Sinusoidal Steady State Analysis (AC Analysis) Part I Sinusoidal Steady State Analysis (AC Analysis)...

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Sinusoidal Steady State Analysis

(AC Analysis)

Part I

Dr. Mohamed Refky Amin

Electronics and Electrical Communications Engineering Department (EECE)

Cairo University

elc.n102.eng@gmail.com

http://scholar.cu.edu.eg/refky/

OUTLINE

โ€ข Previously on ELCN102

โ€ข Solution of AC Circuits

Simplification Method

Loop Analysis Method

Node Analysis Method

Superposition Method

Dr. Mohamed Refky 2

Previously on ELCN102

Dr. Mohamed Refky

Phasor Relationships for Circuit Elements

The impedance ๐‘ of a circuit is the ratio of the phasor voltage ๐‘‰to the phasor current ๐ผ, measured in ฮฉ.

Resistor Inductor Capacitor

๐‘ฃ๐‘… ๐‘ก = ๐‘…๐‘–๐‘… ๐‘ก ๐‘ฃ๐ฟ ๐‘ก = ๐ฟ๐‘‘๐‘–๐ฟ ๐‘ก

๐‘‘๐‘ก๐‘–๐ถ ๐‘ก = ๐ถ

๐‘‘๐‘ฃ๐ถ ๐‘ก

๐‘‘๐‘ก

๐‘‰๐‘… = ๐‘… ร— ๐ผ๐‘…๐‘‰๐ฟ = ๐œ”๐ฟ๐ผ๐ฟโˆ 90

๐‘œ

= ๐‘—๐œ”๐ฟ ร— ๐ผ๐ฟ

๐ผ๐ถ = ๐œ”๐ถ๐‘‰๐ถโˆ 90๐‘œ

= ๐‘—๐œ”๐ถ ร— ๐‘‰๐ถ

๐‘๐‘… = ๐‘… ๐‘๐ฟ = ๐‘—๐œ”L ๐‘๐ถ =1

๐‘—๐œ”๐ถ= โˆ’

๐‘—

๐œ”๐ถ

Previously on ELCN102

Dr. Mohamed Refky

Phasor Relationships for Circuit Elements

The admittance ๐‘Œ of a circuit is the ratio of the phasor current ๐ผto the phasor voltage ๐‘‰, measured in ฮฉโˆ’1.

Resistor Inductor Capacitor

๐‘ฃ๐‘… ๐‘ก = ๐‘…๐‘–๐‘… ๐‘ก ๐‘ฃ๐ฟ ๐‘ก = ๐ฟ๐‘‘๐‘–๐ฟ ๐‘ก

๐‘‘๐‘ก๐‘–๐ถ ๐‘ก = ๐ถ

๐‘‘๐‘ฃ๐ถ ๐‘ก

๐‘‘๐‘ก

๐‘‰๐‘… = ๐‘… ร— ๐ผ๐‘…๐‘‰๐ฟ = ๐œ”๐ฟ๐ผ๐ฟโˆ 90

๐‘œ

= ๐‘—๐œ”๐ฟ ร— ๐ผ๐ฟ

๐ผ๐ถ = ๐œ”๐ถ๐‘‰๐ถโˆ 90๐‘œ

= ๐‘—๐œ”๐ถ ร— ๐‘‰๐ถ

๐‘Œ๐‘… =1

๐‘…๐‘Œ๐ฟ =

1

๐‘—๐œ”L= โˆ’

๐‘—

๐œ”L๐‘Œ๐ถ = ๐‘—๐œ”๐ถ

Previously on ELCN102

Dr. Mohamed Refky

Impedance and Admittance

The impedance ๐‘ of a circuit is the ratio of the phasor voltage ๐‘‰to the phasor current ๐ผ, measured in ฮฉ.

๐‘ = ๐‘… + ๐‘—๐‘‹

๐‘… is the resistance & ๐‘‹ is the reactance

๐‘ is inductive if ๐‘‹ is +๐‘ฃ๐‘’.

๐‘ is capacitive if ๐‘‹ is โˆ’๐‘ฃ๐‘’.

