1. The “bite-sized chunk” technique is a good one for any problem. Break a complex system into simple components and do them one at a time. Your problem-solving prowess will be tested most severely when you encounter a problem for which this technique won’t work. We are about to encounter such a problem, in an electrical circuit. Due to the lack of time, I am going to once again set a bad example and just “cover” the technique.

2. Kirchoff’s Rules No, it is pronounced “KIRKOFF’s” rules. The ch sounds like “k,” not like “ch.” Analyze this circuit for me, please. Find the currents I 1, I 2, and I 3. Be flexible: I could give I 1 instead of  2, and ask for I 2, I 3, &  2. 1  1 = 80 V 1  2 = 45 V 20  30  40  a b c d e g f h I3I3 I2I2 I1I1

3. 1  1 = 80 V 1  2 = 45 V 20  40  a b c d e g f h I3I3 I2I2 I1I1 I see two sets of resistors in series.This.And this. You know how to analyze those. series 2 series 1 Further analysis is difficult. For example, series 1 seems to be in parallel with the 30  resistor, but what about  2 ? You don’t know how to analyze that. 30 

4. A new technique is needed to analyze this, and far more complex circuits. Kirchoff’s Rules Kirchoff’s Junction Rule: at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction. Also called Kirchoff’s First Rule.* Kirchoff’s Loop Rule: the sum of the changes of potential around any closed path of a circuit must be zero. Also called Kirchoff’s Second Rule. ** *This is just conservation of charge: charge in = charge out. **This is just conservation of energy: a charge ending up where it started out neither gains nor loses energy (E i = E f ).

5. Solving Problems with Kirchoff’s Rules Just as we had a litany for force problems in our Mechanics section, we have a litany for circuit problems. Litany for Circuit Problems 1. Draw the circuit (if not already done). 2. Label + and – for each battery (the short side is -). 3. Label the current in each branch of the circuit with a symbol and an arrow. You may choose whichever direction you wish for the arrow. 4. Apply Kirchoff’s Junction Rule at each junction. The direction of the current arrows tell you whether current is flowing in (+) or out (-). Step 4 will probably give you fewer equations than variables. Proceed to step 5 go get additional equations.

6. 5. Apply Kirchoff’s Loop Rule for as many loops as necessary to get enough equations to solve for your unknowns. Follow each loop in one direction only— your choice. 5a. For a resistor, the sign of the potential difference is negative if your chosen loop direction is the same as the chosen current direction through that resistor; positive if opposite. 5b. For a battery, the sign of the potential difference is positive if your chosen loop direction moves from the negative terminal towards the positive; negative if opposite. 6. Collect equations, solve, and check results.

7. We need a shortened version of the litany for quick reference. 1. Draw the circuit. 2. Label + and – for each battery. 3. Label the current in each branch of the circuit with a symbol and an arrow. 4. Apply Kirchoff’s Junction Rule at each junction. Current in is +. Brief litany for Circuit Problems 5. Apply Kirchoff’s Loop Rule for as many loops as necessary. Follow each loop in one direction only. 5a. Resistor: + - I loop V is - 5b. Battery: loop V is + 6. Solve.

8. 1  1 = 80 V 1  2 = 45 V 20  40  a b c d e g f h I3I3 I2I2 I1I1 Back to our circuit: we have 3 unknowns (I 1, I 2, and I 3 ), so we will need 3 equations. We begin with the junctions. Junction a:I 3 – I 1 – I 2 = 0--eq. 1 Junction d: -I 3 + I 1 + I 2 = 0 Junction d gave no new information, so we still need two more equations. 30  ad

9. 1  1 = 80 V 1  2 = 45 V 20  40  a b c d e g f h I3I3 I2I2 I1I1 There are three loops. Loop 1.Loop 2.Loop 3. Any two loops will produce independent equations. Using the third loop will provide no new information. 30 

10. Reminders: I loop V is - + - loop V is + The “green” loop: (- 30 I 1 ) + (+45) + (-1 I 3 ) + (- 40 I 3 ) = 0 - 30 I 1 + 45 - 41 I 3 = 0 --eq. 2 The “blue” loop: (+ 40 I 3 ) + ( +1 I 3 ) + (-45) + (+20 I 2 ) + (+1 I 2 ) + (-85) = 0 41 I 3 -130 + 21 I 2 = 0 --eq. 3 Three equations, three unknowns; the rest is “algebra.” Make sure to use voltages in V and resistances in . Then currents will be A.

