EE201 Lecture 8 P 1 Nodal Analysis cont Mesh Analysis Nodal Analysis Good and Bad Choices Consider the following circuit 2 3 5 5V 3V 4 2A 5 EE201 Lecture 8 P 2 B A E 2A 5 5V C 3V 4 2 3 5 D Good choices 1 labelled nodes 2 ground at E Therefore only 2 variables and not four EE201 Lecture 8 P 3 Apply KCL to find node voltages 2 unknowns VA and VD KCL A 5 VA 5 2VA 2 0 7 VA 23 KCL D 5VD 4 VD 3 2 0 9 VD 14 Therefore VD 14 9 V VA 23 7 V The intelligent choices for nodal analysis led to reduction of a 5 node circuit to two unknowns and no coupled equations to solve simultaneously EE201 Lecture 8 P 4 B Supernode 5V A E 2A 5 C 3V 4 2 3 5 D Bad choices 1 ground at D The importance of choosing the right node as ground you do not apply KCL at ground EE201 Lecture 8 P 5 Apply KCL to find node voltages 2 unknowns VA and VB KCL A 5 VA VB 2 VA VE 2 0 But VE VB 5 7VA 7VB 12 0 KCL SN 5 VB VA 2 VE VA 5VE 4VC 0 But VC VB 2 7VA 16VB 43 0 Simultaneous equations 7VA 7VB 12 0 7VA 16VB 43 0 EE201 Lecture 8 P 6 Are solutions the same Look at voltage drop across circuit elements Good choice VD 14 9 Bad VA 23 7 VB 217 63 VA 109 63 But what is drop across 0 2 resistor Good choice V5mho VA VB 23 7 5 12 7 V Bad V5mho VA VB 109 63 217 63 12 7 V Node voltages are different because of the different reference nodes used It is the voltage difference between nodes that is physical and must be the same regardless of choice of ground EE201 Lecture 8 P 7 Loop Mesh Analysis A circuit analysis technique based on KVL Ohm s law and the existence of circulation currents Consider the circuit below iA R iA iB iB Terminology There are 3 loops and 2 meshes in this circuit A loop is any closed path around a circuit but passing through a node only once A mesh is a loop which contains no other loops EE201 Lecture 8 P 8 Mesh currents can be defined within the circuit as shown below R1 R2 R3 V1 i1 i2 V2 Basis of Mesh Analysis Write KVL equations for each mesh then solve the coupled simultaneous equations KVL1 V1 R1i1 R3 i1 i2 0 KVL2 V2 R2i2 R3 i2 i1 0 Once equations are solved for i1 and i2 the behavior of the circuit is known EE201 Lecture 8 P 9 Mesh Analysis guidelines 1 Draw a clear circuit diagram indicating meshes 2 Assign a current to each mesh For n meshes there will be n equations 3 Apply KVL around each mesh If dependent voltage source present relate controlling variable to the mesh current 4 If circuit contains independent or dependent current sources apply KVL to a supermesh to bypass the current source Relate currents through sources to the assigned mesh currents 5 Do not define any voltage variables EE201 Lecture 8 P 10 Example Solve by mesh analysis 1 5 5V 3 3A 10V 2 4 3 meshes for KVL but the mesh equations for meshes 2 and 3 contain a current source Therefore define supermesh or loop that contains 2 and 3 EE201 Lecture 8 5 P 11 Supermesh 1 5V 10V 3 3A 2 i3 i2 i1 4 Solution Use KVL to obtain equations KVL 1 KVL SM I source 10 5i1 3 i1 i2 2 i1 i3 0 5 4i3 2 i3 i1 3 i2 i1 1i2 0 i3 i2 3A 3 equations and 3 unknowns solve for i1 i2 i3 Once the currents are known the circuit behavior is known
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