๐‘, ๐‘…, and ๐‘‹ are in units of ฮฉ

Impedance

๐‘๐ฟ = ๐‘—๐œ”L

๐‘๐ถ =1

๐‘—๐œ”๐ถ= โˆ’

๐‘—

๐œ”๐ถ

Previously on ELCN102

Dr. Mohamed Refky

Impedance and Admittance

The admittance ๐‘Œ of a circuit is the ratio of the phasor current ๐ผ to

the phasor voltage ๐‘‰, measured in ฮฉโˆ’1.

๐‘Œ = ๐บ + ๐‘—๐ต

๐บ is the conductance & ๐ต is the susceptance.

๐‘Œ is inductive if ๐ต is โˆ’๐‘ฃ๐‘’.

๐‘Œ is capacitive if ๐ต is +๐‘ฃ๐‘’.

๐‘Œ, ๐บ, and ๐ต are in units of ฮฉโˆ’1

Admittance

๐‘Œ๐ฟ =1

๐‘—๐œ”L= โˆ’

๐‘—

๐œ”L

๐‘Œ๐ถ = ๐‘—๐œ”๐ถ

Previously on ELCN102

Dr. Mohamed Refky

Impedance Combination

๐‘๐‘’๐‘ž = ๐‘1 + ๐‘2 +โ‹ฏ+ ๐‘๐‘

Series Combination

Previously on ELCN102

Dr. Mohamed Refky

Impedance Combination

1

๐‘๐‘’๐‘ž=

1

๐‘1+

1

๐‘2+โ‹ฏ+

1

๐‘๐‘

Parallel Combination

Previously on ELCN102

Dr. Mohamed Refky

Admittance Combination

1

๐‘Œ๐‘’๐‘ž=

1

๐‘Œ1+1

๐‘Œ2+โ‹ฏ+

1

๐‘Œ๐‘

Series Combination

Previously on ELCN102

Dr. Mohamed Refky

Admittance Combination

๐‘Œ๐‘’๐‘ž = ๐‘Œ1 + ๐‘Œ2 +โ‹ฏ+ ๐‘Œ๐‘

Parallel Combination

Previously on ELCN102

Dr. Mohamed Refky

Star-Delta Transformation

๐‘๐ด๐ต = ๐‘๐ด + ๐‘๐ต +๐‘๐ด๐‘๐ต๐‘๐ถ

๐‘๐ด๐ถ = ๐‘๐ด + ๐‘๐ถ +๐‘๐ด๐‘๐ถ๐‘๐ต

๐‘๐ต๐ถ = ๐‘๐ต + ๐‘๐ถ +๐‘๐ต๐‘๐ถ๐‘๐ด

๐‘๐ด =๐‘๐ด๐ต๐‘๐ด๐ถ

๐‘๐ด๐ถ + ๐‘๐ต๐ถ + ๐‘๐ด๐ต๐‘๐ถ =

๐‘๐ต๐ถ๐‘๐ด๐ถ๐‘๐ด๐ถ + ๐‘๐ต๐ถ + ๐‘๐ด๐ต

๐‘๐ต =๐‘๐ด๐ต๐‘๐ต๐ถ

๐‘๐ด๐ถ + ๐‘๐ต๐ถ + ๐‘๐ด๐ต

Solution of AC Circuits

Dr. Mohamed Refky

DefinitionA circuit is said to be solved when the voltage across and the

current in every element have been determined due to input

excitation (voltage and/or current sources).

Solution of AC Circuits

Dr. Mohamed Refky

Methods of Solution of AC CircuitsTo solve a AC circuit you can use one or more of the following

methods:

โ€ข Simplification Method

โ€ข Loop Analysis Method

โ€ข Node Analysis Method

โ€ข Superposition Method

โ€ข Thevenin equivalent circuit

โ€ข Norton equivalent circuit

Solution of AC Circuits

Dr. Mohamed Refky

Simplification Method In step by step simplification we can use:

โ€ข Source transformation

โ€ข Combination of active elements

โ€ข Combination of series and parallel elements

โ€ข Star-delta & delta-star transformation

Simplification Method

Dr. Mohamed Refky

Source Transformationโ€œA voltage source ๐‘‰๐ด๐ถ with a series impedance ๐‘ can be

transformed into a current source ๐ผ๐ด๐ถ = ๐‘‰๐ด๐ถ/๐‘ and a parallel

impedance ๐‘โ€

โ€œ A current source ๐ผ๐ด๐ถ with a parallel impedance ๐‘ can be

transformed into a voltage source ๐‘‰๐ด๐ถ = ๐ผ๐ด๐ถ ร— ๐‘ and a series

impedance ๐‘โ€

Simplification Method

Dr. Mohamed Refky

Example (1)Use simplification method to find ๐‘‰๐‘ฅ for the circuit shown.

