1. 1 Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 16: 22 Nov. 2004 Web page: http://hep.ucsb.edu/people/claudio/ph3-04/

2. 2 In both cases these pieces of a circuit can be thought of as "equivalent" to a single resistor  Equivalent means that for a given V ab the total current flowing would be the same as if it was a single resistor of resistance R eq, i.e., I=V ab /R Resistor in parallel or in series In parallel: In series R1R1 R2R2 R eq R1R1 R2R2 a a a b b b

3. 3 R eq for resistors in series The current flowing through the two resistors is the same The voltage drops are  V ac = I R 1 and V cb = I R 2  V ab = V a – V b = V a – V c + V c – V b = V ac + V cb  V ab = I R 1 + I R 2 = I (R 1 + R 2 ) Same as the current flowing through equivalent resistance R eq = R 1 + R 2 R1R1 R2R2 a b II c R eq a b R eq = R 1 + R 2

4. 4 Resistors in series, comments It makes sense that the resistances add Remember, resistance is something that impedes the flow of current If the current has to go through both resistors, the current has to overcome two "obstacles" to its flow

5. 5 R eq for resistors in parallel The currents in the two resistors are different But the voltage drops across the two resistors are the same  V ab = I 1 R 1 = I 2 R 2  I = I 1 + I 2 = V ab (1/R 1 + 1/R 2 ) Same as the current flowing through equivalent resistance R eq R1R1 R2R2 a b II=I 1 + I 2 I1I1 I2I2 R eq a b

6. 6 How does the current split between two resistors in parallel? V ab = I 1 R 1 = I 2 R 2  The current wants to flow through the smaller resistor  Makes sense! R1R1 R2R2 a b II=I 1 + I 2 I1I1 I2I2

7. 7 Resistors in parallel, comments R eq is always < R 1 and < R 2 This also makes sense When the current encounters two resistors in parallel, there are two possible paths for the current to flow It makes sense that the current will have an easier time going past the "obstacle" than it would have if only one resistance was present R1R1 R2R2 a b II=I 1 + I 2 I1I1 I2I2

8. 8 Summary and contrast with capacitors

9. 9 A big resistor and a small resistor... R eq =1  + 1 M  = 1,000,001  The big resistor wins..... R eq =0.999999  The small resistor wins.....

10. 10 Find the equivalent resistance These three are in series. R = 3 + 6 + 9 = 18  These two are in parallel R = 3.9  These three are in series R + 2 + 3.9 + 8 = 13.9 

11. 11 These two are in parallel R = 3.1  Three resistors in series R = 1 + 3.1 + 7 = 11.1  R eq = 11.1  Note – for two resistances in parallel, R eq < than each individual one. For two resistances in sereis, R eq is greater

12. 12 Another example I ?

13. 13 Another example (k means k  ) "Short", R=0 Series: R = 3 + 5 = 8 k 

14. 14 Kirchoff's rules We have already applied them, at least implicitely First rule: at a node (or junction)  I = 0 This is basically a statement that charge is conserved Careful about the signs! It is a good idea to always draw the arrows!

15. 15 Kirchoff's rules (continued) Second rule: the total voltage drop across a closed loop is zero For example: V ab + V bc + V ca = 0 (V a – V b ) + (V b – V c ) + (V c – V a ) = 0 But this holds for any loop, e.g. a-b-d-c-a or b-a-f-e-d-b,......

16. 16 Careful about signs: When you apply Kirchoff's 2 nd rule, the absolute value of the voltage drop across a resistor is IR and across a source of emf is  You must keep the sign straight! For emf, it is easy. For resistors, depends on the sign of I a b a b V ab = +  a b V ab = -  a b I I V ab = IR V ab = -IR These signs are such that current always runs from high to low potential. Note algebraically I can be +ve or –ve. What matters is your convention, i.e., the direction of the arrow

17. 17 Another example: find R eq Add a fictitious source of current, say I=1A, across the terminals. Then if I can calculate the voltage across the terminals, R = V/I. Also, label everything!!!!! I I1I1 I1I1 I3I3 I3I3 I4I4 I2I2 I5I5 a b c de f Kirchoff law for current: Node a: I = I 1 + I 5 Node e: I 5 = I 4 + I 2 Node d: I 4 + I 1 = I 3 Node b: I 2 + I 3 = I 4 equations, 5 unknowns (I 1, I 2, I 3, I 4, I 5 )

