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Copyright © 2009 Pearson Education, Inc. Chapter 26 DC Circuits.

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1 Copyright © 2009 Pearson Education, Inc. Chapter 26 DC Circuits

2 Copyright © 2009 Pearson Education, Inc. This gives the reciprocal of the equivalent resistance: 26-2 Resistors in Series and in Parallel

3 Copyright © 2009 Pearson Education, Inc. An analogy using water may be helpful in visualizing parallel circuits. The water (current) splits into two streams; each falls the same height, and the total current is the sum of the two currents. With two pipes open, the resistance to water flow is half what it is with one pipe open Resistors in Series and in Parallel

4 ConcepTest 26.1Series Resistors 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V

5 ConcepTest 26.1Series Resistors 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2.V 1 = 8 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6  = 2 A, and then use Ohm’s law to get voltages. Follow-up: What happens if the voltage is doubled?

6 ConcepTest 26.2Parallel Resistors 1) increases 2) remains the same 3) decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

7 ConcepTest 26.2Parallel Resistors 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? Follow-up: What happens to the current through each resistor?

8 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The lightbulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current.

9 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Example 26-4: Circuit with series and parallel resistors. How much current is drawn from the battery shown?

10 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-3: An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature).

11 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Example 26-5: Current in one branch. What is the current through the 500-Ω resistor shown?

12 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-6: Bulb brightness in a circuit. The circuit shown has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C ? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.

13 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Example 26-7: A two-speed fan. One way a multiple-speed ventilation fan for a car can be designed is to put resistors in series with the fan motor. The resistors reduce the current through the motor and make it run more slowly. Suppose the current in the motor is 5.0 A when it is connected directly across a 12-V battery. (a) What series resistor should be used to reduce the current to 2.0 A for low-speed operation? (b) What power rating should the resistor have?

14 Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Example 26-8: Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) What is the current in the 6.0-Ω resistor?

15 ConcepTest 26.4Circuits twice as much 1) twice as much the same 2) the same 1/2 as much 3) 1/2 as much 1/4 as much 4) 1/4 as much 4 times as much 5) 4 times as much 10 V The three lightbulbs in the circuit all have the same resistance of 1  By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power)

16 ConcepTest 26.4Circuits twice as much 1) twice as much the same 2) the same 1/2 as much 3) 1/2 as much 1/4 as much 4) 1/4 as much 4 times as much 5) 4 times as much 10 V We can use P = V 2 /R to compare the power: P A = (= 100 W P A = (V A ) 2 /R A = (10 V) 2 /1  = 100 W P B = (= 25 W P B = (V B ) 2 /R B = (5 V) 2 /1  = 25 W The three lightbulbs in the circuit all have the same resistance of 1  By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power) Follow-up: What is the total current in the circuit?

17 ConcepTest 26.5More Circuits increases 1) increases decreases 2) decreases stays the same 3) stays the same V R1R1 R3R3 R4R4 R2R2 S What happens to the voltage across the resistor R 4 when the switch is closed?

18 V R1R1 R3R3 R4R4 R2R2 S A B C increase in the voltage across R 1 V AB constant V AB increasesV BC must decrease We just saw that closing the switch causes an increase in the voltage across R 1 (which is V AB ). The voltage of the battery is constant, so if V AB increases, then V BC must decrease! What happens to the voltage across the resistor R 4 when the switch is closed? increases 1) increases decreases 2) decreases stays the same 3) stays the same ConcepTest 26.5More Circuits Follow-up: ? Follow-up: What happens to the current through R 4 ?

19 Copyright © 2009 Pearson Education, Inc. Some circuits cannot be broken down into series and parallel connections. For these circuits we use Kirchhoff’s rules Kirchhoff’s Rules

20 Copyright © 2009 Pearson Education, Inc. Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it Kirchhoff’s Rules

21 Copyright © 2009 Pearson Education, Inc. Loop rule: The sum of the changes in potential around a closed loop is zero Kirchhoff’s Rules

22 Copyright © 2009 Pearson Education, Inc. Problem Solving: Kirchhoff’s Rules 1. Label each current, including its direction. 2. Identify unknowns. 3. Apply junction and loop rules; you will need as many independent equations as there are unknowns. 4.Solve the equations, being careful with signs. If the solution for a current is negative, that current is in the opposite direction from the one you have chosen Kirchhoff’s Rules

23 Copyright © 2009 Pearson Education, Inc Kirchhoff’s Rules Example 26-9: Using Kirchhoff’s rules. Calculate the currents I 1, I 2, and I 3 in the three branches of the circuit in the figure.

24 ConcepTest 26.9Junction Rule ConcepTest 26.9 Junction Rule 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A 5 A 8 A 2 A P What is the current in branch P?

25 red One exiting branch has 2 A other branch (at P) must have 6 A The current entering the junction in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. 5 A 8 A 2 A P Junction 6 A S 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A What is the current in branch P? ConcepTest 26.9Junction Rule

26 ConcepTest 26.10Kirchhoff’s Rules ConcepTest Kirchhoff’s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes

27 the point between the bulbs is at 12 VBut so is the point between the batteries When the switch is open, the point between the bulbs is at 12 V. But so is the point between the batteries. If there is no potential difference, then no current will flow once the switch is closed!! Thus, nothing changes. The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes ConcepTest 26.10Kirchhoff’s Rules ConcepTest Kirchhoff’s Rules 24 V Follow-up: Follow-up: What happens if the bottom battery is replaced by a 24 V battery?

