EE14 303 ELECTRIC CIRCUIT THEORY

MODULE I (Part 1)

Prepared by: Muhammedali Shafeeque K

shafeeq@meaengg.in

onlinemas.weebly.com

References:

1. Edminister, Electric Circuits – Schaum’s Outline Series, McGraw-Hill.

2. A.Chakrabarti, Circuit Theory, Dhanapat Rai.

3.Valkenberg, Network Analysis, Prentice-Hall of India.

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Electric Circuit: An electric circuit is a closed path consisting of electrical elements.

Network Elements:

Three main electric network elements are (i) Resistor (ii) Inductor (iii) Capacitor

Resistor & Resistance (R):

An electrical element, which opposes flow current is called a resistor. The property of an element

to oppose the flow of current is called Resistance. Unit is ohm (Ω).

Ohm’s law defines the relation between current flowing through a resistor,I, and voltage across it

,V as follows:

𝑅 =𝑉

𝐼

All electrical devices that consume energy is called a resistor.

Inductor & Inductance (L)

An electrical element, which opposes change in current is called a inductor. The property of an

element to oppose the change in current is called Inductance. Unit is Henry (H).

Relation between current flowing through a inductor,i, and voltage across it ,v is as follows:

Also, the circuit element that stores energy in a magnetic field is defined as an inductor.

Capacitor & Capacitance (C)

An electrical element, which opposes change in voltage is called a Capacitor. The property of an

element to oppose the change in voltage is called Capacitance. Unit is Farad(F).

Relation between current flowing through a capacitor, i, and voltage across it ,v is as follows:

The circuit element that stores energy in an electric field is called a capacitor.

Active & Passive elements:

Active elements are electric elements, which are able to supply energy to the network. Voltage or

current sources are active elements.

Passive elements are electric elements, which are able to either convert energy to another form

or store it in an electric or magnetic field. Resistors, inductors, and capacitors are passive

elements.

Lumped Parameters:

The circuit diagrams of resistor, inductor or capacitor are termed as lumped-parameter circuits,

since a single element in one location is used to represent a distributed resistance, inductance, or

capacitance. For example, a coil consisting of a large number of turns of insulated wire has

resistance throughout the entire length of the wire. But, a single resistance lumped at one place

represents the distributed resistance.

Series Connection of resistors:

If and only if, any of the two ends of a resistor is connected to any of the two ends of other

resistor, it is called series connection.

For series connection equivalent resistance,

Req= R1+R2

Parallel Connection of resistors:

If two ends of a resistor is connected to two ends of other resistor, it is called parallel connection

For parallel connection, equivalent resistance, Req is such that

1

𝑅𝑒𝑞=

1

𝑅1+

1

𝑅2

Kirchhoff’s Voltage Law (KVL):

Kirchhoff's voltage law (KVL) states that, for any closed path in a network, the algebraic sum of

the voltages is zero.

Kirchhoff’s Current Law (KCL):

The connection of two or more circuit elements creates a junction called a node.

Kirchhoff’s current law (KCL) states that the algebraic sum of the currents at a node is zero.

It may be stated alternatively that the sum of the currents entering a node is equal to the sum of

the currents leaving that node.

The basis for the law is the conservation of electric charge.

Voltage division rule:

Total voltage across two series resistors will be divided between each as per the voltage division

rule.

𝑉𝑅1 = 𝐼𝑅1 =𝑉

𝑅1 + 𝑅2𝑅1

Similarly,

𝑉𝑅2 =𝑉

𝑅1 + 𝑅2𝑅2

Current division rule:

In a parallel circuit, total current will be divided between each resistors as per the current

division rule.

𝐼𝑅1 =𝑉

𝑅1=

𝐼𝑅𝑒𝑞

𝑅1=

𝐼𝑅1𝑅2

𝑅1 + 𝑅2

𝑅1=

𝐼𝑅2

𝑅1 + 𝑅2

Similarly,

𝐼𝑅2 =𝐼𝑅1

𝑅1 + 𝑅2

Electric Sources:

Electric sources are broadly classified into two types- Voltage Source and current source.

An electric source, which supplies voltage, is a voltage source and one, which supplies current, is

called current source.

Independent Sources:

Independent Voltage Source: A voltage source of which the strength of voltage supplied by it is

constant with change in circuit parameters is called Independent Voltage Source.

Independent Current Source: A current source of which the strength of current supplied by it is

constant with change in circuit parameters is called Independent current Source.

