Today in Physics 217: circuits! Review of DC circuits: Kirchhoff s rules! Solving equations from Kirchhoff s rules for simple DC circuits 2 December 2002 Physics 217, Fall 2002 1
Lumped circuit elements: Kirchhoff s rules C R (archetype: parallel plates) (archetype: serpentine wire) L (archetype: solenoid) 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. 2 December 2002 Physics 217, Fall 2002 2
Kirchhoff 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: ρ dq! J = =! J da= ( = 0 ). t dt 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.: 2 1 3 4 1 2 3 4 = 0 2 December 2002 Physics 217, Fall 2002 3
Kirchhoff s rules (continued)! Kirchhoff rule #2: V = R for resistors; V = 0 around a circuit with no EMFs. This, as you know, is an expression of! =! E d", and means that the sum of the potential differences and EMFs around any complete loop, accounting for polarity (i.e. the algebraic sum), is zero. Writing these down for a circuit with N independent currents will generate substantially more than N equations relating the currents. Selection of N linearly independent equations from among them allows one to solve for the N unknowns. This is best illustrated by example. 2 December 2002 Physics 217, Fall 2002 4
Example: A simple DC circuit What are the currents in this circuit, in terms of the EMFs and resistances?! 1 1 R 2 2 R 1 3 4 R 6 R 5! 2 5 R 4 6 R 3 2 December 2002 Physics 217, Fall 2002 5
A 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 give 1 2 3 = 0 2 4 5 = 0 = 0 3 4 6 The lower node gives a relation that isn t independent of the other three: add the three we have and you get = which 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.! 1 R 6 1 R 2 2 R 1 3 4 R 5! 2 5 R 4 6 R 3 1 5 6 0, 2 December 2002 Physics 217, Fall 2002 6
A simple DC circuit (continued) Now Rule #2. Take the three obvious circuits in the picture. Suppose that the current is carried by positive charges, so that the polarity of the potential differences are as indicated by the signs. Then!!! R R R = 0 R R R = 0 R R R = 0 1 1 2 4 5 6 1 2 5 4 1 2 2 3 2 5 4 4 5 3 6! 1 R 6 1 R 2 2 R 1 3 4 R 5! 2 5 R 4 6 R 3 They are apparently independent. Now we have six equations for our six unknowns. 2 December 2002 Physics 217, Fall 2002 7
A simple DC circuit (continued) There are many other loops in the figure, but they don t give us any information that the first three didn t. For instance:! R R R = 0 ( 1 1 2 4 5 6 1 ) (! 2 R 5 4 R 1 2 R 2 3 0) (! R R R 0) = = 2 5 4 4 5 3 6! 1 R 2 3 R 3 6 R 6 1 = 0, the loop equation for the circuit s perimeter. n general, a circuit with N nodes gives N 1 independent node equations, and (usually way) more than N loop equations, most of which will not be independent. 2 December 2002 Physics 217, Fall 2002 8! 1 R 6 1 R 2 2 R 1 3 4 R 5! 2 5 R 4 6 R 3
A simple DC circuit (continued) So our six equations in six unknowns are = 0 1 2 3 = 0 2 4 5 = 0 3 4 6 R R R =! 6 1 1 2 4 5 1 R R R =! 1 2 2 3 5 4 2 R R R =! 5 4 4 5 3 6 2 This can be written conveniently as a matrix equation: 2 December 2002 Physics 217, Fall 2002 9
or A simple DC circuit (continued) 1 1 1 0 0 0 1 0 0 1 0 1 1 0 2 0 0 0 1 1 0 1 3 0 = R6 R1 0 0 R4 0 4! 1 0 R! 1 R2 R5 0 0 5 2 0 0 0 R5 R4 R3 6! 2 M[ n ] = [! n ]., Solving for the s just involves inverting the matrix, for which there are many good methods: 1 [ ] = M [! ] n 2 December 2002 Physics 217, Fall 2002 10 n.
A simple DC circuit (continued) For small matrices, it s not even too bad to use Cramer s Rule to solve such equations. f a matrix equation is written as # α v = β, where v and β are column matrices (vectors!) and α is a square matrix, then the components of v are given by v i αi =, α where denotes a determinant, and is the determinant of the matrix made by replacing the ith column of α with the elements of β. α i 2 December 2002 Physics 217, Fall 2002 11
A 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 be 1 = RR 5 3 RR 5 4 RR 1 3 RR 1 4! 1 RR 2 3 RR 2 4 RR 5 2 RR 5 1 ( RR 1 3 RR 2 4)! 2 RRR 5 4 2 RRR 5 2 1 RRR 5 3 4 RRR 5 3 1 RRR 5 2 6 RRR 5 6 1 RRR 5 6 4 RRR 5 3 6 RRR 3 2 4 RRR 3 2 1 RRR RRR RRR RRR RRR RRR 6 4 1 6 4 2 3 1 4 3 2 6 2 4 1 3 1 6 1 Others available by request. 2 December 2002 Physics 217, Fall 2002 12
Recipe for Kirchhoff-rule problems! dentify each independent branch of the circuit, and define a current that runs in that branch. t 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.! dentify 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 K equations simultaneously by some expedient means. Do it on a computer if K is larger than a few. Use Cramer s Rule if it s not. 2 December 2002 Physics 217, Fall 2002 13