Download presentation

Presentation is loading. Please wait.

Published byBruno Parks Modified about 1 year ago

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

2
Copyright © 2009 Pearson Education, Inc. Electric circuit needs battery or generator to produce current – these are called sources of emf. Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf: 26-1 EMF and Terminal Voltage

3
Copyright © 2009 Pearson Education, Inc. This resistance behaves as though it were in series with the emf. 26-1 EMF and Terminal Voltage

4
Copyright © 2009 Pearson Education, Inc. 26-1 EMF and Terminal Voltage Example 26-1: Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, V ab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.

5
Copyright © 2009 Pearson Education, Inc. A series connection has a single path from the battery, through each circuit element in turn, then back to the battery. 26-2 Resistors in Series and in Parallel

6
Copyright © 2009 Pearson Education, Inc. The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage: 26-2 Resistors in Series and in Parallel

7
Copyright © 2009 Pearson Education, Inc. From this we get the equivalent resistance (that single resistance that gives the same current in the circuit): 26-2 Resistors in Series and in Parallel

8
Copyright © 2009 Pearson Education, Inc. A parallel connection splits the current; the voltage across each resistor is the same: 26-2 Resistors in Series and in Parallel

9
Copyright © 2009 Pearson Education, Inc. The total current is the sum of the currents across each resistor: 26-2 Resistors in Series and in Parallel,

10
Copyright © 2009 Pearson Education, Inc. This gives the reciprocal of the equivalent resistance: 26-2 Resistors in Series and in Parallel

11
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. 26-2 Resistors in Series and in Parallel

12
Copyright © 2009 Pearson Education, Inc. 26-2 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.

13
Copyright © 2009 Pearson Education, Inc. 26-2 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).

14
Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example 26-4: Circuit with series and parallel resistors. How much current is drawn from the battery shown?

15
Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example 26-5: Current in one branch. What is the current through the 500-Ω resistor shown? (Note: This is the same circuit as in the previous problem.) The total current in the circuit was found to be 17 mA.

16
Copyright © 2009 Pearson Education, Inc. 26-2 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.

17
Copyright © 2009 Pearson Education, Inc. 26-2 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?

18
Copyright © 2009 Pearson Education, Inc. 26-2 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?

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. 26-3 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. 26-3 Kirchhoff’s Rules

21
Copyright © 2009 Pearson Education, Inc. Loop rule: The sum of the changes in potential around a closed loop is zero. 26-3 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. 26-3 Kirchhoff’s Rules

23
Copyright © 2009 Pearson Education, Inc. 26-3 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
Copyright © 2009 Pearson Education, Inc. EMFs in series in the same direction: total voltage is the sum of the separate voltages. 26-4 Series and Parallel EMFs; Battery Charging

25
Copyright © 2009 Pearson Education, Inc. EMFs in series, opposite direction: total voltage is the difference, but the lower- voltage battery is charged. 26-4 Series and Parallel EMFs; Battery Charging

26
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. 26-4 Series and Parallel EMFs; Battery Charging

27
Copyright © 2009 Pearson Education, Inc. 26-4 Series and Parallel EMFs; Battery Charging Example 26-10: Jump starting a car. A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and internal resistance 0.020 Ω. Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 Ω. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor can be represented as a resistor R s = 0.15 Ω. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected.

28
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. 26-5 Circuits Containing Resistor and Capacitor (RC Circuits)

29
Copyright © 2009 Pearson Education, Inc. 26-5 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:

30
Copyright © 2009 Pearson Education, Inc. 26-5 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:

31
Copyright © 2009 Pearson Education, Inc. 26-5 Circuits Containing Resistor and Capacitor (RC Circuits) The current at any time t can be found by differentiating the charge:

32
Copyright © 2009 Pearson Education, Inc. 26-5 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.

33
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)

34
Copyright © 2009 Pearson Education, Inc. 26-5 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

35
Copyright © 2009 Pearson Education, Inc. 26-5 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?

36
Copyright © 2009 Pearson Education, Inc. 26-5 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.

37
Copyright © 2009 Pearson Education, Inc. 26-5 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.

38
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

39
Copyright © 2009 Pearson Education, Inc. A person receiving a shock has become part of a complete circuit. 26-6 Electric Hazards

40
Copyright © 2009 Pearson Education, Inc. Faulty wiring and improper grounding can be hazardous. Make sure electrical work is done by a professional. 26-6 Electric Hazards

41
Copyright © 2009 Pearson Education, Inc. The safest plugs are those with three prongs; they have a separate ground line. Here is an example of household wiring – colors can vary, though! Be sure you know which is the hot wire before you do anything. 26-6 Electric Hazards

42
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. 26-7 Ammeters and Voltmeters

43
Copyright © 2009 Pearson Education, Inc. 26-7 Ammeters and Voltmeters Example 26-15: Ammeter design. Design an ammeter to read 1.0 A at full scale using a galvanometer with a full-scale sensitivity of 50 μA and a resistance r = 30 Ω. Check if the scale is linear.