Dependent Sources:

Dependent Voltage source: A voltage source of which the strength of voltage supplied by it

changes with change in circuit parameters is called dependent voltage Source.

Dependent current source: A current source of which the strength of current supplied by it

changes with change in circuit parameters is called dependent current Source.

Dependent Sources are classified as follows:

i. Current Controlled Voltage Source (CCVS): A voltage source of which the strength of

voltage supplied by it changes with change in current through any circuit element.

ii. Voltage Controlled Voltage Source (VCVS): A voltage source of which the strength of

voltage supplied by it changes with change in voltage across any circuit element.

iii. Current Controlled Current Source (CCCS): A current source of which the strength of

current supplied by it changes with change in current through any circuit element.

iv. Voltage Controlled Current Source (VCCS): A current source of which the strength of

current supplied by it changes with change in voltage across any circuit element.

Representations of Electric Sources:

a) Independent Voltage source

b) Dependent Voltage Source

c) Independent Current Source

d) Dependent Current Source

Practical Sources:

Practical Voltage source:

A practical voltage source is represented by an ideal voltage source in series with a

resistance (called as internal resistance).

Practical Current source:

A practical current source is represented by an ideal current source in parallel with a

resistance (called as internal resistance).

Source Transformation:

Voltage Source to Current Source:

Condition for source transformation: While transforming voltage source to current

source power delivered to RL by voltage source should be equal to power delivered to RL

by current source.

Power delivered to RL by voltage source =𝐼𝐿2𝑅𝐿

= 𝑉

𝑅 + 𝑅𝐿

2

𝑅𝐿

=𝑉2𝑅𝐿

𝑅 + 𝑅𝐿 2

Power delivered to RL by current source =𝐼𝐿2𝑅𝐿

= 𝐼𝑆𝑅𝑠ℎ

𝑅𝑠ℎ + 𝑅𝐿

2

𝑅𝐿

(Using Current division rule for finding IL)

=𝐼𝑆

2𝑅𝑠ℎ2𝑅𝐿

𝑅𝑠ℎ + 𝑅𝐿 2

But, Power delivered to RL by voltage source= Power delivered to RL by current source

Therefore ,

𝑉2𝑅𝐿

𝑅 + 𝑅𝐿 2=

𝐼𝑆2𝑅𝑠ℎ

2𝑅𝐿

𝑅𝑠ℎ + 𝑅𝐿 2

or

𝑉

𝑅 + 𝑅𝐿=

𝐼𝑆𝑅𝑠ℎ

𝑅𝑠ℎ + 𝑅𝐿

Equating numerator and denominator,

Rsh=R

and Is=V/Rsh=V/R

Current Source to Voltage Source:

Condition for source transformation: While transforming current source to voltage

source, power delivered to RL by current source should be equal to power delivered to RL

by voltage source.

Power delivered to RL by current source=𝐼𝐿2𝑅𝐿

= 𝐼𝑅

𝑅 + 𝑅𝐿

2

𝑅𝐿

(Using Current division rule for finding I )

=𝐼2𝑅2𝑅𝐿

𝑅 + 𝑅𝐿 2

Power delivered to RL by voltage source =𝐼𝐿2𝑅𝐿

= 𝑉𝑆

𝑅𝑠𝑒 + 𝑅𝐿

2

𝑅𝐿

=𝑉𝑆

2𝑅𝐿

𝑅𝑠𝑒 + 𝑅𝐿 2

But, Power delivered to RL by current source = Power delivered to RL by voltage source

Therefore ,

𝐼2𝑅2𝑅𝐿

𝑅 + 𝑅𝐿 2=

𝑉𝑆2𝑅𝐿

𝑅𝑠𝑒 + 𝑅𝐿 2

or 𝐼𝑅

𝑅 + 𝑅𝐿=

𝑉𝑆𝑅𝑠𝑒 + 𝑅𝐿

Equating numerator and denominator,

Rse=R

and VS=IR

Mesh Analysis:

Loop: Any closed path in an electric circuit is called a loop.

Mesh: Any loop having no inner loop is called a mesh.

Steps for solving Electric circuits using mesh analysis:

1) Identify the meshes in the given circuit.

2) Mark the mesh current in each mesh (preferably clockwise direction)

3) Write KVL equation of each meshes, which is called mesh equations.

4) Solve mesh equations for mesh currents.

5) Draw circuit with actual mesh currents and direction.

