Unit – 1.pptx

ELE101: Fundamentals of Electrical and

Electronics Engineering

Unit – I: Contents



Ohm’s Law



KCL



KVL



Node Voltage Analysis



Mesh Current



Circuits with Independent Sources



Thevenin and Norton Equivalent



Maximum Power Transfer



Superposition Theorem

Ohm’s Law

 _{Ohm's law states that the current through a conductor between two points}

is directly proportional to the potential difference across the two points.

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.

 _{Ohm's law holds for circuits containing only resistive elements (no}

capacitances or inductances)

 _{Resistors which are in series or in parallel may be grouped together into a}

single "equivalent resistance" in order to apply Ohm's law in analyzing the circuit

 _{When reactive elements such as capacitors, inductors, or transmission lines}

are involved in a circuit to which AC or time-varying voltage or current is applied, the relationship between voltage and current becomes the solution to a differential equation, so Ohm's law does not directly apply

 _{Material that obeys Ohm's Law is called "ohmic" or "linear" because the}

potential difference across it varies linearly with the current.

 _{Problem1 1.}

An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-free. What is the resistance of the lamp?

 Solution :By conservation of energy, the potential that was

gained (V=6.0V) must be lost in the resistor. So, by Ohm's Law:

V = I R R=V/I R = 3.0 

PROBLEMS

ON OHMS LAW

Q2. If a current of 3 A is divided by the following circuit, the current flowing through the 4 Ohm resistor is …

Q3. The diagram at the right shows part of a circuit into which a current I is flowing. Which ammeter shows the highest reading?

Q4. The diagram to the right represents a part of a circuit containing an ohmic resistor, a voltmeter and an ammeter. If the reading on the ammeter A increases the reading on voltmeter V …

Q5. A battery is connected to two identical light bulbs in parallel as well as another identical bulb in series. An ammeter and a

voltmeter are also connected as shown in the circuit diagram

below. One of the light bulbs connected in parallel is unscrewed. How will the ammeter and the

voltmeter readings change?

Q6. In the circuit to the right B₁, B₂ and B₃ are identical light bulbs. The internal resistance of the battery can be ignored.

Which statement is true regarding the relative brightness of the bulbs?

The three bulbs glow with the same brightness, OR B₂ and B₃ glow with the same brightness but brighter than B₁. OR B₂ and B₃ glow with the same brightness but less brightly than B₁ OR B₁ glows brighter than B₂ while B₂ in turn glows brighter than B₃

Q7. A student connects three identical resistors as shown in the

sketch to the right.

The potential difference across the battery is 12 Volt. What are the readings on V₁ and V₂ respectively?

Kirchoff’s Current Law (KCL)

 _{This law is also called Kirchhoff's first law, Kirchhoff's point rule,}

or Kirchhoff's junction rule (or nodal rule).

 _{At any node (junction) in an electrical circuit, the sum}

of currents flowing into that node is equal to the sum of currents flowing out of that node, or:

 _{The algebraic sum of currents in a network of conductors meeting at a}

point is zero.

Kirchoff’s Voltage Law (KVL)

mesh) rule, and Kirchhoff's second rule.

 Similarly to KCL, it can be stated as:

v1 + v2 + v3 – v4 = 0

WRITING KVL EQUATION

 -V_B + V₁ + V₂ = 0

 _{For the second loop (Element #2, Element #3, Element #4). Note, you}

have to be careful with this one because you might not expect the voltage across Element #3 to be defined the way it is.

 -V₂ - V₃ + V₄ = 0

 _{For the third loop (Battery, Element #1, Element #3, Element #4)}

 -V_B + V₁ - V₃ + V₄ = 0

PROBLEMS ON KVL, KCL: Problem 8



For circuit below, find value of i1,i2,i3

assuming

R1 = 100 ohm, R2 = 200 ohm, R3 = 300

ohm

E1= 3 V, E2 = 4 V



Write the KVL equations for the following

loops:

Problem 9

Problem 10: Find I

₁

, I

₂

, I

₃

Solution



The Loop/Mesh-Current Method (based

on KVL):

Solution (Continued)



The Node-Voltage Method (based on

KCL):

Problem 11



Find currents in all branches

Classification of Sources

 _{Independent Sources: Ideal Independent Source maintains same}

voltage or current regardless of the other elements present in the circuit.Its value is either constant (DC) or sinusoidal (AC).

 _{The strength of voltage or current is not changed by any variation in}

connected network.

 _{Dependent Sources: A dependent source is a voltage source or}

a current source whose value depends on a voltage or current somewhere else in the network. The voltage or current values is proportional to some other voltage or current in the circuit

Classification of Dependent Sources

 _{Voltage-controlled voltage source: The source delivers the voltage as}

per the voltage of the dependent element

 _{Voltage-controlled current source: The source delivers the current as}

per the voltage of the dependent element

 _{Current-controlled current source: The source delivers the current}

as per the current of the dependent element

 _{Current-controlled voltage source: The source delivers the voltage as}

per the current of the dependent element.

Independent Sources

Thevenin Theorem

 _{Any linear electrical network with voltage and current sources}

and resistances can be replaced at terminals A-B by an equivalent voltage source V_th in series connection with an equivalent resistance R_th.