44
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. 26-7 Ammeters and Voltmeters

45
Copyright © 2009 Pearson Education, Inc. 26-7 Ammeters and Voltmeters Example 26-16: Voltmeter design. Using a galvanometer with internal resistance 30 Ω and full-scale current sensitivity of 50 μA, design a voltmeter that reads from 0 to 15 V. Is the scale linear?

46
Copyright © 2009 Pearson Education, Inc. An ohmmeter measures resistance; it requires a battery to provide a current. 26-7 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. 26-7 Ammeters and Voltmeters

48
Copyright © 2009 Pearson Education, Inc. 26-7 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. Review Questions

50
ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? A) 12 V B) zero C) 3 V D) 4 V E) you need to know the actual value of R

51
equal evenly 3 V drop Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? A) 12 V B) zero C) 3 V D) 4 V E) you need to know the actual value of R Follow-up: What would be the potential difference if R = Follow-up: What would be the potential difference if R = 1

52
ConcepTest 26.1bSeries Resistors II 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 ? A) 12 V B) zero C) 6 V D) 8 V E) 4 V

53
ConcepTest 26.1bSeries Resistors II 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 ? A) 12 V B) zero C) 6 V D) 8 V E) 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?

54
ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5 R 2 = 2 A) 10 A B) zero C) 5 A D) 2 A E) 7 A

55
voltagesame V 1 = I 1 R 1 I 1 = 2 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s law, V 1 = I 1 R 1 to find the current I 1 = 2 A. ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5 R 2 = 2 A) 10 A B) zero C) 5 A D) 2 A E) 7 A Follow-up: What is the total current through the battery?

56
ConcepTest 26.2bParallel Resistors II A) increases B) remains the same C) decreases D) 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?

57
ConcepTest 26.2bParallel Resistors II A) increases B) remains the same C) decreases D) 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?

58
ConcepTest 26.3aShort Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb A) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb B) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire C) all the current flows through the wire none of the above D) none of the above

59
zeroALL The current divides based on the ratio of the resistances. If one of the resistances is zero, then ALL of the current will flow through that path. ConcepTest 26.3aShort Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb A) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb B) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire C) all the current flows through the wire none of the above D) none of the above Follow-up: Doesn’t the wire have SOME resistance?

60
ConcepTest 26.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before A) glow brighter than before glow just the same as before B) glow just the same as before glow dimmer than before C) glow dimmer than before D) go out completely E) explode

61
total resistance of the circuit decreasescurrent through bulb A increases Since bulb B is bypassed by the wire, the total resistance of the circuit decreases. This means that the current through bulb A increases. ConcepTest 26.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before A) glow brighter than before glow just the same as before B) glow just the same as before glow dimmer than before C) glow dimmer than before D) go out completely E) explode Follow-up: What happens to bulb B?

62
ConcepTest 26.4aCircuits I circuit I A) circuit I circuit II B) circuit II both the same C) both the same it depends on R D) it depends on R The lightbulbs in the circuits below are identical with the same resistance R. Which circuit produces more light? (brightness power)

63
ConcepTest 26.4aCircuits I circuit I A) circuit I circuit II B) circuit II both the same C) both the same it depends on R D) it depends on R The lightbulbs in the circuits below are identical with the same resistance R. Which circuit produces more light? (brightness power) parallellowering the total resistance I willdraw a higher current P = IV In circuit I, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit I will draw a higher current, which leads to more light, because P = IV.

64
ConcepTest 26.4bCircuits II twice as much A) twice as much the same B) the same 1/2 as much C) 1/2 as much 1/4 as much D) 1/4 as much 4 times as much E) 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)

65
ConcepTest 26.4bCircuits II twice as much A) twice as much the same B) the same 1/2 as much C) 1/2 as much 1/4 as much D) 1/4 as much 4 times as much E) 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?

66
ConcepTest 26.5aMore Circuits I increase A) increase decrease B) decrease stay the same C) stay the same What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: V R1R1 R3R3 R2R2 S

67
ConcepTest 26.5aMore Circuits I increase A) increase decrease B) decrease stay the same C) stay the same What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: decreases the equivalent resistancecurrent from the battery increases increase in the voltage across R 1 With the switch closed, the addition of R 2 to R 3 decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R 1. V R1R1 R3R3 R2R2 S Follow-up: ? Follow-up: What happens to the current through R 3 ?

68
ConcepTest 26.5bMore Circuits II increases A) increases decreases B) decreases stays the same C) stays the same V R1R1 R3R3 R4R4 R2R2 S What happens to the voltage across the resistor R 4 when the switch is closed?

69
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 A) increases decreases B) decreases stays the same C) stays the same ConcepTest 26.5bMore Circuits II Follow-up: ? Follow-up: What happens to the current through R 4 ?

