EE110-elementary-circuit-and-network-theory-(a)

Elementary Circuit and Network Theory
(EE110 – Classes 1 to 4)
R S Ananda Murthy
Associate Professor
Department of Electrical & Electronics Engineering,
JSS Science & Technology University
Sri Jayachamarajendra College of Engineering,
Mysore 570 006
R S Ananda Murthy Elementary Circuit and Network Theory

Learning Outcomes
After completing these lectures the student should be able to –
State Ohm’s law and its limitations.
Apply Kirchhoff’s Laws to solve network problems.
Explain the concepts of voltage source and current source.
Determine the response of a resistive network excited by
D.C. source/s by Loop Current Method (Mesh Analysis).
Determine the response of a resistive network excited by
D.C. source/s by Node Voltage Method.

Statement of Ohm’s Law
At moderate constant temperature with other physical
conditions remaining unaltered, the magnitude of electric
current I through a conductor is directly proportional to the
potential difference V applied across its ends.
Mathematically this law can be expressed as
I ∝ V or V = RI I = GV or R =
V
I
where R is the constant of proportionality known as
resistance and G = 1/R is known as the conductance.
R is measured in Ohms (Ω) and G is measured in Mho
( ).

Limitation of Ohm’s Law
Ohm’s law holds good only for metallic conductors at
moderate temperatures.
At very low and very high temperatures even metallic
conductors do not obey Ohm’s law.
This law is not obeyed by vacuum tubes, discharge tubes,
semiconductors and electrolytes.
Devices that obey Ohm’s law are known as ohmic devices
and those which do not are called non-ohmic devices.

Ohmic and Non-ohmic Device Characteristics
A B
A
B
A graph of current passing through a device versus voltage
applied across it is known as volt-ampere characteristic, or
v-i characteristic.
Which of the devices whose characteristics shown above
obeys Ohm’s law?

Factors Affecting Resistance of a Conductor
The value of resistance depends upon the material and its
dimensions. If l is the length and A is the cross-sectional
area of a wire, then, the resistance of the wire is given by
R =
ρl
A
where ρ is the speciﬁc resistance expressed in Ω-m which
is a property of the material of the wire.

Temperature Affects Resistance
As ρ and dimensions of the wire change with temperature,
the resistance R also changes with temperature.
The resistance at a temperature T is given by
Rt = R0(1+α0T)
where R0 is the resistance at 0◦C and α0 is the
temperature co-efﬁcient of resistance of the material at
0◦C.

Power Loss in a Resistance
The power loss in the resistance is given by
P = I2
R or P =
V2
R
where V is the voltage across it and I is the current
through it.
Practical resistances have a power rating.
If the power dissipated in a resistor exceeds its power
rating, it will burn out.

Practice Problem Set 1
1 A potential difference of 2.5 V causes a current of 250µA
to ﬂow in a conductor. Calculate the resistance of the
conductor.
2 What is the voltage across an electric heater of resistance
10Ω through which passes a current of 23 A?
3 Calculate the current in a circuit due to a potential
difference of 10 V applied to a 10 kΩ resistor. If the supply
voltage is doubled while the circuit resistance is trebled,
what is the new current in the circuit?
4 A potential difference of 10 V is applied to a 3.3 kΩ.
Calculate the circuit current assuming that the circuit obeys
Ohm’s Law.

Practice Problem Set 2
1 A current in a circuit is due to a potential difference of 30 V
applied to a resistor of resistance 300 Ω. What resistance
would permit the same current to ﬂow if the supply voltage
were 300 V.
2 A coil has 2000 turns of copper wire having a
cross-sectional area of 0.8 mm2. The mean length per turn
is 80 cm and the resistivity of copper is 0.02 µΩ-m at
normal working temperature. Calculate the resistance of
the coil and the power dissipated when the coil is
connected to a 110 V D.C. supply.
3 A resistor made of copper wire has a resitance of 120 Ω at
25◦C. Find the resistance of the resistor at a temperature
of 55◦ C. Take temperature coefﬁcient of copper as
4.2×10−3 at 0◦.