6) Find current through each resistors.

Problem: Find current through all resistors using mesh analysis.

Solution:

Step1: Identify the meshes in the given circuit.

Step2: Mark the mesh current in each mesh (in clockwise direction)

Step3: Write KVL equation of each meshes, (mesh equations).

Mesh equation of Mesh1,

Mesh equation of Mesh2,

Rearranging,

………………………………(1)

………………………….(2)

Step 4: Solve mesh equations for mesh currents.

Solving (1) and (2),

I1= 2A and I2=1A

Step5: Draw circuit with actual mesh currents and direction.

Step6: Find current through each resistors.

I5Ω= I1= 2A

I10Ω= I1-I2= 2-1=1A

I2Ω= I2= 1A

Super Mesh Analysis:

If there is any current source in the circuit, it cannot be solved by conventional mesh analysis.

Rather, super mesh analysis can be used for solving such circuits.

Super Mesh Case (i): Circuits with current source related to a single mesh.

Problem: Solve the following circuit using mesh analysis.

Solution: Since there is current source, it can be solved by super mesh analysis only

Step1 :Identify meshes.

There are two meshes.

Step2: Identify super mesh (mesh having current source);

Mesh 2 is a super mesh (since it have 2A current source)

Step3: write equation of super mesh 2 with current passing through current source

-I2= 2A …………………………(1)

(-ve sign since current source oppossit to mesh current)

Step 4: write mesh equation of other meshes as usual method

For mesh 1

-20 + I1+2(I1-I2)=0

Rearranging,

3I1-2I2=20 ……………………………(2)

Step 5: Solve mesh equations to solve for mesh currents.

Solving (1) and (2)

I1=16/3 A and I2=2A

Step 6: Draw circuit with real mesh currents and find current through all resistors.

Note that second mesh current is marked as anticlockwise direction as we got I2 as negative.

I1Ω=16/3 A

I2Ω=16/3+2=22/3 A

I4Ω=2 A

Super Mesh Case (ii): Circuits with current source related to two meshes.

Problem: Solve the following circuit using mesh analysis.

Step1 :Identify meshes.

There are two meshes.

Step2: Identify super mesh (mesh having current source);

Mesh 1 &2 are a super mesh case ii (since 2A current source is linking both meshes)

Step3: write equation of super mesh 1 &2 with current passing through current source

I2 –I1 = 2A …………………………(1)

Step4: Consider a loop which is outer to both super meshes 1 &2 as in fig.

Step5: write the KVL equation of this outer loop

-10 + I1+2I2+3I2=0

Rearranging,

I1+5I2=10 …………………………(2)

Step6 : Solve the mesh equations for mesh currents

Solving (1) & (2)

I1=0

I2=2A

Step7 : Draw circuit with real mesh currents and find current through all resistors.

I1Ω=0A

I2Ω=2A

I3Ω=2A

Nodal Analysis:

Node: A node is an intersection point of two or more elements in a circuit.

Problem: Solve following circuit with nodal analysis.

Step1: Identify the nodes in the circuit.

Step2: Select a node as reference node and its voltage as 0.Mark the voltages of nodes of which

node voltages are known. Mark the unknown voltage nodes with variables like V1,V2 …….

(Node voltage is the voltage across the node and reference)

Step3: Write the KCL equations at each unknown voltage nodes.

At node V1

+10−𝑉1

2−

𝑉1−0

1−

𝑉1−𝑉2

2= 0 ……………………………………(1)

At node V2

+𝑉1−𝑉2

2+ 2 = 0 ……………………………………….(2)

Step4: Solve KCL equations to find node voltages.

Multiply (1) by 2

10-V1-2V1-V1-V2=0

Rearranging,

4V1+V2=10 ………………………………..(3)

Multiply (2) by 2

V1-V2+4=0

Rearranging,

V1-V2=-4 …………………..(4)

Solving (3) & (4),

V1=6/5 V

V2=26/5 V

Step5: Find the current through all resistors.

I2Ω =10 −

65

2=

22

5A

I1Ω =

65 − 0

1=

6

5A

I2Ω =

65 −

265

2= −

10

5A

Super Nodal Analysis:

If in a circuit there is a voltage source between two unknown voltage node, conventional nodal

analysis fails. Such circuits can be solved by super node analysis.

Step1: Identify the nodes and mark node voltages as usual nodal analysis.