70
ConcepTest 26.6ircuits ConcepTest 26.6Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal. V R1R1 R2R2 R3R3 R5R5 R4R4 A) R 1 and B) both R 1 and R 2 equally and C) R 3 and R 4 D) R 5 E) all the same

71
I 1 = I 2 R 5 R 3 + R 4, branch containing R 5 has less resistance The same current must flow through the left and right combinations of resistors. On the LEFT, the current splits equally, so I 1 = I 2. On the RIGHT, more current will go through R 5 than R 3 + R 4, since the branch containing R 5 has less resistance. ConcepTest 26.6ircuits ConcepTest 26.6Even More Circuits A) R 1 and B) both R 1 and R 2 equally and C) R 3 and R 4 D) R 5 E) all the same Which resistor has the greatest current going through it? Assume that all the resistors are equal. V R1R1 R2R2 R3R3 R5R5 R4R4 Follow-up: ? Follow-up: Which one has the smallest voltage drop?

72
ConcepTest 26.7Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? A) the power B) the current C) the voltage D) both (A) and (B) E) both (B) and (C)

73
increases the resistancedecreases the current The voltage is provided at 120 V from the outside. The light dimmer increases the resistance and therefore decreases the current that flows through the lightbulb. ConcepTest 26.7Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? A) the power B) the current C) the voltage D) both (A) and (B) E) both (B) and (C) Follow-up: Why does the voltage not change?

74
ConcepTest 26.8aLightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? A) the 25 W bulb B) the 100 W bulb C) both have the same D) this has nothing to do with resistance

75
P = V 2 / R, lower power ratinghigher resistance Since P = V 2 / R, the bulb with the lower power rating has to have the higher resistance. ConcepTest 26.8aLightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? A) the 25 W bulb B) the 100 W bulb C) both have the same D) this has nothing to do with resistance Follow-up: Which one carries the greater current?

76
ConcepTest 26.8bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? A) heater 1 B) heater 2 C) both equally

77
P = V 2 / R, smaller resistance larger power Using P = V 2 / R, the heater with the smaller resistance will have the larger power output. Thus, heater 2 will give off more heat. ConcepTest 26.8bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? A) heater 1 B) heater 2 C) both equally Follow-up: Which one carries the greater current?

78
ConcepTest 26.9Junction Rule ConcepTest 26.9 Junction Rule A) 2 A B) 3 A C) 5 A D) 6 A E) 10 A 5 A 8 A 2 A P What is the current in branch P?

79
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 A) 2 A B) 3 A C) 5 A D) 6 A E) 10 A What is the current in branch P? ConcepTest 26.9Junction Rule

80
ConcepTest 26.10Kirchhoff’s Rules ConcepTest 26.10 Kirchhoff’s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? A) both bulbs go out B) intensity of both bulbs increases C) intensity of both bulbs decreases D) A gets brighter and B gets dimmer E) nothing changes

81
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? A) both bulbs go out B) intensity of both bulbs increases C) intensity of both bulbs decreases D) A gets brighter and B gets dimmer E) nothing changes ConcepTest 26.10Kirchhoff’s Rules ConcepTest 26.10 Kirchhoff’s Rules 24 V Follow-up: Follow-up: What happens if the bottom battery is replaced by a 24 V battery?

82
ConcepTest 26.11Wheatstone Bridge A) l B) l/2 C) l/3 D) l/4 E) zero An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is: I V

83
resistors are identical voltage drops are the same potentialsab same Since all resistors are identical, the voltage drops are the same across the upper branch and the lower branch. Thus, the potentials at points a and b are also the same. Therefore, no current flows. ConcepTest 26.11Wheatstone Bridge A) l B) l/2 C) l/3 D) l/4 E) zero An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is: I V

84
ConcepTest 26.12Kirchhoff’s Rules ConcepTest 26.12 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? A) 2 – I 1 – 2I 2 = 0 B) 2 – 2I 1 – 2I 2 – 4I 3 = 0 C) 2 – I 1 – 4 – 2I 2 = 0 D) I 3 – 4 – 2I 2 + 6 = 0 E) 2 – I 1 – 3I 3 – 6 = 0

85
ConcepTest 26.12Kirchhoff’s Rules ConcepTest 26.12 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? A) 2 – I 1 – 2I 2 = 0 B) 2 – 2I 1 – 2I 2 – 4I 3 = 0 C) 2 – I 1 – 4 – 2I 2 = 0 D) I 3 – 4 – 2I 2 + 6 = 0 E) 2 – I 1 – 3I 3 – 6 = 0

86
Copyright © 2009 Pearson Education, Inc. Ch. 26 Homework Finish reading Ch. 26. Begin looking over Ch. 27. Group problems #’s 6, 8, 18 Homework problems #’s 3, 7, 11, 19, 31, 33, 45

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google