Simplification Method

Dr. Mohamed Refky

Example (2)Use simplification method to find ๐ผ๐‘ฅ for the circuit shown.

Loop Analysis Method

Dr. Mohamed Refky

Definition

The Loop Analysis Method (Mesh Method) uses KVL to generate

a set of simultaneous equations.

1) Convert the independent current sources into equivalent

voltage sources

2) Identify the number of independent loop (๐ฟ) on the circuit

3) Label a loop current on each loop.

4) Write an expression for the KVL around each loop.

5) Solve the simultaneous equations to get the loop currents.

Loop Analysis Method

Dr. Mohamed Refky

Matrix Form

๐‘11 โˆ’๐‘12 โ‹ฏ โˆ’๐‘1๐‘โˆ’๐‘21 ๐‘22 โˆ’๐‘2๐‘โ‹ฎ

โˆ’๐‘๐‘1

โ‹ฎโˆ’๐‘๐‘2

โ‹ฑ โ‹ฎโ‹ฏ ๐‘๐‘๐‘

๐ผ1๐ผ2โ‹ฎ๐ผ๐‘

=

๐‘‰1๐‘‰2โ‹ฎ๐‘‰๐‘

๐‘๐‘–๐‘– =๐‘–๐‘š๐‘๐‘’๐‘‘๐‘Ž๐‘›๐‘๐‘’ ๐‘–๐‘› ๐‘™๐‘œ๐‘œ๐‘ ๐‘–

๐‘๐‘–๐‘— =๐ถ๐‘œ๐‘š๐‘š๐‘œ๐‘› ๐‘–๐‘š๐‘๐‘’๐‘‘๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘™๐‘œ๐‘œ๐‘๐‘  ๐‘– ๐‘Ž๐‘›๐‘‘ ๐‘— = ๐‘๐‘—๐‘–

๐‘‰๐‘– =๐‘ฃ๐‘œ๐‘™๐‘ก๐‘Ž๐‘”๐‘’ ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’๐‘  ๐‘–๐‘› ๐‘™๐‘œ๐‘œ๐‘ ๐‘–๐‘‰ is +ve if it supplies

current in the direction

of the loop current

Loop Analysis Method

Dr. Mohamed Refky

Example (3)Use loop analysis to find ๐ผ๐‘ฅ for the circuit shown.

Loop Analysis Method

Dr. Mohamed Refky

Example (4)Use loop analysis to find ๐ผ๐‘ฅ for the circuit shown.

Loop Analysis Method

Dr. Mohamed Refky

Example (5)Use loop analysis to find ๐‘‰๐‘ฅ for the circuit shown.

Node Analysis Method

Dr. Mohamed Refky 23

Definition

The Node Analysis Method (Nodal Analysis) uses KCL to

generate a set of simultaneous equations.

1) Convert independent voltage sources into equivalent current

sources.

2) Identify the number of non simple nodes (๐‘) of the circuit.

3) Write an expression for the KCL at each ๐‘ โˆ’ 1 Node

(exclude the ground node).

4) Solve the resultant simultaneous equations to get the voltages.

Node Analysis Method

Dr. Mohamed Refky

Matrix Form

๐‘Œ11 โˆ’๐‘Œ12 โ‹ฏ โˆ’๐‘Œ1๐‘โˆ’๐‘Œ21 ๐‘Œ22 โˆ’๐‘Œ2๐‘โ‹ฎ

โˆ’๐‘Œ๐‘1

โ‹ฎโˆ’๐‘Œ๐‘2

โ‹ฑ โ‹ฎโ‹ฏ ๐‘Œ๐‘๐‘

๐‘‰1๐‘‰2โ‹ฎ๐‘‰๐‘

=

๐ผ1๐ผ2โ‹ฎ๐ผ๐‘

๐‘Œ๐‘–๐‘– =๐‘Ž๐‘‘๐‘š๐‘–๐‘ก๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘œ๐‘“ ๐‘›๐‘œ๐‘‘๐‘’ ๐‘–

