Download presentation

Presentation is loading. Please wait.

Published byJoshua Wilkinson Modified over 4 years ago

1
Circuit Analysis

2
Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source. 2.Redraw the circuit with the equivalent resistance for the combination found in step 1. 3.Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance. 4.Solve for the currents and voltages in the final equivalent circuit.

4
Working Backward

5
Find current flowing each resistor

6
Voltage Division

7
Application of the Voltage- Division Principle

9
Current Division

11
Application of the Current-Division Principle

12
Voltage division Voltage division and current division

13
Current division

14
Although they are very important concepts, series/parallel equivalents and the current/voltage division principles are not sufficient to solve all circuits.

15
Mesh Current Analysis Sources

16
Definition of a loop Definition of a mesh

17
Choosing the Mesh Currents When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.

19
Writing Equations to Solve for Mesh Currents If a network contains only resistors and independent voltage sources, we can write the required equations by following each current around its mesh and applying KVL.

20
For mesh 1, we have For mesh 2, we obtain For mesh 3, we have

21
Determine the two mesh currents, i 1 and i 2, in the circuit below. For the left-hand mesh, -42 + 6 i 1 + 3 ( i 1 - i 2 ) = 0 For the right-hand mesh, 3 ( i 2 - i 1 ) + 4 i 2 - 10 = 0 Solving, we find that i 1 = 6 A and i 2 = 4 A. (The current flowing downward through the 3- resistor is therefore i 1 - i 2 = 2 A. )

22
Mesh Currents in Circuits Containing Current Sources *A common mistake is to assume the voltages across current sources are zero. Therefore, loop equation cannot be set up at mesh one due to the voltage across the current source is unknown Anyway, the problem is still solvable.

23
As the current source common to two mesh, combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined. Mesh 3: It is the supermesh. Three linear equations and three unknown

24
Find the three mesh currents in the circuit below. Creating a “supermesh” from meshes 1 and 3: -7 + 1 ( i 1 - i 2 ) + 3 ( i 3 - i 2 ) + 1 i 3 = 0 [1] Around mesh 2: 1 ( i 2 - i 1 ) + 2 i 2 + 3 ( i 2 - i 3 ) = 0[2] Rearranging, i 1 - 4 i 2 + 4 i 3 = 7[1] -i 1 + 6 i 2 - 3 i 3 = 0[2] i 1 - i 3 = 7[3] Solving, i 1 = 9 A, i 2 = 2.5 A, and i 3 = 2 A. Finally, we relate the currents in meshes 1 and 3: i 1 - i 3 = 7[3]

25
supermesh of mesh1 and mesh2 current source branch current

26
Three equations and three unknown.

27
Mesh-Current Analysis 1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement. 2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh. 3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns. 4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means. 5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.

28
Superposition Superposition Theorem – the response of a circuit to more than one source can be determined by analyzing the circuit’s response to each source (alone) and then combining the results Insert Figure 7.2

29
Superposition Insert Figure 7.3

30
Superposition Analyze Separately, then Combine Results

31
Use superposition to find the current i x. Current source is zero – open circuit as I = 0 and solve i Xv Voltage source is zero – short circuit as V= 0 and solve i Xv

32
Use superposition to find the current i x. The controlled voltage source is included in all cases as it is controlled by the current i x.

33
Voltage and Current Sources Insert Figure 7.7

34
Voltage and Current Sources Insert Figure 7.8

35
Voltage and Current Sources Insert Figure 7.9

36
Source Transformation Under what condition, the voltage and current of the load is the same when operating at the two practical sources? For voltage source, For current source We have,

38
Voltage and Current Sources Equivalent Voltage and Current Sources – for every voltage source, there exists an equivalent current source, and vice versa

40
Thevenin’s Theorem Thevenin’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a voltage source in series with a source resistance

41
Thevenin’s Theorem Thevenin Voltage (V TH ) – the voltage present at the output terminals of the circuit when the load is removed Insert Figure 7.18

42
Thevenin’s Theorem Thevenin Resistance (R TH ) – the resistance measured across the output terminals with the load removed

43
Thévenin Equivalent Circuits

48
Finding the Thévenin Resistance Directly When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit. We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.

51
Computation of Thévenin resistance

52
Equivalence of open-circuit and Thévenin voltage

53
A circuit and its Thévenin equivalent

54
Superposition As the voltage source does not contribute any output voltage, Only the current source has the effect.

55
Determine the Thévenin and Norton Equivalents of Network A in (a). Source transformation

56
Find the Thévenin equivalent of the circuit shown in (a). As i = -1, therefore, the controlled voltage source is -1.5V. Use nodal analysis at node v, v Thus, R th =v/I = 0.6/1 = 0.6 ohms

58
Applications of Thevenin’s Theorem Load Voltage Ranges – Thevenin’s theorem is most commonly used to predict the change in load voltage that will result from a change in load resistance

59
Applications of Thevenin’s Theorem Maximum Power Transfer –Maximum power transfer from a circuit to a variable load occurs when the load resistance equals the source resistance –For a series-parallel circuit, maximum power occurs when R L = R TH

60
Applications of Thevenin’s Theorem Multiload Circuits Insert Figure 7.30

61
Norton’s Theorem Norton’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a current source in parallel with a source resistance

62
Norton’s Theorem Norton Current (I N ) – the current through the shorted load terminals Insert Figure 7.35

63
Computation of Norton current

64
Norton’s Theorem Norton Resistance (R N ) – the resistance measured across the open load terminals (measured and calculated exactly like R TH )

65
Norton’s Theorem Norton-to-Thevenin and Thevenin-to-Norton Conversions Insert Figure 7.39

66
Step-by-step Thévenin/Norton- Equivalent-Circuit Analysis 1. Perform two of these: a. Determine the open-circuit voltage V t = v oc. b. Determine the short-circuit current I n = i sc. c. Zero the sources and find the Thévenin resistance R t looking back into the terminals.

67
2. Use the equation V t = R t I n to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source V t in series with R t. 4. The Norton equivalent consists of a current source I n in parallel with R t.

69
Maximum Power Transfer The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.

70
Graphical representation of maximum power transfer Power transfer between source and load

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google