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ROCHESTER PHY 217 - Lecture 37B - Circuits

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2 December 2002 Physics 217, Fall 2002 1Today in Physics 217: circuits! Review of DC circuits: Kirchhoff’s rules ! Solving equations from Kirchhoff’s rules for simple DC circuits2 December 2002 Physics 217, Fall 2002 2Kirchhoff’s rulesLumped circuit elements:are devices with a lot of capacitance, resistance and inductance concentrated in them, which can be connected together in various ways, using wires with negligible resistance, to form circuits which, apart from any lumped inductance, have negligible inductance. Add EMFs and one can create many solvable physics problems. C(archetype: parallel plates)R(archetype: serpentine wire)L(archetype: solenoid)2 December 2002 Physics 217, Fall 2002 3Kirchhoff’s rules (continued)Physics we know that applies here:!Kirchhoff rule #1: currents are confined to the wires, and charge doesn’t build up anywhere within them. This is simply an expression of charge conservation:What it means is that the total current flowing into any point in a circuit equals the total flowing out. This is especially useful to apply to junctions (a.k.a. nodes), since it gives us equations with the currents in them, e.g.:()ρ∂⋅=− ⇒ = ⋅ =− =∂∫!!!!! 0.dQIdtdtJJa1I3I2I4I⇔−−−=12340IIII2 December 2002 Physics 217, Fall 2002 4Kirchhoff’s rules (continued)!Kirchhoff rule #2: ∆V= IRfor resistors; ∆V= 0 around a circuit with no EMFs. This, as you know, is an expression of and means that the sum of the potential differences and EMFs around any complete loop, accounting for polarity (i.e. the algebraicsum), is zero. Writing these down for a circuit with Nindependent currents will generate substantially more than Nequations relating the currents. Selection of Nlinearly independent equations from among them allows one to solve for the Nunknowns. This is best illustrated by example.=− ⋅∫!! ,dE"2 December 2002 Physics 217, Fall 2002 5Example: A simple DC circuit!16R1R2R3R4R5R!21I2I3I4I5I6I++++++++What are the currents in this circuit, in terms of the EMFs and resistances?2 December 2002 Physics 217, Fall 2002 6A simple DC circuit (continued)To solve, apply Kirchhoff’s rules and solve the system of equations that results. First Rule #1, the node equation. The upper three nodes giveThe lower node gives a relation that isn’t independent of the other three: add the three we have and you getwhich is just the equation for the lower node. Only three of the four node equations are independent; we can use any three of them. We now have three equations, six unknowns.!16R1R2R3R4R5R!21I2I3I4I5I6I++++++++−−=+−=−−=123245346000IIIIIIIII−−=1560,III2 December 2002 Physics 217, Fall 2002 7A simple DC circuit (continued)Now Rule #2. Take the three obviouscircuits in the picture. Suppose thatthe current is carried by positivecharges, so that the polarity of thepotential differences are as indicatedby the + signs. ThenThey are apparently independent. Now we have six equations for our six unknowns.!16R1R2R3R4R5R!21I2I3I4I5I6I++++++++−−−=−+−=−−+=!!!112456125412232544536000RI RI RIRI RI RIRI RI RIIIIIII2 December 2002 Physics 217, Fall 2002 8A simple DC circuit (continued)There are many other loops in thefigure, but they don’t give us anyinformation that the first three didn’t. For instance:the loop equation for the circuit’s perimeter. In general, a circuit with N nodes gives N–1 independent node equations, and (usually way) more than Nloop equations, most of which will not be independent.!16R1R2R3R4R5R!21I2I3I4I5I6I++++++++()()()−−−=+− + − =−−−+=−−−=!!!!11245612541223254453612336610000,RI RI RIRI RI RIRI RI RIRI RI RI2 December 2002 Physics 217, Fall 2002 9A simple DC circuit (continued)So our six equations in six unknowns areThis can be written conveniently as a matrix equation:−− =+− =−−=++=−++ =+−=!!!12324534 661 12 45 112 23 54 254 45 36 2000IIIIIIII IRI RI RIRI RI RIRI RI RI2 December 2002 Physics 217, Fall 2002 10A simple DC circuit (continued)orSolving for the Is “just” involves inverting the matrix, for which there are many good methods:−−−−−=−−!!!12361 4 4 1125 5 254 36 2111000 0010110 0001101 0,00 0000000IIIRR R IRRR IRR RI[][]=!.nnMI[] []−=!1.nnIM2 December 2002 Physics 217, Fall 2002 11A simple DC circuit (continued)For small matrices, it’s not even too bad to use Cramer’s Ruleto solve such equations. If a matrix equation is written aswhere v and ββββare column matrices (vectors!) and ααααis a square matrix, then the components of v are given bywhere |…| denotes a determinant, and is the determinant of the matrix made by replacing the ith column of ααααwith the elements of ββββ.⋅=#vαβ,αβ,αβ,αβ,αα=,iivαi2 December 2002 Physics 217, Fall 2002 12A simple DC circuit (continued)By any means, the solutions get ugly if there are more than a few equations. For instance, in the present case, the current turns out to beOthers available by request.()+++ ++ +++−=++++++ + + +++++++!!53 54 13 14123 24 52 5113 24 21542 521 534 531 526561 564 536 324 321641 642 314 326 241 316RR RR RR RRRR RR RR RRRR RRIRRR RRR RRR RRR RRRRRR RRR RRR RRR RRRRRR RRR RRR RRR RRR RRR1I2 December 2002 Physics 217, Fall 2002 13Recipe for Kirchhoff-rule problems! Identify each independent branch of the circuit, and define a current that runs in that branch. It doesn’t matter which way you define the current to run – your algebra will tell you, with a minus sign, when you get it wrong. Count the number of currents; say, K.! Count the number of nodes, say, N. Write down the node equations for N–1 of them.! Identify K–N+1 loops, and generate their loop equations. The polarity of all the voltage differences is determined by the directions you defined for the unknown currents.! Solve these Kequations simultaneously by some expedient means. Do it on a computer if Kis larger than a few. Use Cramer’s Rule if it’s


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