Objects can be charged by rubbing

Transcription

1 Electrostatics

2 Objects can be charged by rubbing

3 Charge comes in two types, positive and negative; like charges repel and opposite charges attract

4 Electric charge is conserved the arithmetic sum of the total charge cannot change in any interaction.

5 Electric Charge in the Atom Atom: Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge)

6 Atom is electrically neutral. Rubbing charges objects by moving electrons from one to the other.

7 Polar molecule: neutral overall, but charge not evenly distributed

8 Insulators and Conductors Conductor: Insulator: Charge flows freely Almost no charge flows Metals Most other materials Some materials are semiconductors.

9 Charging by using Friction It is possible to charge objects by using friction. Some materials have a greater affinity for electrons than other materials and rubbing them together will often transfer electrons from one material to the other. The table shows the triboelectic series. Generally using materials further apart in the series will enable a transfer of a greater amount of charge.

10 Induced Charge; the Electroscope Metal objects can be charged by conduction:

11 They can also be charged by induction:

12 The electroscope can be used for detecting charge:

13 The electroscope can be charged either by conduction or by induction.

14 The charged electroscope can then be used to determine the sign of an unknown charge.

15 Coulomb s Law Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them.

16 Coulomb s law: This equation gives the magnitude of the force.

17 The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same.

18 Unit of charge: coulomb, C The proportionality constant in Coulomb s law is then: k = 9.00x10 9 N m 2 C 2 Charges produced by rubbing are typically around a microcoulomb:

19 The magnitude of the charge on the electron (the elementary charge) : Electric charge is quantized in units of the electron charge. The elementary charge was first measured by Millikan (1909) in his famous oil drop experiment.

20 The proportionality constant k can also be written in terms of, the permittivity of free space:

21 Coulomb s law strictly applies only to point charges. Superposition: for multiple point charges, the forces on each charge from every other charge can be calculated and then added as vectors.

22 The net electrostatic force on a charge is the vector sum of all the electrostatic forces acting on it. When solving Coulomb s law (2-d) problems, use the Coulomb s law formula to calculate a magnitude of force (a + value), then use the principle of like and unlike charges to obtain the direction of the force vectors. At this point one can add the forces to find the net electrostatic force.

23 Example: a) What is the electrostatic force between two electrons nm apart? ( 1 nm=1x10-9 m) b) What acceleration would this cause on each electron if they were free to move? (me= 9.11x10-31 kg) Solution: a) We are really just interested in a magnitude. The force would be repulsive. F e = k q 1 q 2 r 2 = 9x x x10 19 = ( 1x10 9 ) x10-10 N

24 b) F Net = ma, a = F Net m = 2.53x10 20 m/s 2

25 Example: Determine the net electrostatic force exerted by q 1, and q 2 on charge q 3. (q 1 = +3.00µC,q 2 = -4.00µC, q 3 = +5.00µC) q 3 10 cm q cm q 2

26 Solution: Draw the FBD showing the forces acting on the +5µC charge. The direction of forces can be determined by knowing like charges repel, and unlike charges attract. The magnitudes of the forces can be found by using coulomb s law (drop signs). F e1 F e1 = (9x109 )(3x10 6 )(5x10 6 ).1 2 =13.5N q 3 F e2 F e2 = (9x109 )(5x10 6 )(4x10 6 ).05 2 = 72N

27 You can now add forces to find the net electrostatic force acting on q3. The vector relation ship is: The final result is: F = F + F Net e1 e2 F Net = 73.3 N, S of W

28 The Electric Field The electric field is the electrostatic force on a small charge, divided by the charge: E = F e q Comparison to gravitational field: The units of electric Field: N/C, and also V/m The electric field is a vector g = F g m

29 The electrostatic force on a point charge in an electric field: F e = q E A positive charge placed in an electric field will experience a force in the same direction, while a negative charge will experience a force in the opposite direction.

30 Field Lines-Point Charges The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge. The electric field is stronger where the field lines are closer together. The field can also be deduced by the behaviour of a positive test charge. Recall that a positive charge will experience a force in the direction of the field.

31 Below are diagrams showing the electric field line created by the presence of two charges equal in magnitude. Two oppositely Two + charges charged charges Field lines always leave the positive charge and enter a negative charge. What happens if the charges are unequal??

32 Field Lines Charged Parallel Plates The electric field between two closely spaced, oppositely charged parallel plates is constant. This means that if you were to place a positive test charge anywhere between the plates it would experience the same force. For the field to be a constant value the lines have to be evenly spaced. The field is not uniform at the edges (very top and bottom in this case).

33 Summary of field lines: 1. Field lines indicate the direction of the field; the field is tangent to the line. 2. The magnitude of the field is proportional to the density of the lines. 3. Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

34 Electric Field due to Point Charges The force Fe between the two charges q and q 2 can be stated by Coulomb's Law: F e = kqq 2 r 2 The electric field E, created by charge q at q 2 's location is: E = F e q 2 = kqq 2 r 2 q 2 E = k q r 2 Using this formula one can calculate the electric field a distance r from point charge q.

