Chat with us, powered by LiveChat Use the Delta-Wye transformation to find the resistance as seen at ?the input. Label each Delta/Wye you identify on the circuit diagram and ?discuss your approach an - Essayabode

Use the Delta-Wye transformation to find the resistance as seen at ?the input. Label each Delta/Wye you identify on the circuit diagram and ?discuss your approach an

 

  • Use the Delta-Wye transformation to find the resistance as seen at  the input. Label each Delta/Wye you identify on the circuit diagram and  discuss your approach and your reasoning for the solution.
  • Why do industrial plants use power factor correction? Here is a good resource to read. Find other examples and list your resources.

Delta to wye transformation

Eet310 circuit analysis

∆-to-Y and Y-to-∆ conversion (1 of 2)

The ∆-to-Y and Y-to-∆ conversion formulas allow a three terminal resistive network to be replaced with an equivalent network.

For the ∆-to-Y conversion, each resistor in the Y is equal to the product of the resistors in the two adjacent ∆ branches divided by the sum of all three ∆ resistors.

For example,

2

∆-to-Y and Y-to-∆ conversion (2 of 2)

The ∆-to-Y and Y-to-∆ conversion formulas allow a three terminal resistive network to be replaced with an equivalent network.

For the Y-to-∆ conversion, each resistor in the ∆ is equal to the sum of all products of Y resistors, taken two at a time divided by the opposite Y resistor.

For example,

3

Wye – Delta Conversions

4

Wye – Delta Conversions

5

Wye – Delta Conversions

Determine the total resistance of the circuit shown below.

6

Wye – Delta Conversions

Convert the delta configuration to wye

7

Wye – Delta Conversions

Determine the equivalent resistance of the following circuit

8

Wye – Delta Conversions

Determine Rab for the following circuit

9

Wye – Delta Conversions

Determine the total resistance Rab for the circuit shown below. All the resistors are equal

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Analysis of ac circuits

EET310 Circuit analysis

Objective of the Lecture

After successful completion of the lecture you will be able to:

Determine the reactance of both a capacitor and an inductor.

Analyze an AC circuit using either NVA or LCA.

Determine the frequency response an AC circuit

2

Introduction

3

A large

complex AC

circuits

Simplify

circuit analysis

Circuit Theorems

‧Thevenin’s theorem ‧ Norton theorem

‧Circuit linearity ‧ Superposition

‧source transformation ‧ max. power transfer

Superposition theorem

The superposition theorem that you studied in dc circuits can be applied to ac circuits by using complex numbers. Recall that it is applied to circuits with multiple independent sources to solve for the current in any element.

In a linear circuit with multiple independent sources, the current in any element is the algebraic sum of the currents produced by each source acting alone.

One way to summarize the superposition theorem is

4

Summary

Superposition theorem

Leave one of the sources in the circuit, and replace all others with their internal impedance. For ideal voltage sources, the internal impedance is zero. For ideal current sources, the internal impedance is infinite. We will call this procedure zeroing the source.

Find the current in the branch of interest produced by the one remaining source.

Repeat Steps 1 and 2 for each source in turn. When complete, you will have a number of current values equal to the number of sources in the circuit.

Add the individual current values as phasor quantities.

Steps for applying the superposition theorem:

5

Superposition theorem

Steps from the text are applied in the following example. To simplify this example, reactances are given for the capacitors.

What is the current in C2? The internal source resistances are zero.

Example

-j10 kW

-j20 kW

6.8 kW

C1

C2

R

VS1

VS2

Step 1.

Replace VS2 with its internal impedance (zero) and find IC2 due to VS1 acting alone.

-j10 kW

-j20 kW

6.8 kW

C1

C2

R

VS1

VS2

zeroed

Note: To simplify math equations, units are not shown on the solution. All impedances are in kW, currents in mA, and voltages in V.

continued…

6

Superposition theorem

Step 1. (continued)

-j10 kW

-j20 kW

6.8 kW

C1

C2

R

VS1

VS2

zeroed

Looking from VS1, the total impedance (in kW) is

The total current from VS1 is

Apply current divider to find IC2 due to VS1:

continued…

7

Superposition theorem

Step 2.

