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EE2010 Fundamentals of Electric Circuits Lecture - 6 Voltage Sources, Current Sources, Mesh Analysis.

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Presentation on theme: "EE2010 Fundamentals of Electric Circuits Lecture - 6 Voltage Sources, Current Sources, Mesh Analysis."— Presentation transcript:

1 EE2010 Fundamentals of Electric Circuits Lecture - 6 Voltage Sources, Current Sources, Mesh Analysis

2 Voltage Sources in Series and in Parallel Ideal voltage sources cannot be combined in parallel (unless they have the same voltage)

3 We can connect ideal voltage sources in series. Voltage sources in series can be reduced to a single voltage source Voltage Sources in Series and in Parallel

4 Current Sources in Series and Parallel ideal current sources cannot be combined in series (unless they have the same current)

5 Current sources in parallel can be combined as a single current source Current Sources in Series and Parallel

6 Example

7 Determine the current I 1 and the voltage Vs for the network Example

8 SOURCE CONVERSIONS the equivalence between a current source and a voltage source exists only at their external terminals.

9 Example a) Determine the current I L. b) Convert the voltage source to a current source. c)Using the resulting current source of part (b), calculate the current through the load resistor, and compare your answer to the result of part (a).

10 CURRENT SOURCES IN PARALLEL Voltage sources of different terminal voltages cannot be placed in parallel because of a violation of Kirchhoff’s voltage law. Current sources of different values cannot be placed in series due to a violation of Kirchhoff’s current law. However, current sources can be placed in parallel just as voltage sources can be placed in series.

11 CURRENT SOURCES IN PARALLEL Two or more current sources in parallel can be replaced by a single current source having a magnitude determined by the difference of the sum of the currents in one direction and the sum in the opposite direction. The new parallel internal resistance is the total resistance of the resulting parallel resistive elements.

12 Example Reduce the parallel current sources as shown in Fig. to a single current source. Solution: The net source current is I = 10 A - 6 A = 4 A with the direction of the larger. The net internal resistance is the parallel combination of resistors, R1 and R2:

13 Reduce the parallel current sources in Fig. to a single current source. Solution: The net current is Example

14 MESH ANALYSIS The currents to be defined are called mesh or loop currents A closed“loop” around the inside of each window; these loops are similar to the loops defined in the wire mesh fence in Fig. The number of mesh currents required to analyze a network will equal the number of “windows” of the configuration.

15 Mesh Analysis Procedure 1.Assign a distinct current in the clockwise direction to each independent, closed loop of the network. 2.Indicate the polarities within each loop for each resistor as determined by the assumed direction of loop current for that loop. 3.Apply Kirchhoff’s voltage law around each closed loop in the clockwise direction. 4.Solve the resulting simultaneous linear equations for the assumed loop currents MESH ANALYSIS

16 EXAMPLE Solution: Step 1: Two loop currents (I 1 and I 2 ) are assigned in the clockwise direction in the windows of the network. Step 2: Polarities are drawn within each window to agree with assumed current directions. Step 3: Kirchhoff’s voltage law is applied around each loop in the clockwise direction.

17 loop 1: + E1 - V1 - V3 = 0 (clockwise starting at point a) EXAMPLE

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21 Supermesh Currents Start as before and assign a mesh current to each independent loop, including the current sources, as if they were resistors or voltage sources. Then mentally (redraw the network if necessary) remove the current sources (replace with open-circuit equivalents), and apply Kirchhoff’s voltage law to all the remaining independent paths of the network using the mesh currents just defined.

22 Supermesh Currents

23 Example Node a is then used to relate the mesh currents and the current source using Kirchhoff’s current law:


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