx in the circuit of Figure
We first combine the 6 and 3 resistors,
replacing them with (6)(3)/(6 + 3) = 2Ω
vx= (12 sin t)* 2 /4 + 2 = 4 sin t volts

60. Quiz
Use voltage division to determine vx

61. Current division rule

62. Problem
• Write an expression for the current through
the 3 resistor in the circuit

63. Kirchhoff's Law
• The I-V relationship for a device tells us how current
and voltage are related within that device.
• Kirchhoff’s laws tell us how voltages relate to other
voltages in a circuit, and how currents relate to other
currents in a circuit.
• KVL: The sum of voltage drops around a closed path
must equal zero.
• KCL: The sum of currents leaving a closed surface or
point must equal zero.

64. KIRCHHOFF’S VOLTAGE LAW
• The algebraic sum of the voltages around any
closed path is zero

65. Cont...
Suppose I add up the potential
drops around the closed path,
from “a” to “b” to “c” and back
to “a”. • Since I end where I
began, the total drop in
potential I encounter along the
path must be zero: Vab + Vbc +
Vca = 0

66. problem
• We know the voltage across two of the three
elements in the circuit. Thus, KVL can be
applied immediately to obtain vx
-5-7+vx =0, vx= 12 V
ix= vx/R = 12/100= 120mA

67. Quiz
Determine ix and vx in the circuit

68. Problem
Find the voltage across R2 and the voltage labeled
vx
KVL equation around the loop on the left, starting at point c: 4 − 36 + vR2 =
0
vR2 = 32 V.
-32 + 12 + 14 + vx = 0=> vx=6V

69. KIRCHHOFF’S CURRENT LAW
• The algebraic sum of the currents entering any node is zero.
• A node is not a circuit element, and it certainly cannot store,
destroy, or generate charge
• equate the sum of the currents having reference arrows
directed into the node to the sum of those directed out of the
node

70. Problem
• Compute the current through resistor R3 if it
is known that the voltage source supplies a
current of 3 A
iR1-2-i+5=0=> i=3-2+5=6A

71. Compute the power absorbed in
each element for the circuit
shown in Figure
KVL around the loop:
−120 + v30 + 2vA − vA = 0
By ohms law
v30 = 30i and vA = −15i
−120 + 30i − 30i + 15i = 0=> i=8A

72. Cont...
p120v = (120)(−8) = −960 W
p30 = (8) ^2 (30) = 1920 W
pdep = (2vA)(8) = 2[(−15)(8)](8) = −1920 W
p15 = (8) ^2 *(15) = 960 W

73. Quiz
• In the circuit of Figure, vs1 = 120 V, vs2 = 30 V,
R1 = 30 , and R2 = 15 . Compute the power
absorbed by each element.

74. Tutorial

76. Tutorial

78. Mesh Analysis
Mesh -closed loop
Mesh-> KVL+ohm’s law
Procedure to solve mesh
1. Find the no. of meshes
2. Assign the mesh currents in clockwise direction
3. Write mesh equations by Using KVL first and
ohm’s law next
4. Solve the equations

79. Problems

81. Problem

86. Supermesh
• a kind of “supermesh” from two meshes that have
a current source as a common element; the current
source is in the interior of the supermesh
• If the current source lies on the perimeter of the
circuit, then the single mesh in which it is found is
ignored.
• Kirchhoff’s voltage law is thus applied only to

89. NODAL ANALYSIS
Node -Group of branch connected at single point
Node-> KCL+ohm’s law
Procedure to solve mesh
1. Find the no. of nodes
2. Assign node voltage with reference to ground node.
3. Write nodal equations by Using KCL first and ohm’s
law next
4. Solve the equations

90. • Apply KCL to nodes 1 and 2
• v1 = 5 V and v2 = 2 V.

91. Determine the current flowing left to right through
the 15 Ω resistor of Figure
• Solving, we find that v1 = 20 V and v2 = 20 V so
that v1 − v2 = 0. In other words, zero current is
flowing through the 15 resistor in this circuit!

93. Tutorial 2
• For the circuit of Figure, determine the nodal
voltages v1, v2 and v3.

96. • Determine the power supplied by the dependent source
of Figure

98. Tutorial
Determine the power supplied by the
dependent source of Fig

100. • For the circuit of Fig., compute the voltage
across each current source

101. • For the circuit of Fig, determine the nodal
voltage v1 if A is (a) 2i1; (b) 2v1.

103. ⓘ Start presenting to display the poll results on this slide.
For the circuit of Fig, determine the nodal
voltage v1 if A is 2v

105. THE SUPERNODE
• The voltage source together as a sort
of supernode and apply KCL to both
nodes at the same time
• if the total current leaving node 2 is
zero and the total current leaving
node 3 is zero, then the total current
leaving the combination of the two
nodes is zero. This concept is
represented graphically

106. Problem
• Determine the value of the unknown node
voltage v1 in the circuit of Fig.

108. For the circuit, compute the voltage across each
current source.

110. ⓘ Start presenting to display the poll results on this slide.
Determine the potential difference between nodes P
and Q i.e. VPQ

111. Summary of Supernode Analysis
Procedure
• Count the number of nodes (N)
• Designate a reference node.
• Label the nodal voltages (there are N − 1 of them).
• If the circuit contains voltage sources, form a supernode about each
one.
• Write a KCL equation for each nonreference node and for each supernode
that does not contain the reference node
• Relate the voltage across each voltage source to nodal voltages Express
any additional unknowns (i.e., currents or voltages other than nodal
voltages) in terms of appropriate nodal voltages
• Organize the equations
• Solve the system of equations for the nodal voltages

112. Determine the node-to-reference voltages in
the circuit of Fig

122. Phase shift
The time taken for any wave to
complete one full cycle is called
the period
The frequency of a wave is defined
as the number of cycles that a
wave completes in one second.
The relation between time period
and frequency is given by f=1/T

123. ⓘ Start presenting to display the poll results on this slide.
The period of a sine wave is 20 milliseconds.
What is the frequency?

