1. DC Electrical Circuits Chapter 28 Electromotive Force Potential Differences Resistors in Parallel and Series Circuits with Capacitors

2. Resistors in Series The pair of resistors, R 1 and R 2, can be replaced by a single equivalent resistor R; one which, given I, has the same total voltage drop as the original pair. I R1R1 R2R2 ab V1V1 V2V2 c Note: the current I is the same, anywhere between a and b, but there is a voltage drop V 1 across R 1, and a voltage drop V 2 across R 2.

3. Resistors in Series V = V 1 +V 2 = I R 1 +I R 2 R1R1 R2R2 ab V1V1 V2V2 I The pair of resistors can be replaced by a single equivalent resistor R eq ; one which, given I, has the same total voltage drop as the original pair.

4. Resistors in Series V = V 1 +V 2 = I R 1 +I R 2 We want to write this as V= I R eq R1R1 R2R2 ab V1V1 V2V2 I The pair of resistors can be replaced by a single equivalent resistor R eq ; one which, given I, has the same total voltage drop as the original pair.

5. Resistors in Series V = V 1 +V 2 = I R 1 +I R 2 We want to write this as V= I R eq hence R eq = R 1 + R 2 R1R1 R2R2 ab V1V1 V2V2 I The pair of resistors can be replaced by a single equivalent resistor R eq ; one which, given I, has the same total voltage drop as the original pair.

6. Resistors in Series R1R1 R2R2 ab V1V1 V2V2 I What is the voltage drop across each resistor ? V 1 = I R 1 but I = V / (R 1 + R 2 )  V 1 = V R 1 / (R 1 + R 2 ) V 2 = I R 2 but I = V / (R 1 + R 2 )  V 2 = V R 2 / (R 1 + R 2 )

7. Resistors in Parallel Again find the equivalent single resistor which has the same V if I is given. ab R1R1 R2R2 I I1I1 I2I2 V

8. Resistors in Parallel Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1 +I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2 ) ab R1R1 R2R2 I I1I1 I2I2 V

9. Resistors in Parallel Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1 +I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2 ) We want to write this as: I = V / R eq ab R1R1 R2R2 I I1I1 I2I2 V

10. Resistors in Parallel Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1 +I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2 ) We want to write this as: I = V / R eq Hence 1 / R eq =1/ R 1 +1/ R 2 ab R1R1 R2R2 I I1I1 I2I2 V

11. Parallel and Series ResistorsCapacitors Parallel 1/R=1/R 1 +1/R 2 C=C 1 +C 2 Series R=R 1 +R 2 1/C=1/C 1 +1/C 2

12. Batteries and Generators Current is produced by applying a potential difference across a conductor (I=V/R) [This is not equilibrium so there is an electric field inside the conductor]. This potential difference is set up by some source, such as a battery or generator [that generates charges, from some other type of energy, i.e. chemical, solar, mechanical]. Conventionally an “applied voltage” is given the symbol E (units: volts). For historical reasons, this applied voltage is often called the “electromotive force” (emf) [even though it’s not a force].

13. The Voltaic Pile +- Volta’s original battery Ag Zn wet cloth electrical converter........converts chemical energy to electrical energy Mixture of Ammonium Chloride& Manganese Dioxide Carbon Zinc case + -

14. Electrical Description of a Battery A battery does work on positive charges in moving them to higher potential (inside the battery). The EMF E, most precisely, is the work per unit charge exerted to move the charges “uphill” (to the + terminal, inside)... but you can just think of it as an “applied voltage.” Current will flow, in the external circuit, from the + terminal, to the – terminal, of the battery. + - symbol for battery symbol for resistance R E I

15. Resistors in Series Given R eq = R 1 + R 2, the current is I= E /(R 1 + R 2 ) One can then work backwards to get the voltage across each resistor: + - E R2R2 I R1R1

16. Resistors in Series Given R eq = R 1 + R 2, the current is I= E /(R 1 + R 2 ) One can then work backwards to get the voltage across each resistor: + - E R2R2 I R1R1

17. The Loop Method + - E R2R2 I R1R1 Go around the circuit in one direction. If you pass a voltage source from – to +,  the voltage increases by E (or V). As you pass a resistor the voltage decreases by V = I R. The total change in voltage after a complete loop is zero.

18. Analyzing Resistor Networks 6  + E - 2  1. Replace resistors step by step. + E - 3  2  + - E R2R2 I R1R1 2. The loop method E – IR 1 – IR 2 = 0 + E - 55 I I = E / R E = I (R 1 + R 2 ) I = E / (R 1 + R 2 )

19. Analyzing Resistor Networks 6  + E - 2  Often you can replace sets of resistors step by step.

20. Analyzing Resistor Networks 6  + E - 2  Often you can replace sets of resistors step by step. 1/6+1/6=1/(3) + E - 3  2  step 1

21. Analyzing Resistor Networks 6  + E - 2  Often you can replace sets of resistors step by step. 1/6+1/6=1/(3) + E - 3  2  + E - 55 3 + 2 = 5 step 1 step 2

22. Internal Resistance of a Battery One important point: batteries actually have an internal resistance r Often we neglect this, but sometimes it is significant. + - R r E battery

23. Effect of Internal Resistance Analyzed by the Loop Method Start at any point in the circuit. Go around the circuit in a loop. Add up (subtract) the potential differences across each element (keep the signs straight!). E - Ir - IR = 0 (using V=IR)  I = E / (R + r) V R = I R = E R / (R + r) = E / [1 + (r/R)] if r << R  V R = E + - r E R I

24. Kirchhoff’s Laws 1. At any circuit junction, currents entering must equal currents leaving. Kirchoff devised two laws that are universally applicable in circuit analysis: 2. Sum of all  V’s across all circuit elements in a loop must be zero. I1I1 I2I2 I 3 = I 1 + I 2 + - r E R I E - Ir - IR = 0