Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 27/02/

Similar presentations


Presentation on theme: "Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 27/02/"— Presentation transcript:

1 Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 27/02/2021 1

2 Chapter 15 – Series & Parallel ac Circuits

3 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 27/02/2021 3

4 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.

5 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

6 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

7 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

8 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.

9 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

10 Determine Y T.

11 Determine Y T.-Network is redrawn with phasor notation

12 Determine Y T.-The admittance Y is

13 Sketch the admittance diagram.

14 Find E and I L.

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

16 Determine the equivalent series circuit as far as the terminal characteristics of the network are concerned.

17

18 Determine the equivalent parallel network from the equivalent series circuit, and calculate the total admittance Y T

19 Determine Y T for the equivalent parallel circuit.

20 Chapter 16 – Series–Parallel ac Networks

21 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

22 EXAMPLE - For Fig : a. Calculate the current I s b. Find the voltage V ab Solution Simplify the circuit and redraw In this case the voltage V ab is lost in the redrawn network, which will be worked backwards later

23 Now we know Z 1 = 5Ω /53.13 0 and Z 2 = 10Ω / -36.87 0 Determine Z T Calculate I or I S

24 Now we know Z 1 = 5Ω /53.13 0 and Z 2 = 10Ω / -36.87 0 and I S = 22.36 A /-26.56 0 Determine branch currents using ohms law

25 Now I 1 and I 2 known Working backwards to original cct

26 To find V ab apply KVL,

27 EXAMPLE - For the network of Fig a. Compute I. b. Find I 1, I 2, and I 3. c. Verify KCL by showing that I = I 1 + I 2 + I 3 d. Find the total impedance of the circuit.

28 EXAMPLE - For the network of Fig Redrawing the circuit reveals a parallel circuit

29 Calculate Impedances to determine currents

30 The total admittance is The current I becomes

31 Since the voltage is same across parallel branches

32 c.Verify by KCL d.Find the total impedance of the circuit

33 A general sinusoidal ac ladder network is as shown. The current I 6 is desired. ac ladder network with Z T, Z T ’, Z T ” and currents I, I 3 defined.

34 Determining impedances and then working backwards calculating currents to finally know the current I 6 as desired.

35 Ch 15-Q. 17, 23, 31 Ch 16-Q. 11, 13 Submission by 09:00 am 23 Apr 2012 27/02/2021 35

36 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

37 27/02/2021 37


Download ppt "Chapter 15 – Series & Parallel ac Circuits Chapter 16 – Series–Parallel ac Networks Lecture 22 by Moeen Ghiyas 27/02/"

Similar presentations


Ads by Google