 This equivalent voltage V_th is the voltage obtained at terminals A-B of

the network with terminals A-B open circuited.

 This equivalent resistance R_th is the resistance obtained at terminals

A-B of the network with all its independent current sources open circuited and all its independent voltage sources short circuited.

Problem 12

Problem 13: Use Thevenin Theorem to find I

Problem 13 (Continued)



Vth = -2V



Rth = 2 ohm



I

= -2/5 A

Problem 14

Find V_X by using Thevenin Theorem.

1 2



4



6



2



V

3 0 V

_

+

_

A

B





1 2  4 

6  3 0 V +_

A B  

(30)(6)

10

6 12

V





V



Notice that there is no current flowing in the 4  resistor (A-B) is open. Thus there can be no voltage across the resistor.

We now deactivate the sources to the left of A-B and find the resistance seen looking in these terminals.

1 2



4



6



A

B





R_TH

R_TH = 12||6 + 4 = 8 

8 

1 0 V V _{T H}

R _{T H}

2  V X

+ _ + _ A B  

10 2

2

2 8







( )( )

V

V

For the circuit below, find V_AB using Thevenin Theorem

+_ 2 0 V

5 

2 0 

1 0 

1 7  1 . 5 A

A

B





Problem 15

+_

2 0 V

5 

2 0 

1 0  1 . 5 A

A B  

20(20)

(1.5)(10)

(20 5)

31

OS AB TH

TH

V

V

V

V

V















5 

2 0 

1 0  A

B  

5(20)

10

14

(5 20)

R











1 4 

3 1 V V _{T H}

R _{T H}

1 7  V A B

+ _ + _ A B  

17

V



V

Problem 16

Find the voltage across the 100  load resistor by usingThevenin Theorem.

+_ 8 6 V

5 0 

3 0 

4 0 

1 0 0 

6 I _S I _S





A

B

First remove the 100  load resistor and find V_AB = V_TH to the left of terminals A-B.

+_ 8 6 V

5 0 

3 0 

4 0 

6 I _S I_S





A

B

86 80

6

0

1

6

30

36

S S S

AB S S

I

I

I

A

V

I

I

V

















 

To find R_TH we deactivate all independent sources but retain all dependent sources

5 0 

3 0 

4 0 

6 I _S I_S





A

B R _{T H}

5 0 

3 0 

4 0 

6 I_S I _S

1 A 1 A

I_S + 1 V

50

I



30(

I

 

1) 6

I



0

15

43

S

I





A

5 0 

3 0 

4 0 

6 I_S I_S

1 A = I 1 A

IS + 1 V

15

50 1(40) 0

43 V   _{   }    

57.4

V



volts

57.4

1

V

V

R

I









The Thevenin equivalent circuit tied to the 100  load resistor is shown below.

+

_

R _{T H}

V _{T H}

5 7 . 4 

3 6 V _{1 0 0 }

100

36 100

22.9

57.4

100

x

V





V



Norton’s Theorem

 _{Any linear electrical network with voltage and current sources and}

only resistances can be replaced at terminals A-B by an equivalent current source I_NO in parallel connection with an equivalent resistance R_NO.

 This equivalent current I_NO is the current obtained at terminals A-B of

the network with terminals A-B short circuited.

 This equivalent resistance R_NO is the resistance obtained at terminals

A-B of the network with all its voltage sources short circuited and all its current sources open circuited.

Norton’s Equivalent Circuit

Problem 17

Problem 18

Find the Norton equivalent circuit to the left of terminals A-B for the network shown below. Connect the Norton equivalent circuit to the load and find the current in the 50  resistor.

+_{_}

2 0 

6 0  4 0 

5 0  1 0 A

5 0 V



 A

B

+_{_}

2 0 

6 0 

4 0  1 0 A

5 0 V

I_{S S}

It can be shown by standard circuit analysis that

10.7

SS

I



A

It can also be shown that by deactivating the sources, We find the resistance looking into terminals A-B is

55

R





R_N and R_TH will always be the same value for a given circuit. The Norton equivalent circuit tied to the load is shown below.

1 0 . 7 A _{5 5  5 0 }

Conversion to Thevenin Equivalent

Maximum Power Transfer Theorem

 _{The maximum power transfer theorem states that, to}

obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

 _{It is also referred to as "Jacobi's law"}

 _{The theorem can be extended to AC circuits that include reactance,}

and states that maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance.

53

Problem 19

Find the value of R

for maximum power

transfer to R

.

54

150

V = (360) = 300V

Th

180

(150)(30)

R = = 25Ω

Th

150 + 30

55

R

= 25Ω

L

56

Calculate the maximum power that can be

delivered to R

.

 

300 2

p = i R = (25)

L

50

p = 900W

G D Goenka University

Superposition Theorem

 _{For a linear system the response (Voltage or Current) in any branch of}

a circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are replaced by their internal impedances.

 _{Replacing all other independent voltage sources with a short}

circuit (thereby eliminating difference of potential. i.e. V=0, internal impedance of ideal voltage source is ZERO (short circuit)).

 _{Replacing all other independent current sources with an open}

circuit (thereby eliminating current. i.e. I=0, internal impedance of ideal current source is infinite (open circuit).

Problem 21



Find current in branches:

Solution

Problem 22



Find voltage V and current I

Solution