124. The frequency of a sine wave is 30 Hz. What is its period?

125. A radian is defined as the angular distance measured along the
circumference of a circle which is equal to the radius of the circle.

126. Phase of a Sine Wave
• The sine wave is shifted to
the right by 90° (pi/2 rad)
• The waveform B is lagging
behind waveform A by 90°. In
other words, the sine wave A
is leading the waveform B by
90°
• The sine wave A is lagging
behind the waveform B by
90°. In both cases, the phase
difference is 90°

127. In general, the sine wave is represented by the equation
v(t) =Vmsin 𝝎t
When a sine wave is shifted to the left of the reference wave
by a certain angle, the general expression can be written as
v(t) =Vmsin (𝝎t+Φ)
When a sine wave is shifted to the right of the reference
wave by a certain angle
v(t) =Vmsin (𝝎t-Φ)

128. PHASE RELATION IN A PURE RESISTOR
When a sinusoidal voltage of certain magnitude is applied to a resistor,
a certain amount of sine wave current passes through it. The
voltage/current relation in case of a resistor is linear,
i.e. v (t) = i(t)R
Consider the function
i(t)=Imsin𝜔t=IM[Im ej𝜔t ] or Im∠0o
v(t)= ImR sin𝜔t=Vm sin𝜔t
= IM[Vm ej𝜔t ] or Vm∠0o
Where Vm = ImR
Impedance is defined as the ratio of
voltage to current function

129. PHASE RELATION IN A PURE INDUCTOR
v(t)= Ldi/dt
Consider the function
i(t)=Imsin𝜔t=IM[Im ej𝜔t ] or Im∠0o
v(t)= L d(Imsin𝜔t)/dt=L𝜔 Imcos𝜔t =Vmcos𝜔t or Vmsin(𝜔t+90)
= IM[Vm ej(𝜔t+90) ] or Vm∠90o
Vm= 𝜔 L Im =XL Im and ej90 =j=1 ∠90o
Voltage and current are out of phase.
The current lags behind the voltage by 90° in a
pure inductor

130. Impedance
The impedance which is the ratio of exponential
voltage to the corresponding current, is given by
Impedance= Vm∠90o / Im∠0o
Z= j𝜔L= j XL
Where XL = 𝜔L
Vm= 𝜔 L Im

131. PHASE RELATION IN A PURE CAPACITOR
v(t)=1/C∫ i(t) dt
Consider the function
i(t)=Imsin𝜔t=IM[Im ej𝜔t ] or Im∠0o
v(t)=1/C ∫Imsin𝜔t dt = 1/𝜔C Im [-cos𝜔t]= Im/𝜔C sin(𝜔t-90)
v(t)= Vmsin(𝜔t-90)=IM[Vm ej(𝜔t-90) ] or Vm∠-90o
Where Vm=Im/𝜔C
Current leads the voltage by 90°
Impedance= Vm∠-90o / Im∠0o
Z= -j/𝜔C= -j XC
Where XC =1/𝜔C

132. Quiz
How many cycles does a sine wave go through in
10 seconds when its frequency is 60 Hz?

133. Quiz
If the peak value of a certain sine wave voltage is 10 V, what is the peak-to-
peak value?

134. ⓘ Start presenting to display the poll results on this slide.
Sine wave A has a positive going zero
crossing at 30°. Sine wave B has a positive
going zero crossing at 45°. What is the phase
angle between two signals?

135. Phasor
Positive angles are measured
counterclockwise from 0°,
whereas negative angles are
measured clockwise from 0°
For positive angle 𝜽 , the
corresponding negative
angle is 𝜽-360°. the positive
angle 135° of vector A can be
represented by a negative
angle -225°, (135°-360°).

136. Phasor
The sine wave B lags behind
the sine wave A by 45°
The sine wave C leads the
sine wave A by 30°.

137. Draw the phasor diagram to represent the two
sine waves

138. Series RLC
In the circuit, determine the total impedance, current I,
phase angle 𝜽, and the voltage across each element.

140. A signal generator supplies a sine wave of 20 V, 5
kHz to the circuit. Determine the total current
IT, the phase angle and total impedance in the
circuit.
Parallel RC

142. Parallel RL
A 50 V resistor is connected in parallel with an
inductive reactance of 30 V with angle 90. A 20 V
signal is applied to the circuit. Find the total
impedance and line current in the circuit

144. Problem
Determine the equivalent impedance for the
circuit

146. Nodal analysis- AC circuit

148. DUALITY
• In an electrical circuit itself there are pairs of
terms which can be interchanged to get new
circuits. Such pair of dual terms are given
below

150. Duality
To draw the dual of any network, the following steps are to
be followed.
1. In each loop of a network place a node; and place an
extra node, called the reference node, outside the
network.
2. Draw the lines connecting adjacent nodes passing
through each element, and also to the reference node, by
placing the dual of each element in the line passing through
original elements.

151. Problem
Draw the dual network for the given network

153. Draw the dual network for the given network

154. Solution