Lecture 7 notes Nodal Analysis
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1 Lecture 7 notes Nodl Anlysis Generl Network Anlysis In mny cses you hve multiple unknowns in circuit, sy the voltges cross multiple resistors. Network nlysis is systemtic wy to generte multiple equtions which cn be solved to find the multiple unknowns. These equtions re bsed on bsic Kirchoff's nd Ohm's lws. A. Stolp 9/10/08, 1/7/13 Loop or Mesh Anlysis You my hve used these methods in previous clsses, prticulrly in Physics. The best thing to do now is to forget ll tht. Loop nlysis is rrely the esiest wy to nlyze circuit nd is inherently confusing. Hopefully I've brought you to stge where you hve some intuitive feeling for how currents flow in circuits. I don't wnt to ruin tht now by screwing round with loop currents tht don't relly exist. Nodl nlysis This is much better method. It's just s powerful, usully esier, nd helps you develop your intuitive feeling for how circuits work. Nodl Anlysis Node ll points connected by wire, ll t sme voltge (potentil) Ground: One node in the circuit which will be our reference node. Ground, by definition, will be the zero voltge node. All other node voltges will be referenced to ground nd my be positive or negtive. Think of gge pressure in fluid system. In tht cse tmospheric pressure is considered zero. If there is no ground in the circuit, define one for yourself. Try to chose node which is hooked to one side of voltge source. Nodl oltge: The voltge of node referenced to ground. The objective of nodl nlysis is to find ll the nodl voltges. If you know the voltge t node then it's "known" node. Ground is known node (duh, it's zero). If one end of known voltge source hooked to ground, then the node on the other end is lso known (nother duh). Method: Lbel ll the unknown nodes s; "", "b", "c", etc. Then the unknown nodl voltges become;, b, c, etc. Write KCL eqution for ech unknown node, defining currents s necessry. Replce ech unknown current with n Ohm's lw reltionship using the nodl voltges. Now you hve just s mny equtions s unknowns. Solve. Nodl Anlysis Steps 1) If the circuit doesn't lredy hve ground, lbel one node s ground (zero voltge). If the ground cn be defined s one side of voltge source, tht will mke the following steps esier. Lbel the remining node, either with known voltges or with letters,, b,... ) Lbel unknown node voltges s, b,... nd lbel the current in ech resistor s I 1, I,... 3) Write Kirchoff's current equtions for ech unknown node. 4) Replce the currents in your KCL equtions with expressions like this. 5) Solve the multiple equtions for the multiple unknown voltges. b Ohm's lw reltionship using the nodl voltges. Nodl Anlysis Exmples Ex 1 Use nodl nlysis to find the voltge cross ( R1 ).. 1 R1 R. I R3. S 10 R ma 1) See next pge Lbel one node s ground (zero voltge). By choosing the negtive side of voltge source s ground, the upper-left node is known (10). Lbel the remining nodes, either with known voltges or with letters,, b,... Lecture 7 notes p1
2 ) Lbel unknown node voltges s, b,... Lecture 7 notes p nd lbel the current in ech resistor s I 1, I,... I 1 R 10. 3) Write Kirchoff's current equtions for node. R1 I 1 + I R3 I R3 4) Replce the currents in the KCL equtions with Ohm's lw reltionships. 10 b b ma 5) Solve: Usully it's esier to put in the numbers t this point ma Multiply both sides by vlue tht will cler the denomintors Either wy, you still hve to find R1 from. 4. ma ma R1 R1 0.5 b doesn't mtter in this cse b) Find the current through (I R3 ). I R3 3.5 ma Ex Sme circuit used in Thévenin exmple, where ws R L S 9 I. S 50 ma R ) Define ground nd nodes:. 9 b unknown nodes ---> will need equtions Lecture 7 notes p
3 ) Lbel unknown node voltges s, b,... nd lbel the current in ech resistor s I 1, I,... It doesn't mtter if these currents re in the correct directions. Lecture 7 notes p3 I 1 I b R R I 9. 3) Write Kirchoff's current equtions for ech unknown node. 1 S node I 1 I + I 4 I 4 I 3 node b I I 3 + 4) Replce the currents in your KCL equtions with expressions like this. b node I 1 I + I 4 node b I I 3 + b +. 0 b Now you hve just s mny equtions s unknowns. b ) Solve the multiple equtions for the multiple unknown voltges. Solve by ny method you like: Or, with numbers node b + R. b b b + b 1. R. 1 R. node b b b b b ma b 5.. b... b 48 ma b \ \ \. b. b b 6. \ / b b b 6 <-- substitute for b b b 7.5. b b 3. b 7.5. b 13.5 b.5. b 1. b b Sme s L of Ex 4 of Thévenin exmples: Lecture 7 notes p3
4 Lecture 7 notes p4. 1 Ex 3 Like Superposition Ex. ) Use nodl nlysis to find the voltge cross ( R ). You MUST show ll the steps of nodl nlysis work to get credit, including drwing pproprite symbols nd lbels on the circuit shown. 1) Define ground nd nodes: ) Lbel unknown node voltges s, b,... nd lbel the current in ech resistor s I 1, I, I R3 +. R _. 0 1 I 1 3 I R3 I. 1 node :. ma 3) Write Kirchoff's current equtions for ech unknown node. I + I R3 4) Replce the currents in the KCL equtions with Ohm's lw reltionships. ma 0 5) Solve the eqution for the unknown voltge. Usully it's esier to put in the numbers t this point ma Multiply both sides by vlue tht will cler the denomintors. R ma Remember, we needed to find the voltge cross ( R ). R b) Find the current through (I R3 ). 0 I R3 1.6 ma ctully flows the other wy Lecture 7 notes p4
5 Lecture 7 notes p5 Ex 4 Use nodl nlysis to find the voltge cross R 5 ( R5 ) nd the current through (I R1 ). From exm 1, F09 You MUST show ll the steps of nodl nlysis work to get credit, including drwing pproprite symbols nd lbels on the circuit shown I R1 R R R R5. 63 ma I R R R5 63 ma node :. 6 I R1 + I R + I R ạ ma ạ ạ multiply both sides by ạ ạ 6. ạ 6. ma R5 9 1 I R1 30 ma Lecture 7 notes p5
6 Lecture 7 notes p6 Wht if one side of voltge source isn't ground? I 1 + I 3 I 1 1 +? 1 b b I 3 Wht do you put in for?? I Go to the other side of b Only problem is tht you get the sme eqution t node b! Where does the second eqution come from? Use something like this: b Similr Circuit, but no 1. If the ground is lredy t the bottom, use the sme method s bove. I 1 I 3 b b I If you cn chose your ground, you cn mke life little simpler. I 1 I 3. 0 I Lecture 7 notes p6 b b
I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
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