๐‘Œ๐‘–๐‘— =๐‘๐‘œ๐‘š๐‘š๐‘œ๐‘› ๐‘Ž๐‘‘๐‘š๐‘–๐‘ก๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘›๐‘œ๐‘‘๐‘’ ๐‘– ๐‘Ž๐‘›๐‘‘ ๐‘— = ๐‘Œ๐‘—๐‘–

๐ผ๐‘– =๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’๐‘  ๐‘Ž๐‘ก ๐‘›๐‘œ๐‘‘๐‘’ ๐‘– ๐ผ is +ve if it supply

current into the node

Node Analysis Method

Dr. Mohamed Refky

Example (6)Use node analysis to find ๐‘‰1 & ๐‘‰2 for the circuit shown.

Superposition Theorem

Dr. Mohamed Refky

DefinitionFor a linear circuit containing multiple independent sources, the

voltage across (or current through) any of its elements is the

algebraic sum of the voltages across (or currents through) that

element due to each independent source acting alone.

10โˆ 30๐‘œV ๐ผ๐‘Ž

5โˆ 0๐‘œA ๐ผ๐‘

Total ๐ผ = ๐ผ๐‘Ž + ๐ผ๐‘

Superposition Theorem

Dr. Mohamed Refky

Example (7)Use superposition theorem to find ๐ผ๐‘ฅ for the circuit shown.

Theveninโ€™s Theorem

Dr. Mohamed Refky

DefinitionA linear two-terminal circuit, can be replaced by an equivalent

circuit consisting of a voltage source ๐‘‰๐‘กโ„Ž in series with a

impedance ๐‘๐‘กโ„Ž.

Theveninโ€™s Theorem

Dr. Mohamed Refky

Solution Steps1) Identify the load impedance and introduce two nodes ๐‘Ž and ๐‘

2) Remove the load impedance between node ๐‘Ž and ๐‘

3) Calculate the open circuit voltage between nodes ๐‘Ž and ๐‘.This voltage is ๐‘‰๐‘กโ„Ž of the Thevenin equivalent circuit.

4) Set all the independent sources to zero (voltage sources are

SC and current sources are OC) and calculate the impedance

seen between nodes ๐‘Ž and ๐‘. This impedance is ๐‘๐‘กโ„Ž of the

Thevenin equivalent circuit.

Theveninโ€™s Theorem

Dr. Mohamed Refky

Example (8)Obtain the Thevenin equivalent at terminals ๐‘Ž and ๐‘ of the circuit

shown.

Nortonโ€™s Theorem

Dr. Mohamed Refky

DefinitionA linear two-terminal circuit can be replaced by equivalent circuit

consisting of a current source ๐ผ๐‘ in parallel with a impedance ๐‘๐‘

Nortonโ€™s Theorem

Dr. Mohamed Refky

Solution Steps1) Identify the load impedance and introduce two nodes ๐‘Ž and ๐‘

2) Remove the load impedance between node ๐‘Ž and ๐‘ and set all

the independent sources to zero (voltage sources are SC and

current sources are OC) and calculate the impedance seen

between nodes ๐‘Ž and ๐‘. This resistance is ๐‘๐‘ of the Norton

equivalent circuit.

3) Replace the load impedance with a short circuit and calculate

the short circuit current between nodes ๐‘Ž and ๐‘. This current

is ๐ผ๐‘ of the Norton equivalent circuit.

Nortonโ€™s Theorem

Dr. Mohamed Refky

Thevenin and Norton equivalent circuits

Thevenin equivalent circuit must be equivalent to Norton

equivalent circuit

๐‘๐‘ = ๐‘๐‘กโ„Ž, ๐‘‰๐‘กโ„Ž = ๐ผ๐‘๐‘๐‘, ๐ผ๐‘ =๐‘‰๐‘กโ„Ž๐‘๐‘กโ„Ž

โ†’ ๐‘๐‘กโ„Ž =๐‘‰๐‘กโ„Ž๐ผ๐‘

Theveninโ€™s Theorem

Dr. Mohamed Refky

Example (9)Obtain the Norton equivalent at terminals ๐‘Ž and ๐‘ of the circuit

shown.