1 §18.1 Electric Current e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- A metal wire. Assume electrons flow to the right. Current is a measure of the amount of charge that passes though an area perpendicular to the flow of charge. Current units: 1C/sec = 1 amp

2 A current will flow until there is no potential difference. The direction of current flow in a wire is opposite the flow of the electrons. (In the previous drawing the current is to the left.)

3 §18.2 EMF and Circuits An ideal battery maintains a constant potential difference. This potential difference is called the battery’s EMF(  ). The work done by an ideal battery in pumping a charge q is W = q .

4 The circuit symbol for a battery (EMF source) is +  At high potential At low potential Batteries do work by converting chemical energy into electrical energy. A battery dies when it can no longer sustain its chemical reactions and so can do no more work to move charges.

5 §18.3 Microscopic View of Current in a Metal Electrons in a metal might have a speed of ~10 6 m/s, but since the direction of travel is random, an electron has v drift = 0.

6 Only when the ends of a wire are at different potentials (E  0) will there be a net flow of electrons along the wire (v drift  0). Typically, v drift < 1 mm/sec.

7 Calculate the number of charges (N e ) that pass through the shaded region in a time  t: The current in the wire is: l

8 §18.4 Resistance and Resistivity A material is considered ohmic if  V  I, where The proportionality constant R is called resistance and is measured in ohms (  ; and 1  = 1 V/A).

9 §18.5 Kirchhoff’s Rules Junction rule: The current that flows into a junction is the same as the current that flows out. (Charge is conserved) A junction is a place where two or more wires (or other components) meet. Loop rule: The sum of the voltage dropped around a closed loop is zero. (Energy is conserved.)

10 For a resistor: If you cross a resistor in the direction of the current flow, the voltage drops by an amount IR (write as  IR). There is a voltage rise if you cross the other way (write as +IR). B A I If the current flows from A to B, then the potential decreases from A to B. The potential difference between A and B is < 0 (  V =  IR).

11 For batteries (or other sources of EMF): If you move from the positive to the negative terminal the potential drops by  (write as  ). The potential rises if you cross in the other direction (write as +  ). +  At high potential At low potential 

12 A current will only flow around a closed loop. B V AB is the terminal voltage. A Applying the loop rule:

13 In a circuit, if the current always flows in the same direction it is called a direct current (DC) circuit.

14 § 18.6 Series and Parallel Circuits The current through the two resistors is the same. It is not “used up” as it flows around the circuit! These resistors are in series. Apply Kirchhoff’s loop rule: Resistors:

15 The pair of resistors R 1 and R 2 can be replaced with a single equivalent resistor provided that R eq = R 1 + R 2. In general, for resistors in series

16 Current only flows around closed loops. When the current reaches point A it splits into two currents. R 1 and R 2 do not have the same current through them, they are in parallel. Apply Kirchhoff’s loop rule: The potential drop across each resistor is the same.

17 Applying the junction rule at A: I = I 1 + I 2. From the loop rules: Substituting for I 1 and I 2 in the junction rule:

18 In general, for resistors in parallel The pair of resistors R 1 and R 2 can be replaced with a single equivalent resistor provided that

19 Example (text problem 18.40): In the given circuit, what is the total resistance between points A and B? R 2 and R 3 are in parallel. Replace with an equivalent resistor R 23. R 3 = 24  R 2 = 12  R 1 = 15  A B

20 R 23 = 8  R 1 = 15  A B The resistors R 23 and R 1 are in series: The circuit can now be redrawn: B R 123 = 23  A Is the equivalent circuit and the total resistance is 23 . Example continued:

21 Capacitors: C2C2 C1C1  For capacitors in series the charge on the plates is the same. Apply Kirchhoff’s loop rule:

22 In general, for capacitors in series The pair of capacitors C 1 and C 2 can be replaced with a single equivalent capacitor provided that

23 C2C2 C1C1  For capacitors in parallel the charge on the plates may be different. Here Apply Kirchhoff’s loop rule:

24 In general, for capacitors in parallel The pair of capacitors C 1 and C 2 can be replaced with a single equivalent capacitor provided that C eq = C 1 + C 2.

