EE201 Lecture 7 P 1 Nodal Analysis Nodal Analysis A systematic technique for computing the node voltages of a circuit Once the node voltages are known the CE voltages can be found This yields all branch currents if resistances are known A vin t From KCL A veq Req iin t iin vA Req This is the simplest use of nodal analysis Once vA is known current through the resistor can be calculated EE201 Lecture 7 P 2 Nodal Analysis guidelines 1 Draw a clear circuit diagram indicating nodes 2 Define a reference node ground This should usually be the node with the most branches 3 Assign node voltages to all nodes other than ground 4 If circuit contains voltage sources include them in a supernode This reduces the number of independent variables 5 Apply KCL at each node to obtain node voltages 6 Define no current variables EE201 Lecture 7 P 3 Circuit solution using nodal analysis V 2 A I V1 R1 R2 R3 B V3 C Given I R1 R2 R3 3 nodes in this circuit Node C is ground Voltages across resistors V1 VAC VA 0 V2 VAB VA VB V3 VBC VB 0 EE201 Lecture 7 P 4 Apply KCL basis of nodal analysis at each node A I I1 I2 V1 R1 V2 R2 B I2 I3 V2 R2 V3 R3 Insert nodal voltages into the above equations A I VA R1 VA VB R2 B 0 VB VA R2 VB R3 Which are 2 equations with 2 unknowns VA VB Solve these simultaneous equations to yield VA and VB and the circuit behavior is known EE201 Lecture 7 P 5 Example Find vx and iy 5A iy 5 2 5V 2iy 2 3vx 2 2 vx 10V 3A EE201 Lecture 7 P 6 1 Clear circuit diagram with nodes labelled 5A B 2iy A 5 2 5V 3vx 2 2 vx 10V D C 2 iy 3A 2 Define reference node ground 3 Assign node voltages VA VB VC VD Trouble how do you find current through a voltage source Lecture 7 P 7 5A iy B 2iy A 5 2 5V 3vx 2 2 10V D C vx SN1 2 EE201 3A 4 Define supernode to eliminate difficulty with finding current through voltage sources Supernodes enclose two or more nodes and are an extension of the Guassian surface concept EE201 Lecture 7 P 8 5 Use KCL at each node to obtain set of equations that will yield node voltages How many unknown node voltages are there 2 unknowns VB and VC or VD KCL B 5 3vx iy 2 vx 0 5 3 VB 5 2 VB VC 2 VB 5 0 5 VB 5 2 VB VC 0 VB 2VC 0 KCL SN1 3 5 2 VC VB 5VC 2VC 2iy 2 VC 10 0 Insert iy 2 VB VC and reduce vx 6V VB B 15 5 V C7 V18 iy 2 VB VC 12 A EE201 Lecture 7 P 9 Node Analysis with Dependent Voltage Sources Generally dependent voltage sources are included in a supernode 50 E A 40 ix 100 110V ix B 30V C 500 800 D EE201 Lecture 7 50 E A 40 ix 100 110V ix B P 10 30V C 500 800 D Define supernode to include independent and dependent source EE201 Lecture 7 P 11 KCL supernode VA VE 50 VB 500 VC 800 0 But VE 110 V VC VB 30 VA 110 50 VB 500 VB 30 800 0 Current through 100 resistor VA VB 100 ix VA VC 40 ix VA VB 100 VA VB 30 40 Simultaneous equations to solve VA 110 50 VB 500 VB 30 800 0 VA VB 100 VA VB 30 40 0 Solve these for VA and VB then substitute into equation for current ix
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