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Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 08/06/22 1

Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

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Page 1: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Chapter 15 – Series & Parallel ac Circuits

Chapter 16 – Series–Parallel ac Networks

Lecture 22

by Moeen Ghiyas

19/04/23 1

Page 2: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Chapter 15 – Series & Parallel ac Circuits

Page 3: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Chapter 15 - Series & Parallel ac Circuits

Equivalent Circuits

Chapter 16 - Series-Parallel ac Networks

Reduction Methods

Ladder Networks

Assignment # 4 - Submission by 10:30 am 23 Apr

19/04/23 3

Page 4: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

The term equivalent refers only to the fact that for the same

applied potential, the same impedance and input current will

result (in the equivalent circuit).

Whether a series or parallel ac circuit, the total impedance of

two or more elements in series is often equivalent to an

impedance that can be achieved with fewer elements of

different values

However, the equivalent elements and their values are

determined by the frequency applied.

Page 5: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

For example, for circuit of fig

The total impedance at the frequency

applied is equivalent to a capacitor

with a reactance of 10Ω , as shown

Page 6: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Always keep in mind that this equivalence is true only at

the applied frequency. If the frequency changes, the

reactance of each element changes, and the equivalent

circuit will change — perhaps from capacitive to inductive

in the above example

Page 7: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Another interesting example,

which is the impedance of a

series circuit with a resistor of

1.92Ω and an inductive

reactance of 1.44Ω , as shown

Page 8: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Current I will be same in equivalent circuit for same input voltage E

For a parallel circuit of one resistive element and one reactive

element, the series circuit with the same input impedance will

always be composed of one resistive and one reactive element.

The impedance of each element of the series circuit will be different

from that of the parallel circuit, but the reactive elements will always

be of the same type; i.e., an R-L circuit and an R-C parallel circuit

will have an equivalent R-L and R-C series circuit respectively

The same is true when converting from a series to a parallel circuit.

Page 9: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

The equivalent series circuit for a resistor and reactance in parallel

can be found by determining total impedance in rectangular form;

The equivalent parallel circuit for a resistor and reactance in series

can be found by determining total admittance in rectangular form;

See proof in book

Page 10: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine YT.

Page 11: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine YT. - Network is redrawn with phasor notation

Page 12: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine YT. - The admittance Y is

Page 13: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Sketch the admittance

diagram.

Page 14: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Find E and IL.

Page 15: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Compute the power factor of the network and the power delivered.

Page 16: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine the equivalent series circuit as far as the terminal

characteristics of the network are concerned.

Page 17: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine the equivalent series circuit as far as the terminal

characteristics of the network are concerned.

Page 18: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine the equivalent parallel network from the equivalent

series circuit, and calculate the total admittance YT

Page 19: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determine YT for the equivalent parallel circuit.

Page 20: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Chapter 16 – Series–Parallel ac Networks

Page 21: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Reduce the network to the fundamental structure preferably

towards source to determine the total impedance of the network

and redraw network by combining series and parallel elements.

The source current and voltages can then be determined.

Later work back (Expand) from the source through the network to

find specific quantities.

When you have arrived at a solution, check to see that it is

reasonable by considering the magnitudes. If not, either solve the

network using another approach, or check over your work very

carefully

Page 22: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

EXAMPLE - For Fig :

a. Calculate the current Is

b. Find the voltage Vab

Solution

Simplify the circuit and redraw

In this case the voltage Vab is lost in

the redrawn network, which will be

worked backwards later

Page 23: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Now we know Z1 = 5Ω /53.130

and Z2 = 10Ω / -36.870

Determine ZT

Calculate I or IS

Page 24: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Now we know Z1 = 5Ω /53.130

and Z2 = 10Ω / -36.870

and IS = 22.36 A /-26.560

Determine branch currents using ohms law

Page 25: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Now I1 and I2 known

Working backwards to original cct

Page 26: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

To find Vab apply KVL,

Page 27: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

EXAMPLE - For the network of Fig

a. Compute I.

b. Find I1, I2, and I3.

c. Verify KCL by showing that I = I1 + I2 + I3

d. Find the total impedance of the circuit.

Page 28: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

EXAMPLE - For the network of Fig

Redrawing the circuit reveals a parallel circuit

Page 29: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Calculate Impedances to determine currents

Page 30: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

The total admittance is

The current I becomes

Page 31: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Since the voltage is same across parallel branches

Page 32: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

c. Verify by KCL

d. Find the total impedance of the circuit

Page 33: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

A general sinusoidal ac ladder network is as shown. The current I6 is

desired.

ac ladder network with ZT, ZT’, ZT” and currents I, I3 defined.

Page 34: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Determining impedances and then working backwards calculating

currents to finally know the current I6 as desired.

Page 35: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Ch 15 - Q. 17, 23, 31

Ch 16 - Q. 11, 13

Submission by 09:00 am 23 Apr 2012

19/04/23 35

Page 36: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

Chapter 15 - Series & Parallel ac Circuits

Equivalent Circuits

Chapter 16 - Series-Parallel ac Networks

Reduction Methods

Ladder Networks

Assignment # 4 - Submission by 10:30 am 23 Apr

Page 37: Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 21/08/2015 1

19/04/23 37