SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING. Self Study Course

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1 SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Stud Course MODULE 27 FURTHER APPLICATIONS TO ELECTRICAL CIRCUITS Module Topics 1. Inverse of a matri using elimination 2. Mesh analsis of circuits 3. Nodal analsis of circuits A: Work Scheme based on JAMES (FOURTH EDITION) 1. The first topic uses results from module 17 on matrices. Before starting this module ou should make sure that ou know how to calculate the inverse of a matri. This is described in module 17 and in section 5.4 and section 5.5 of James. The most useful method is to use elimination. This method is described in sections 6 and 7 of module 17. Eample A: Use the elimination method to solve the matri equation ( ) 3 4 Multipl row 1 b 2 ( Form a new row 2 b adding old row 1 and old row 2 ) Hence B backward substitution in row Hence the solution is 17 and 10. ***Do Eercise 74 on p.370 using the elimination method.*** 1

2 2.Mesh Analsis Mesh analsis is a sstematic procedure which uses Kirchhoff s voltage law (KVL) to find all the currents in a circuit. The method of analsis is as follows: 1. Clearl label all the known quantities in the circuit. 2. Identif all the meshes in the circuit. A mesh is defined as a loop with no sub-loops. 3. Assign mesh currents using the convention that the are measured in a clockwise direction. 4. Appl KVL for each mesh and epress the voltages in terms of the mesh currents. 5. Write the resulting sstem of equations in matri form. Solve these equations using the elimination method to obtain the mesh currents. 6. Now the mesh currents are known the voltages ma be obtained from Ohm s law. The method is best illustrated using a worked eample. Eample A: Use mesh analsis to calculate the currents and voltages in the circuit shown in figure 1 below V This circuit has two meshes. Mesh 1: KVL gives Mesh 2: KVL gives We can write this as a pair of simultaneous equations Fig I 1 + 4(I 1 I 2 ) 6 4(I 2 I 1 ) + 2I 2 10I 1 4I I 1 + 6I 2 6 or in matri notation as ( 10 ) ( 4 I1 4 6 Multipl row 1 b 2 and row 2 b 5 I 2 ( 20 8 ) ( I I 2 ) ) Form a new row 2 b adding old row 1 and old row I I

3 Hence B backward substitution in row 1 I I 1 8I I I 2 Hence the solution is I A and I A. Note that the current flowing through the resistor is I 2 I A. The corresponding voltages are given b Ohm s law V IR and are therefore given b v V, v V and v V. Note that we can check that we have done the calculation correctl b appling KVL to each mesh and checking that it gives the correct voltage. In this eample v 1 v which is the voltage source in mesh 1, while v 2 + v which is the voltage source in mesh 2. ***Do Eercise A: 2 below. Use mesh analsis to calculate the currents and voltages in the circuit shown in figure V Fig Nodal Analsis An alternative to mesh analsis is nodal analsis. This is a sstematic procedure which uses Kirchhoff s current law (KCL) to find all the voltages in a circuit. It should be used in preference to mesh analsis for circuits that have fewer nodes than meshes. It also has the advantage that it can be used to analse an circuits not simpl planar ones. The method of analsis is as follows: 1. Clearl label all the known quantities in the circuit. 2. Identif all the nodes in the circuit. 3. Select a node as the reference node and assign it a potential of 0 volts. All other voltages in the circuit are measured with respect to the reference node. 4. Label the voltages on all the other nodes. 5. Assign and label the currents in the circuit (including the polarities). 6. Appl KCL at each node to obtain equations for the voltages. 7. Write the resulting sstem of equations in matri form. Solve these equations using the elimination method to obtain the node voltages. 8. Now the voltages are known the currents ma be obtained from Ohm s law. 3

4 We start b looking at a simple eample with no voltage sources. Eample B: Use nodal analsis to calculate the currents voltages in the circuit shown in figure 3 below. A B 3A C Fig. 3. This eample has 3 meshes and 3 nodes (note that what appears to be a fourth node to the right of node C is reall part of C since there is simpl wire with no resistance between them). We take node C to be the reference node and solve for KCL at node A and node B. Node A: KCL gives Node B: KCL gives 6A 3 + (V A V C )/2 + (V A V B )/ (V B V A )/4 + (V B V C )/4 0 Rearranging and recalling that V C 0 we can write this as a pair of simultaneous equations 3 4 V A 1 4 V B 3, 1 4 V A V B 6. Multipling both equations b 4 we ma write this in matri notation as 3 1 VA Multipl row 2 b 3 ( 3 1 VA 3 6 V B V B ) Form a new row 2 b adding old row 1 and old row VA 0 5 V B Hence B backward substitution in row 1 V B V A V B 12 3V A 12 + V B

5 Hence the solution is V A 0V and V B 12V. ***Do Eercise B: Use nodal analsis to calculate the voltages in the circuit shown in figure 4 below. 2A 4A Fig. 4. B: Work Scheme based on STROUD (SIXTH EDITION) This module is not covered b S., so work through A: Work scheme based on JAMES (FOURTH EDITION), presented above. 5

6 Specimen Test Use the elimination method to solve the following matri equations (i) ( ) 5 5 (ii) ( ) Use mesh analsis to calculate the currents and voltages in the circuit shown in figure 5 below. Verif that the resulting voltages satisf KVL for each mesh. 3Ω 1 2 Fig Use nodal analsis to calculate the voltages in the circuit shown in figure 6 below. Use Ohm s law to calculate the currents flowing in each component of the circuit. A B 4A 3Ω 2A C Fig. 6. 6

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