Step 2: Identify super nodes (unknown voltage nodes having a voltage source across it)

Step3: Neglect any resistor connecting super nodes.

Step4: Write KVL equation between super nodes

Step5:Write KCL equation of super nodes.(write eqn as in usual nodal analysis, but write eqn for

both nodes in a single equation)

Step6: Write node eqns for any other nodes other than super nodes as usual nodal analysis.

Step7:Solve node eqns for node voltages

Step8:findcurrent through all resistors using node voltages.

Problem: Solve the following circuit using nodal analysis.

Step1: Identify the nodes and mark node voltages as usual nodal analysis.

There are three unknown voltage nodes.,V1,V2,V3

Step 2: Identify super nodes

V1 &V2 are super nodes (2V voltage is connecting V1 & V2.)

Step3: Neglect resistor,5Ω, connecting super nodes, V1 & V2.

Step4: Write KVL equation between super nodes, V1 & V2.

V1-V2=2 ……………(1)

Step5: Write KCL equation of super nodes.(write eqn as in usual nodal analysis, but write eqn

for both nodes in a single equation)

For super nodes, V1 & V2.

+20−𝑉1

1−

𝑉1−0

2−

𝑉2−0

2−

𝑉2−𝑉3

1= 0 ……………………….(2)

Step6: Write node eqns for any other nodes other than super nodes as usual nodal analysis.

For node, V3.

+𝑉2−𝑉3

1−

𝑉3−0

2= 0 ………………………………(3)

Step7:Solve node eqns for node voltages

Solve (1),(2) & (3)

From (1),

V1-V2=2 ……………(4)

Multiply (2) by 2

40-2V1-V1-V2-2V2+2V3=0

Rearranging,

3V1+3V2-2V3=40 ……………………(5)

Multiply (3) by 2

2V2-2V3-V3=0

Rearranging, 2V2-3V3=0 ……………………………(6)

Solving (4),(5),(6)

V1=65/7 V

V2=51/7 V

V3=34/7 V

Step8: Find current through all resistors using node voltages.

𝐼1Ω =20 −

657

1𝐴

𝐼5Ω =

657 −

517

5= 2/5𝐴

𝐼2Ω =

657 − 0

2= 130/7𝐴

𝐼2Ω =

517 − 0

2= 102/7𝐴

𝐼1Ω =

517 −

347

1𝐴

𝐼2Ω =

347 − 0

2=

68

7𝐴

Superposition Theorem:

A linear network which contains two or more independent sources can be analyzed to obtain the

various voltages and branch currents by allowing the sources to act one at a time, then

superposing the results.

This principle applies because of the linear relationship between current and voltage.

Steps:

Find current through required resistor independently by only one source, keeping all other

current source open circuit and voltage source by short circuit at this time. Then add all the

currents obtained when acting the sources independently.

Problem: Compute the current in the 23Ω resistor by applying the superposition principle.

Solution:

With the 200-V source acting alone, (Now, the 20-A current source is replaced by an open

circuit. )

When the 20-A source acts alone, the 200-V source is replaced by a short circuit.

The equivalent resistance to the left of the source is

The total current in the 23Ω resistor is

Thevenin’s Theorem:

A linear, active, resistive network which contains one or more voltage or current sources can be

replaced by a single voltage source and a series resistance.

The voltage is called the Thevenin equivalent voltage (simply Thevenin Volatge) and series

resistance is called Thevenin equivalent resistance ( simply Thevenin Resistance).The equivalent

circuit thus formed is called Thevenin Equivalent circuit.

Steps for analyzing circuits:

1) Remove load resistor across terminal ab about which Thevenin equivalent circuit is

determined.

2) Determine RThevnin

a. Keep all voltage source as short circuit and all current source as open circuit

b. Find the equivalent resistance across ab which is RThevnin

3) Determine VThevnin

a. Find the open circuit voltage across ab which is VThevnin

4) Draw Thevenin equivalent circuit by connecting VThevnin and RThevnin in series.

5) Re-connect the removed load resistor at ab and find the current through it.

Norton’s Theorem:

A linear, active, resistive network which contains one or more voltage or current sources can be

replaced by a single current source and a parallel resistance.

The current is called the Norton equivalent current(simply Norton t current) and parallel

resistance is called Norton equivalent current equivalent resistance ( simply Norton equivalent

current Resistance).The equivalent circuit thus formed is called Norton Equivalent circuit.

Steps for analyzing circuits:

1) Remove load resistor across terminal ab about which Norton equivalent circuit is

determined.