Kirchhoff’s Current Law (KCL)
Algebraic sum of currents at a node is equal to zero at all
instants of time.
Here the word ‘node’ means a junction point where two or
more branches are connected.
KCL can be mathematically expressed as
n
∑
k=1
ik = i1 +i2 +i3 +···+ik = 0
where n =number of branches meeting at the node.
This law is a consequence of the basic law of conservation
of matter.
While applying this law follow a consistent sign convention.

Kirchhoff’s Voltage Law (KVL)
Algebraic sum of voltages in a closed path (loop) taken in a
direction is equal to zero at all instant of time.
KVL can be mathematically expressed as
n
∑
k=1
vk = v1 +v2 +v3 +···+vk = 0
where n =number of voltages encountered while tracing a
closed path.
This law is a consequence of the law of conservation of
energy.
While applying this law follow a consistent sign convention.

KCL and KVL Example
(b)
A
B
CD
E
A B
C
(a)
In Fig-(a) at Node A, we get KCL equation
I1 +I2 −I3 −I4 = 0
In Fig-(b) applying KVL in the path A-C-B-A we get
V1 +R1I1 −I2R2 −V2 +R3I3 = 0
Can you tell what sign convention is followed here?

Practice Problem on KCL and KVL
A
B C
In the circuit shown above, R1 = 5Ω, R2 = 20Ω, V1 = 20 V,
V2 = 40 V, and V = 60 V. Find the value of R.

Series Connection of Resistors
By KVL and Ohm’s Law applied to the circuit on the left we get
v = v1 +v2 +v3 +···+vn = i(R1 +R2 +R3 +···+Rn)
If the resistance Req is the equivalent resistance, then,
iReq = i(R1 +R2 +R3 +···+Rn) =⇒ Req =
n
∑
k=1
Rk

Parallel Connection of Resistors
By KCL and Ohm’s Law applied to the circuit shown on the left
we get
i = i1 +i2 +i3 +···+in = v(G1 +G2 +G3 +···+Gn)
If Geq is the equivalent conductance, then,
vGeq = v(G1 +G2 +G3 +···+Gn) =⇒ Geq =
n
∑
k=1
Gk

Practice Problem-1
a b
c
An ohm meter is an instrument that measures the value of
resistance connected between its terminals. What is the
reading shown by the ohm meter when connected to the
network shown above across points (i) a and b; (ii) a and c; (iii)
b and c.

Practice Problem-2
A B
A
B
(a) (b)
20 A
A current of 20 A ﬂows through two ammeters A and B
connected in series as shown in Fig-(a). Across A, the potential
difference is 0.2 V and across B it is 0.3 V. Find how the same
will divide between A and B when they are connected in parallel
as shown in Fig-(b).

Source and Sink (Load)
Load
or
Sink
Source of
Electric
Power
Source — a device which supplies electric energy.
Sink (or load) — a device which consumes electric energy.
Power delivered by the source to sink is given by p = vi
For the sake of standardization, v or i is maintained
constant. Eg. In India, power supply at dometic wall socket
is available at constant nominal value of 240 V.

Types of Sources
D.C. sources – Examples of D.C. sources are battery, solar
photovoltaic panel, laboratory power supplies, and
switched mode power supply (SMPS) found in computers.
A.C. sources – Examples of A.C. sources are domestic
wall outlet, stand-by generators, home uninterruptible
power supplies (UPS).

Ideal Voltage Source
−
+
V v
++
−
i
v
V
i
An ideal voltage source can deliver any current and power
to the load at constant voltage.
For example, the domestic wall socket power supply has a
constant nominal value of 240 V.

Ideal Current Source
v
i
v
++
−
I
i = I
I
An ideal current source can supply any voltage and power
to the load at constant current.
Current sources are typically used in analog ICs
(integrated circuits).

Voltage Source Model of Practical Source
Practical sources of electric energy have a drooping v-i
characteristic approximately taken linear as shown above. Here
Voc = Open circuit voltage at the terminals of the source and
Isc = current delivered by the source when terminals are
shorted. This linear v-i characteristic is represented by
v = −Rini +Voc
which is satisﬁed by the equivalent circuit shown above as
veriﬁed by KVL.