-j10 kW

-j20 kW

6.8 kW

C1

C2

R

VS1

zeroed

Find IC2 due to VS2 by zeroing VS1. The total impedance (in kW) is

The total current from VS2 is

Notice that IS2 = IC2 for this source and is opposite to IC2 due to VS1.

continued…

VS2

8

Superposition theorem

Step 3.

Find IC2 due to both sources by combining the results from steps 1 and 2.

In polar form, the total current in IC2 is

-j10 kW

-j20 kW

6.8 kW

C1

C2

R

VS1

VS2

IC2 due to VS1 is

IC2 due to VS2 is

IC2 due to both sources is

Subtract because currents oppose.

The result is in the 3rd quadrant, meaning that the actual direction is reversed from the assumed positive direction. It is due mainly to VS2.

Problems like this example can be easily solved by Multisim, a popular computer simulation. The Multisim circuit is shown on the following slide.

9

Superposition theorem

Voltage sources are specified in Multisim as peak voltages.

10

Superposition Theorem

Using the superposition theorem, find the current I through the 4ohm reactance (XL2)

11

Superposition Theorem

12

=

+

Determine Io using superposition theorem

Superposition Theorem

13

Superposition Theorem

Determine Vo using superposition theorem

14

Superposition Theorem

Due to the 5 V DC source

15

Superposition Theorem

Due to the 10 V Peak Voltage source

16

Superposition Theorem

Due to the AC current source

17

Source Transformation

18

Source Transformation

Find Vx by applying source transformation

19

Thevenin’s theorem

Thevenin’s theorem can be applied to ac circuits. As applied to ac circuits, Thevenin’s theorem can be stated as:

The Thevenin equivalent circuit is:

Vth

Zth

Any two-terminal linear ac circuit can be reduced to an equivalent circuit that consists of an ac voltage source in series with an equivalent impedance.

20

Thevenin’s theorem

Vth

Zth

ac circuit

Equivalency means that when the same value of load is connected to both the original circuit and Thevenin’s equivalent circuit, the load voltages and currents are the same for both.

R

R

For example, if the same R is connected to both circuits, the voltage across R will be identical for each circuit.

21

Thevenin’s theorem

Vth

Zth

Thevenin’s equivalent voltage (Vth) is the open-circuit voltage between two specified terminals in a circuit.

With load removed, this is the Thevenin voltage.

ac circuit

With load removed, this is the Thevenin voltage.

RL

RL

22

Thevenin’s theorem

Thevenin’s equivalent impedance (Zth) is the total impedance appearing between two specified terminals in a given circuit with all sources replaced by their internal impedances.

Vth

Zth

With sources replaced with their internal impedance, you could measure Zth here.

ac circuit

With sources replaced with their internal impedance, you could measure Zth here.

Replace with internal resistance

23

Thevenin’s theorem

Steps for applying Thevenin’s theorem:

Open the two terminals between which you want to find the Thevenin circuit. This is done by removing the component from which the circuit is to be viewed.

Determine the voltage across the two open terminals.

Determine the impedance viewed from the two open terminals with ideal voltage sources replaced with shorts and ideal current sources replaced with opens.

Connect Vth and Zth in series to produce the complete Thevenin equivalent circuit.

24

Norton’s theorem

Norton’s theorem is also an equivalent circuit that provides a way to reduce complicated circuits to a simpler form. As applied to ac circuits, Norton’s theorem can be stated as:

The Norton equivalent circuit is:

Any two-terminal linear ac circuit can be reduced to an equivalent circuit that consists of an ac current source in parallel with an equivalent impedance.

In

Zn

25

Norton’s theorem

With the load replaced with a short, the current in the short is Norton’s current.

ac circuit

Norton’s equivalent current (In) is the short-circuit current (ISL) between two specified terminals in a circuit.

In

Zn

With the load replaced with a short, the current in the short is Norton’s current.

RL

RL

26

Norton’s theorem

Norton’s equivalent impedance (Zn) is the total impedance appearing between two specified terminals in a given circuit with all sources replaced by their internal impedances.

An ideal current source is replaced with an open circuit.

Notice that Zn = Zth.

With sources replaced with their internal impedance, you could measure Zth here.

ac circuit

With sources replaced with their internal impedance, you could measure Zth here.

In

Zn

27

Norton’s theorem

Steps for applying Norton’s theorem:

Replace the load connected to the two terminals between which the Norton circuit is to be determined with a short.