2) Determine RNorton

a. Keep all voltage source as short circuit and all current source as open circuit

b. Find the equivalent resistance across ab which is RNorton

3) Determine INorton

a. Short circuit terminals ab

b. Find the short circuit current through ab which is INorton

4) Draw Norton equivalent circuit by connecting INorton and RNorton in parallel.

5) Re-connect the removed load resistor at ab and find the current through it.

Problem: Find the current through 15Ω using Thevenin’s Theorem.

Step1:Remove load resistor 15Ω.

Step2: Determine RThevenin

Short 20 V and 10 V voltage sources.

RTh=Rab=3+6||3 =5Ω

Step3: Determine VThevnin

VTh=Vab

From circuit, by KVL eqn, -20+3I+6-10=0

I=30/9 A

Therefore, VTh=Vab=V6Ω-10V

= 𝐼 × 6 − 10

=30

9 × 6 − 10 = 20 𝑉

Step4: Draw Thevenin equivalent circuit by connecting VThevnin and RThevnin in series.

Step5: Re-connect the removed load resistor, 15Ω , at ab and find the current through it.

I15Ω =10

5 + 15= 0.5A

Problem: Find the current through 15Ω using Norton’s Theorem.

Step1:Remove load resistor 15Ω.

Step2: Determine RNorton

Short 20 V and 10 V voltage sources.

RNor=Rab=3+6||3 =5Ω

Step3: Determine INortonn

Short ab, and find SC current = INorton

Eqn of Mesh 1:

-20 +3 I1+6(I1-I2)-10=0

Eqn of Mesh 2:

10+6(I2-I1)+3I2=0

Rearranging,

9I1-6I2=30 ……………………………………(A)

6I1-9I2= 10 ……………………………………(B)

Solving I2=2 A

Therefore, INorton=ISC= I2=2 A

Step4: Draw Norton equivalent circuit by connecting INorton and RNorton in parallel.

Step5: Re-connect the removed load resistor at ab and find the current through it.

𝐼15Ω = 25

5+15= 0.5𝐴 ………(using current division rule)

Maximum Power transfer Theorem:

Maximum transfer theorem is meant for finding value load RL for that the network is delivering

maximum power to load. According Thevenin any network can be reduced as a voltage source in

series with a resistance. Then for analysis such a circuit is taken

Power delivered to load, P=I2RL

ie 𝑃 = 𝑉

𝑅+𝑅𝐿

2𝑅𝐿 =

𝑉2𝑅𝐿

𝑅+𝑅𝐿 2

For P to be maximum by varying RL, 𝑑

𝑑𝑅𝐿 𝑃 = 0

ie, 𝑑

𝑑𝑅𝐿

𝑉2𝑅𝐿

𝑅+𝑅𝐿 2 = 0

ie 𝑉2 𝑅+𝑅𝐿

2 .1−𝑅𝐿 .2 𝑅+𝑅𝐿 .1

𝑅+𝑅𝐿 4

= 0

ie 𝑅 + 𝑅𝐿 2 − 𝑅𝐿 . 2 𝑅 + 𝑅𝐿 = 0

ie 𝑅𝐿 = 𝑅

Hence Maximum Power transfer theorem states that power delivered to load is maxium when

value of load resistance RL is equal to Thevenin equivalent Resistance of the network.

Maximum Power , Pmax= 𝑉

𝑅+𝑅

2𝑅 =

𝑉2

4𝑅

Problem: Find the value of RL for maximum power transfer and find the maximum power.

For applying Maximum Power transfer theorem, the given circuit should be reduced to its

Thevenin equivalent circuit,

Step1:Remove load resistor 15Ω.

Step2: Determine RThevenin

Short 20 V and 10 V voltage sources.

RTh=Rab=3+6||3 =5Ω

Step3: Determine VThevnin

VTh=Vab

From circuit, by KVL eqn, -20+3I+6-10=0

I=30/9 A

Therefore, VTh=Vab=V6Ω-10V

= 𝐼 × 6 − 10

=30

9 × 6 − 10 = 20 𝑉

Step4: Draw Thevenin equivalent circuit by connecting VThevnin and RThevnin in series.

Hence maximum power transfer occur when RL=5Ω

And Pmax = I2 × 5 = 10

5+5

2

× 5 = 5𝑊

Star –Delta Transformation

Delta to Star transformation

Star to Delta transformation