28 ConcepTest 26.12Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0

29 ConcepTest 26.12Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Eq. 3 is valid for the left loop Eq. 3 is valid for the left loop: The left battery gives +2 V, then there is a drop through a 1  resistor with current I 1 flowing. Then we go through the middle battery (but from + to – !), which gives –4 V. Finally, there is a drop through a 2  resistor with current I 2. Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0

30 Copyright © 2009 Pearson Education, Inc. EMFs in series in the same direction: total voltage is the sum of the separate voltages Series and Parallel EMFs; Battery Charging

31 Copyright © 2009 Pearson Education, Inc. EMFs in series, opposite direction: total voltage is the difference, but the lower- voltage battery is charged Series and Parallel EMFs; Battery Charging

32 Copyright © 2009 Pearson Education, Inc. EMFs in parallel only make sense if the voltages are the same; this arrangement can produce more current than a single emf Series and Parallel EMFs; Battery Charging

33 Copyright © 2009 Pearson Education, Inc. When the switch is closed, the capacitor will begin to charge. As it does, the voltage across it increases, and the current through the resistor decreases Circuits Containing Resistor and Capacitor (RC Circuits)

34 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) To find the voltage as a function of time, we write the equation for the voltage changes around the loop: Since Q = dI/dt, we can integrate to find the charge as a function of time:

35 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) The voltage across the capacitor is V C = Q/C : The quantity RC that appears in the exponent is called the time constant of the circuit:

36 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) The current at any time t can be found by differentiating the charge:

37 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) Example 26-11: RC circuit, with emf. The capacitance in the circuit shown is C = 0.30 μ F, the total resistance is 20 kΩ, and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value, (e) the maximum current, and (f) the charge Q when the current I is 0.20 its maximum value.

38 Copyright © 2009 Pearson Education, Inc. If an isolated charged capacitor is connected across a resistor, it discharges: 26-5 Circuits Containing Resistor and Capacitor (RC Circuits)

39 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) Once again, the voltage and current as a function of time can be found from the charge: and

40 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) Example 26-12: Discharging RC circuit. In the RC circuit shown, the battery has fully charged the capacitor, so Q 0 = C E. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 μF. The current I is observed to decrease to 0.50 of its initial value in 40 μs. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R ? (c) What is Q at t = 60 μs?

41 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) Conceptual Example 26-13: Bulb in RC circuit. In the circuit shown, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later.

42 Copyright © 2009 Pearson Education, Inc Circuits Containing Resistor and Capacitor (RC Circuits) Example 26-14: Resistor in a turn signal. Estimate the order of magnitude of the resistor in a turn-signal circuit.

43 Copyright © 2009 Pearson Education, Inc. Most people can “feel” a current of 1 mA; a few mA of current begins to be painful. Currents above 10 mA may cause uncontrollable muscle contractions, making rescue difficult. Currents around 100 mA passing through the torso can cause death by ventricular fibrillation. Higher currents may not cause fibrillation, but can cause severe burns. Household voltage can be lethal if you are wet and in good contact with the ground. Be careful! 26-6 Electric Hazards

44 Copyright © 2009 Pearson Education, Inc. An ammeter measures current; a voltmeter measures voltage. Both are based on galvanometers, unless they are digital. The current in a circuit passes through the ammeter; the ammeter should have low resistance so as not to affect the current Ammeters and Voltmeters

45 Copyright © 2009 Pearson Education, Inc. A voltmeter should not affect the voltage across the circuit element it is measuring; therefore its resistance should be very large Ammeters and Voltmeters

46 Copyright © 2009 Pearson Education, Inc. An ohmmeter measures resistance; it requires a battery to provide a current Ammeters and Voltmeters

47 Copyright © 2009 Pearson Education, Inc. Summary: An ammeter must be in series with the current it is to measure; a voltmeter must be in parallel with the voltage it is to measure Ammeters and Voltmeters

48 Copyright © 2009 Pearson Education, Inc Ammeters and Voltmeters Example 26-17: Voltage reading vs. true voltage. Suppose you are testing an electronic circuit which has two resistors, R 1 and R 2, each 15 kΩ, connected in series as shown in part (a) of the figure. The battery maintains 8.0 V across them and has negligible internal resistance. A voltmeter whose sensitivity is 10,000 Ω/V is put on the 5.0-V scale. What voltage does the meter read when connected across R 1, part (b) of the figure, and what error is caused by the finite resistance of the meter?

49 Copyright © 2009 Pearson Education, Inc. A source of emf transforms energy from some other form to electrical energy. A battery is a source of emf in parallel with an internal resistance. Resistors in series: Summary of Chapter 26

50 Copyright © 2009 Pearson Education, Inc. Resistors in parallel: Kirchhoff’s rules: 1.Sum of currents entering a junction equals sum of currents leaving it. 2.Total potential difference around closed loop is zero. Summary of Chapter 26

51 Copyright © 2009 Pearson Education, Inc. RC circuit has a characteristic time constant: To avoid shocks, don’t allow your body to become part of a complete circuit. Ammeter: measures current. Voltmeter: measures voltage. Summary of Chapter 26


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