25 Example (text problem 18.49): Find the value of a single capacitor that replaces the three in the circuit below if C 1 = C 2 = C 3 = 12  F. C2C2 C1C1 C3C3 B A C 2 and C 3 are in parallel

26 The circuit can be redrawn: C1C1 C 23 B A The remaining two capacitors are in series. C 123 B A Is the final, equivalent circuit. Example continued:

27 § 18.7 Circuit Analysis Using Kirchhoff’s Rules To solve multiloop circuit problems: 1.Assign polarity (+/  ) to all EMF sources. 2.Assign currents to each branch of the circuit. 3.Apply Kirchhoff’s rules. 4.Solve for the unknowns.

28 Example (text problem 18.53): Find the three unknown currents (the current in each resistor). I3I3 I2I2 I1I1  + +  R2R2 R3R3 R1R1 22 11

29 Loop EDCFE: Loop AFCBA: Note: Could also use Loop AFEDCBA Junction C: Note: could also use junction F Example continued:

30 The point is to write down three equations for the three unknown currents. (1) (2) (3) Example continued:

31 Substitute (3) into (1): Multiply the top equation by  R 3 and the bottom equation by +R 1, add the equations together, then solve for I 2. (4) (2) Example continued:

32 Substitute I 2 =  amps in to (2): Now substitute the known values of I 2 and I 3 into (3): Example continued:

33 Example continued: The negative sign on I 2 means that instead of the current going from right to left (from point C to point F) in the branch with resistor 2, it really goes from left to right. It is essential to keep the negative sign when evaluating your equations numerically. Make the correction only when the problem is finished.

34 § 18.8 Power and Energy in Circuits The energy dissipation rate is: For an EMF source: For a resistor:

35 Example: Use the results of the example starting on slide 35 to determine the power dissipated by the three resistors in that circuit. Resistance (  ) Current (A)Power (W)

36 § 18.9 Measuring Currents and Voltages Current is measured with an ammeter. An ammeter is placed in series with a circuit component.  R2R2 R1R1 A1A1 A3A3 A2A2 A 1 measures the current through R 1. A 2 measures the current through R 2. An ammeter has a low internal resistance. A 3 measures the current drawn from the EMF.

37 A voltmeter is used to measure the potential drop across a circuit element. It is placed in parallel with the component. A voltmeter has a large internal resistance.  R2R2 R1R1 V The voltmeter measures the voltage drop across R 1.

38 § RC Circuits Close the switch at t = 0 to start the flow of current. The capacitor is being charged. Apply Kirchhoff’s loop rule: C  R Switch +  + 

39 The current I(t) that satisfies Kirchhoff’s loop rule is: where  is the RC time constant and is a measure of the charge (and discharge) rate of a capacitor.

40 The voltage drop across the capacitor is: The voltage drop across the resistor is: The charge on the capacitor is: Note: Kirchhoff’s loop rule must be satisfied for all times.

41 Plots of the voltage drop across the (charging) capacitor and current in the circuit.

42 C  R S1S1 S2S2 +  I While the capacitor is charging S 2 is open. After the capacitor is fully charged S 1 is opened at the same time S 2 is closed: this removes the battery from the circuit. Current will now flow in the right hand loop only, discharging the capacitor. Apply Kirchhoff’s loop rule: The current in the circuit is But the voltage drop across the capacitor is now

43 The voltage drop across the discharging capacitor:

44 Example (text problem 18.85): A capacitor is charged to an initial voltage of V 0 = 9.0 volts. The capacitor is then discharged through a resistor. The current is measured and is shown in the figure.

45 (a) Find C, R, and the total energy dissipated in the resistor. Use the graph to determine . I 0 = 100 mA; the current is I 0 /e = 36.8 mA at t = 13 msec. Since  = V 0 = 9.0 volts, R = 90  and C = 144  F. All of the energy stored in the capacitor is eventually dissipated by the resistor. Example continued:

46 (b) At what time is the energy in the capacitor half of the initial value? Want: Example continued:

47 Solve for t: Example continued:

48 Summary Current & Drift Velocity Resistance & Resistivity Ohm’s Law Kirchhoff’s Rules Series/Parallel Resistors/Capacitors Power Voltmeters & Ammeters RC Circuits