Current Source Model of Practical Source
Taking i along the vertical axis, the drooping i-v characteristic
of a practical source will be as shown above. This linear i-v
characteristic is represented by
i = −
1
Rin
v +Isc
which is satisﬁed by the equivalent circuit shown above as
veriﬁed by KCL.

Transformation of Practical Source
(a) (b)
A practical source of electric energy can be represented as an
ideal voltage source connected in series with the internal
resistance Rin = (Voc/Isc) as shown in Fig-(a) or as an ideal
current source connected in parallel with the internal resistance
Rin as shown in Fig-(b).

Series Connection of Voltage Sources
By applying KVL in this circuit we get
Veq = V1 +V2 +V3 +···+Vn =
n
∑
k=1
Vk
Ideal voltage sources cannot be connected in parallel since the
terminal voltage of that parallel combination will be
indeterminate.

Parallel Connection of Current Sources
By applying KCL in this circuit we get
Ieq = I1 +I2 +I3 +···+In =
n
∑
k=1
Ik
Ideal current sources cannot be connected in series since the
terminal current of that series combination will be
indeterminate.

Methods of Solving Network Problems
To write network equilibrium equations based on KVL/KCL in a
systematic way, the following methods are available –
Loop Current Method (also known as Mesh Analysis)
Node Voltage Method
As the network size increases, Loop Current Method becomes
more cumbersome. But Node Voltage Method does not have
this drawback.

Steps for Loop Current Method
1 Replace practical current sources if any, by equivalent
voltage sources.
2 Find number of independent loops l = b −n +1. An
independent loop is a closed path which has at least one
branch which is not present in other loops. Here b =
number of branches, and n = number of nodes. Each
component can be taken as a branch. A node is a junction
point of two or more branches. Points which are shorted
should be treated as a single node.
3 Mark independent loop currents intelligently to reduce
computation and number them sequentially.
4 Write loop equations in matrix form by inspection.
5 Solve for the required loop currents and ﬁnd the response.

Writing Loop Equations by Inspection
Loop equations in matrix form can be written as
[R][I] = [V]
Here the elements of l ×l matrix [R] can be written by
inspection using the formula Rjj = (sum of all the resistances in
the loop j), and Rjk = ±(sum of all the resistances common to
loops j and k). + sign to be used if Ij and Ik are in same
direction in common branches. Else − sign to be used.
[I] is a l ×1 column matrix containing loop currents Ik which are
to be calculated.
[V] is also a l ×1 column matrix. The elements of this matrix
can be written by inspection using the formula Vk = sum of all
the voltage sources acting in the direction of Ik in loop k.

Steps for Node Voltage Method
1 Replace all practical voltage sources by equivalent current
sources.
2 Find number of independent nodes using N = n −1.
3 Select a datum node, mark it with a ground symbol, and
number sequentially other independent nodes.
4 Write node equations in matrix form by inspection.
5 Solve for the required node voltages and ﬁnd the response.

Writing Node Equations by Inspection
Node equations in matrix form can be written as
[G][V] = [I]
Here the elements of N ×N matrix [G] can be written by
inspection using the formula Gjj = (sum of all the conductances
connected to node j), and Gjk = −(sum of all conductances
connected between nodes j and k).
[V] is a N ×1 column matrix containing node voltages Vk which
are to be calculated.
[I] is also a N ×1 column matrix. The elements of this matrix
can be written by inspection using the formula Ik = sum of all
the current sources directed towards node k.

Practice Problem Set-3
(a) (b)
4 V
1 A
3 V6 V
A
B
2 V
2 V
1 A
1 A
1 Find the equivalent voltage source and current source
connected across the terminals A and B in Fig-(a).
2 Find the current i in the circuit of Fig-(b) by Loop Current
Method and Node Voltage Method.

License
This work is licensed under a
Creative Commons Attribution 4.0 International License.

EE110-elementary-circuit-and-network-theory-(a)

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