11. Collect our three equations: I 3 – I 1 – I 2 = 0 - 30 I 1 + 45 - 41 I 3 = 0 41 I 3 -130 + 21 I 2 = 0 Rearrange to get variables in “right” order: – I 1 – I 2 + I 3 = 0 - 30 I 1 - 41 I 3 + 45 = 0 21 I 2 + 41 I 3 -130 = 0 Use the middle equation to eliminate I 1 : I 1 = (41 I 3 – 45)/(-30) There are many valid sets of steps to solving a system of equations. Any that works is acceptable.

12. Two equations left to solve: – (41 I 3 – 45)/(-30) – I 2 + I 3 = 0 21 I 2 + 41 I 3 -130 = 0 Might as well work out the numbers: 1.37 I 3 – 1.5 – I 2 + I 3 = 0 21 I 2 + 41 I 3 -130 = 0 – I 2 + 2.37 I 3 – 1.5 = 0 21 I 2 + 41 I 3 -130 = 0 Multiply the top equation by 21: – 21 I 2 + 49.8 I 3 – 31.5 = 0 21 I 2 + 41 I 3 -130 = 0

13. Add the two equations to eliminate I 2 : – 21 I 2 + 49.8 I 3 – 31.5 = 0 + ( 21 I 2 + 41 I 3 -130 = 0 ) 90.8 I 3 – 161.5 = 0 Solve for I 3 : I 3 = 161.5 / 90.8 I 3 = 1.78 Go back to the “middle equation” two slides ago for I 1 : I 1 = (41 I 3 – 45)/(-30) I 1 = - 1.37 I 3 + 1.5 I 1 = - (1.37) (1.78) + 1.5 I 1 = - 0.94

14. Go back two slides to get an equation that gives I 2 : – I 2 + 2.37 I 3 – 1.5 = 0 I 2 = 2.37 I 3 – 1.5 I 2 = (2.37) (1.78) – 1.5 I 2 = 2.72 Summarize answers so your lazy prof doesn’t have to go searching for them and get irritated (don’t forget to show units in your answer): I 2 = 2.72 A I 3 = 1.78 A I 1 = - 0.94 A These don’t look “quite like” the text answer. They rounded to 2 digits. Maybe that’s why. Better check anyway.

15. I 3 – I 1 – I 2 = 0 - 30 I 1 + 45 - 41 I 3 = 0 41 I 3 -130 + 21 I 2 = 0 I 2 = 2.72 A I 3 = 1.78 A I 1 = - 0.94 A 1.78 – (-0.94) – 2.72 = 0  - 30 (-0.94) + 45 - 41 (1.78) = 0.22  ? 41 (1.78) -130 + 21 (2.72) = 0.10  ? Are the last two indication of a mistake or just roundoff error? Recalculating while retaining 2 more digits gives I 1 =0.933, I 2 =2.714, I 3 =1.7806, and the last two results are 0.01 or less  roundoff was the culprit.

16. Assorted Topics …I probably won’t have time to get to. You can connect batteries in series and parallel, just like resistors. Voltages add for batteries in series. Not so for batteries in parallel—use Kirchoff’s rules to analyze. You can connect a battery in series the “wrong way” with other batteries—useful for charging a rechargeable. Even a relatively simple circuit can easily produce a 5x5 or much greater system of equations to be solved. See here for an example solved homework problem. Good luck solving a problem like this on an exam!here

17. Go to www.howstuffworks.com to see how batteries work.www.howstuffworks.comhow batteries work They even expose the secret of the 9 volt battery! Click on the picture above only if you are mature enough to handle this graphic exposé.

18. Go to www.howstuffworks.com to see how batteries work.www.howstuffworks.comhow batteries work They even expose the secret of the 9 volt battery! Shocking! Six 1.5 V batteries in series!

19. You can also connect capacitors in series and in parallel. C1C1 C2C2 C2C2 + - V a C eq =  C i (capacitors in parallel) C1C1 C2C2 + - V C3C3 (capacitors in series) Look familiar?