1. Today’s agenda: Resistors in Series and Parallel. You must be able to calculate currents and voltages in circuit components in series and in parallel. Kirchoff’s Rules. You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit components that are not simply in series or in parallel.

2. Resistances in Circuits There are “two” ways to connect circuit elements. Series: A B Put your finger on the wire at A. If you can move along the wires to B without ever having a choice of which wire to follow, the circuit components are connected in series. Truth in advertising: it is possible to have circuit elements that are connected neither in series nor in parallel. See problem 24.73 in the 12 th edition of our text for an example with capacitors.

3. Parallel: AB Put your finger on the wire at A. If in moving along the wires to B you ever have a choice of which wire to follow, the circuit components are connected in parallel.* *Truth in advertising: actually, the circuit components are not connected in series, and may be connected in parallel. ???

4. Are these resistors in series or parallel? It matters where you put the source of emf. + - V parallel

5. Are these resistors in series or parallel? It matters where you put the source of emf. + - V series

6. If resistors “see” the same potential difference, they are in parallel. If resistors “see” the same current, they are in series. It’s difficult to come up with a simple one- or two-sentence rule for series/parallel. + - V series + - V parallel I V

7. Here’s a circuit with three resistors and a battery: R3R3 R2R2 R1R1 + - V I Current flows… …in the steady state, the same current flows through all resistors… III …there is a potential difference (voltage drop) across each resistor. V1V1 V3V3 V2V2

8. Applying conservation of energy allows us to calculate the equivalent resistance of the series resistors. I am including the derivation in these notes, for the benefit of students who want to look at it. In lecture, I will skip ahead past the derivation.skip ahead

9. An electric charge q is given a potential energy qV by the battery. R3R3 R2R2 R1R1 + - V I III V1V1 V3V3 V2V2 As it moves through the circuit, the charge loses potential energy qV 1 as it passes through R 1, etc. The charge ends up where it started, so the total energy lost must equal the initial potential energy input: qV = qV 1 + qV 2 + qV 3.

10. V = V 1 + V 2 + V 3 V = IR 1 + IR 2 + IR 3 Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in series. R3R3 R2R2 R1R1 + - V I III V1V1 V3V3 V2V2 qV = qV 1 + qV 2 + qV 3

11. R eq + - V I V I As above:V = IR eq From before:V = IR 1 + IR 2 + IR 3 Combining:IR eq = IR 1 + IR 2 + IR 3 R eq = R 1 + R 2 + R 3 For resistors in series, the total resistance is the sum of the separate resistances.

12. We can generalize this to any number of resistors: (resistors in series) a consequence of conservation of energy R3R3 R2R2 R1R1 + - V Note: for resistors in parallel, R eq is always greater than any of the R i.

13. V V V R3R3 R2R2 R1R1 + - V I Current flows… …different currents flows through different resistors… …but the voltage drop across each resistor is the same. I3I3 I1I1 I2I2 Here’s another circuit with three resistors and a battery.

14. Applying conservation of charge allows us to calculate the equivalent resistance of the parallel resistors. I am including the derivation in these notes, for the benefit of students who want to look at it. In lecture, I will skip ahead past the derivation.skip ahead

15. V V V R3R3 R2R2 R1R1 + - V I I3I3 I1I1 I2I2 AB In the steady state, the current I “splits” into I 1, I 2, and I 3 at point A. I I 1, I 2, and I 3 “recombine” to make a current I at point B. Therefore, the net current flowing out of A and into B is I = I 1 + I 2 + I 3. Because the voltage drop across each resistor is V:

16. Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in parallel. V R eq + - V I I AB I From above, I = I 1 + I 2 + I 3, and So that

17. Dividing both sides by V gives We can generalize this to any number of resistors: (resistors in parallel) a consequence of conservation of charge Note: for resistors in parallel, R eq is always less than any of the R i.

18. Summary: Series A B same I, V’s add Parallel AB same V, I’s add “just like” capacitors “just like” capacitors NOT