Determine the current through the short. This is In.

Open the terminals and determine the impedance between the two open terminals with all sources replaced with their internal impedances. This is Zn.

Connect In and Zn in parallel.

28

Maximum power transfer theorem

Recall that in resistive circuits, maximum power is transferred from a given source to a load if RL = RS.

Impedance consists of a resistive and reactive part. To transfer maximum power in reactive circuits, the load resistance should still match the source resistance but it should cancel the source reactance by matching the opposite reactance as illustrated.

RS

RL

RS ± jX

29

Maximum power transfer theorem

The impedance that cancels the reactive portion of the source impedance is called the complex conjugate. The complex conjugate of R – jXC is R + jXL. Notice that each is the complex conjugate of the other.

RS ± jX

In effect the reactance portion of the load forms a series resonant circuit with the reactive portion of the source at the resonant frequency. Thus the matching is frequency dependent.

30

Thevenin and Norton Theorems

31

Thevenin and Norton Theorems

Determine the Thevenin voltage and resistance in terminal ab

32

Thevenin and Norton Theorems

Determine the Thevenin equivalent circuit

33

Thevenin and Norton Theorems

34

Thevenin and Norton Theorems

Determine Io using Norton Theorem

35

Thevenin and Norton Theorems

36

Nodal Analysis in AC Circuits

Find ix in the circuit shown below

37

Nodal Analysis in AC Circuits

38

Nodal Analysis in AC Circuits

Determine the voltages V1 and V2 in the circuit below

39

Applying KCL for the supernode gives 1 equations.

Applying KVL at the supernode gives 1 equations.

2 variables solved by 2 equations.

Nodal Analysis in AC Circuits

40

Mesh Analysis in AC Circuits

Using Mesh current analysis determine Io

41

Mesh Analysis in AC Circuits

Using Mesh current analysis determine Vo

42

Applying KVL for mesh 1 & 2 gives 2 equations.

Applying KVL for the supermesh gives 1 equations.

Applying KCL at node A gives 1 equations.

4 variables solved by 4 equations

Mesh Analysis in AC Circuits

43

Superposition theorem

Equivalent circuit

Thevenin’s theorem

A method for the analysis of circuits with more than one source.

A circuit that produces the same voltage and current to a given load as the original circuit that it replaces.

Key Terms

A method for simplifying a two-terminal circuit to an equivalent circuit with only a voltage source in series with an impedance.

44

Norton’s theorem

Complex conjugate

A method for simplifying a two-terminal circuit to an equivalent circuit with only a current source in parallel with an impedance.

A complex number having the same real part and oppositely signed imaginary part; an impedance containing the same resistance and a reactance opposite in phase but equal in magnitude to that of a given impedance.

Key Terms

45

Quiz

Compared to the original circuit that it replaces, an equivalent circuit will always have

a. the same power dissipation.

b. the same number of components

c. the same voltage and current for a specific load

d. all of the above

46

Quiz

2. To apply the superposition theorem, all sources but the one of interest are replaced with

a. a short

b. an open

c. their internal impedance

d. an impedance equal to the load

47

Quiz

3. Compared to the Thevenin impedance for a given circuit, the Norton impedance is

a. the same

b. smaller

c. larger

d. equal to the complex conjugate

48

Quiz

4. If you put a short on a Thevenin circuit, the current in the short is

a. zero

b. smaller than the Norton current

c. larger than the Norton current

d. the same as the Norton current

49

Quiz

5. A Thevenin circuit is a

a. voltage source in parallel with an impedance

b. voltage source in series with an impedance

c. current source in parallel with an impedance

d. current source in parallel with an impedance

50

Quiz

6. A Norton circuit is a

a. voltage source in parallel with an impedance

b. voltage source in series with an impedance

c. current source in parallel with an impedance

d. current source in parallel with an impedance

51

Quiz

A theorem which is algebraically adds the current from several sources acting independently to find the total current is

a. superposition theorem

b. Thevenin’s theorem

c. Norton’s theorem

d. maximum power theorem

52

Quiz

8. The Thevenin equivalent voltage will appear on the output terminals of a circuit if the load impedance is

a. removed

b. equal to the source impedance

c. the complex conjugate of the source impedance

d. a pure resistance

53

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