18. 18 I I1I1 I1I1 I3I3 I3I3 I4I4 I2I2 I5I5 a b c de f Kirchoff law for current: Node a: I = I 1 + I 5 Node e: I 5 = I 4 + I 2 Node d: I 4 + I 1 = I 3 Node b: I 2 + I 3 = I Now I apply Kirchoff law for voltage loops. But let's be smart about it! I want V ab (or V cf, they are the same) V ab = 2I 5 + I 2 V cf = I 1 + I 3 Try to eliminate variables and remain with I 5, I 2 Node b: eliminates I 3 : I 3 = I – I 2 Node a: eliminates I 1 : I 1 = I – I 5 acdea loop: I 1 – I 4 – 2I 5 = 0  I – I 5 – I 4 – 2I 5 = 0  I 4 = I – 3 I 5 Node e: I 5 = (I – 3I 5 ) + I 2  4I 5 – I 2 = I acdfbea loop:I 1 + I 3 – I 2 – 2I 5 = 0 (I – I 5 ) + (I – I 2 ) – 2 I 5 = 0 2I – 2I 2 – 3I 5 = 0

19. 19 I I1I1 I1I1 I3I3 I3I3 I4I4 I2I2 I5I5 a b c de f 4I 5 – I 2 = I 2I – 2I 2 – 3I 5 = 0 Solution: I 2 = 5I/11 and I 5 = 4I/11 Then: V ab = 2I 5 + I 2 V ab = 13I/11 R eq = 13/11 k  = 1.2 k 

20. 20 Note: Example 26.6 in the textbook is essentially the same. (It has resistances of  instead of k  ) The book solves it using a battery rather than a fictitious current source. But the answer is the same, as it should! Check out the alternative method for yourself! I I1I1 I1I1 I3I3 I3I3 I4I4 I2I2 I5I5 a b c de f

21. 21 Find R eq of this circuit: The other side of the 4  resistor is not connected to anything It is as if it was not there! Two resistors in series, R = 7  + 5  = 12 

22. 22 Another problem..... Find the current through the 5  resistor First step: Label everything! a b c de f I1I1 I1I1 I3I3 I4I4 I5I5 I6I6 I6I6 I2I2

23. 23 a b c de f I1I1 I1I1 I3I3 I4I4 I5I5 I6I6 I6I6 I2I2 Use Kirchoff law for current nodes to eliminate some variables. Node c:I 4 = I 2 + I 5  eliminate I 4 Node d: I 6 = I 5 + I 3  eliminate I 6 Node b: I 1 = I 3 + I 4 = I 3 + I 2 + I 5  eliminate I 1 Node f : I 6 + I 2 = I 1 (I 5 + I 3 ) + I 2 = I 3 + I 2 + I 5  no extra information ! Now everything is in terms of I 2, I 3 and I 5 It is easy to eliminate I 3 : Loop bdb: 3I 4 + 4I 5 – 2I 3 =0 3(I 2 +I 5 ) + 4I 5 – 2I 3 = 0  I 3 = ½ (3I 2 + 7I 5 )

24. 24 a b c de f I1I1 I1I1 I3I3 I4I4 I5I5 I6I6 I6I6 I2I2 All currents in terms of I 2 and I 5 : I 1 = ½ (5I 2 + 9I 5 ) I 3 = ½ (3I 2 + 7I 5 ) I 4 = I 2 + I 5 I 6 = ½ (3I 2 + 9I 5 ) Loop fabcf:-12 + 2I 1 + 3I 4 – 5I 2 = 0 -12 + 5I 2 + 9I 5 + 3I 2 + 3I 5 – 5I 2 = 0 12I 5 + 3I 2 = 12 Loop fcdef-5I 2 + 4I 5 + I 6 – 18 = 0 -5I 2 + 4I 5 + 3/2 I 2 + 9/2 I 2 – 18 = 0 13I 5 – 7I 2 = 36 Two equations, two unknowns, can solve, get I 2 = - 2.24 A