35 Example(1-D): What is the electric field 2.50 mm from an electron (q e = -1.6x10-19 C)? Solution: The electric field always points towards a single negative charge (think what would happen to a positive test charge ). q The intensity (magnitude of E field) can be calculated using the formula: E E = k q r = ( 9x109 )( 1.6x10 19 ) = 2 ( 2.50x10 3 ) x10-4 N/C The sign is dropped because one is interested in a magnitude.

36 Example(2-D): Find the net electric field at location p, given that Q 1 = 4.50µC, Q 2 = -3.0 µc,d1= 8.00 cm, and d3 = cm. You have to consider the electric field produced by each charge at location p.

37 Solution: The fields acting a location p can be drawn and all angles calculated: E 2 The magnitudes of E 1 and E 2 can be calculate using the formula: q E = k E 1 = 4.05x10 6 N/C, r 2 E 2 = 7.50x10 6 N/C The net electric field at p is given by : E Net = E 1 + E 2 At this point one could add the vectors head to tail or use the components. The final answer is:e Net = 6.02x10 6 N/C N of E E 1

38 Parallel plates--revisited Recall that the electric field is uniform between the plates. Therefore, *This relationship is only true for a uniform electric field. E = ΔV d

39 Example: A negatively charged mass (m= 15 mg) was suspended between two charged plates with a voltage of 28 µv between them, and separated by a distance of 5.00 cm. It is the balance of the electrostatic force and gravity which keeps the mass suspended. How many excess electrons are on the mass? (Hint: find the total charge 1 st ) positive plate m negative plate As an aside note, this problem is similar to the principles of the famous oil drop experiment by Millikan.

40 Solution: It is a static force problem. Fe m Fg positive plate negative plate The magnitude of the electrostatic force must equal the magnitude of the force of gravity. F g = F e, mg = qe, also E = ΔV d mg = q ΔV d, q = mgd ΔV The total charge q= C The number of electrons that make up this charge: n = q 1.6x10 19 =1.64x1018 electrons

41 Electric Potential and Electric Potential Energy If one were to move a positive charge towards another positive charge it takes work to do so. One would be storing electric potential energy. V is the symbol for electric potential (J/C). It is the electric potential energy per unit charge, at a specific location. In the example below we are moving a charge q from position A to B q + q + fixed charge Let V A = electric potential at point A, and V B = the electric potential at point B(caused by fixed charge ). B A

42 The electric potential energy is related to electric potential by the following: E P =qv, where E P = electric potential energy (J) q= charge (C) V= electric potential (J/C), or (V) Let the electric potential energy at locations A and B are E PA, and E PB. The change in electric potential energy (ΔEp) in moving a charge q from A to B is: ΔEp= E PB - E PA = qv B -qv A =q(v B -V A )=qδv In summary, ΔEp=qΔV The change in electric potential energy is the product of the charge q, and the potential difference (or voltage) the charge moves through.

43 Conservation of Energy In the past we looked at conservation of energy taking into consideration the gravitational force. This principle applies to all forces, including the electrical forces. The general energy conservation (assuming no friction) is: ΔE P + ΔE K = 0 Where ΔE P is the change in electric potential energy, and ΔE K is the change in kinetic energy. We are assuming the energy changes caused by the gravitational force is negligible.

44 Example: An electron at rest is accelerated through a potential difference of 5.00mV. What was the final speed of the electron? Recall that me= 9.11x10-31 kg, and q=-1.6x10-19 C Solution: ΔE p + ΔE k = 0, qδv+ 1 2 mv 2 f 1 2 mv 2 i = 0 Plugging in the numbers and solving for v f, one finds v f = 4.19x10 4 m/s. This is a scalar relationship, so in general you should keep track of all signs such as the charge and potential difference. Sometimes in a problem, a voltage may be stated, but this is may be a magnitude of the potential difference and you may have to use common sense with signs.

45 The Electron-Volt (ev) The electron-volt (ev) is a unit of energy. It is the amount of energy gained by an electron moving through a potential difference of 1 volt. 1.6x10-19 J = 1 ev This can be show with energy conservation: ΔE p +ΔE k =0, -qδv=δe k = -(-1.6x10-19 C)(1V)= 1.6x10-19 J

46 The Cathode Ray Tube (CRT) The CRT consists of a vacuum tube, an electron gun, and a phosphor screen. Electrons are fired from the gun and the electrons strike the screen causing a phosphor pixel to fluoresce. Old TV s used CRT s.

47 The amount of deflection (d) of the electron beam turns out to be proportional to the deflecting voltage (V d ) and inversely proportional to the accelerating voltage (V a ). Or one could state: d V d V a This means that the deflection is equal to a constant (C) multiplied by the ratio Vd/Va. d = C V d C = d V a V a V d

48 Example: A CRT has a deflection of 5.00 cm on the screen. If the accelerating voltage were reduce by two thirds, and the deflecting voltage were doubled, what would be the new deflection?

49 Solution: d1 = initial deflection, d 2 = final deflection V d1 = initial deflecting voltage, V d2 =final deflecting voltage V a1 = initial accelerating voltage, V a2 = final accelerating voltage V a2 = 1/3 V a1 and V d2 = 2V d1 d 1 V a1 V d1 = d 2 V a2 V d 2 d 2 = V d 2 V d1 V a1 V a2 d 1 = (2)(3)d 1 = (6)d 1 = 30 cm

50 Electric Potential Energy of Point Charges Below are two charges separated by a distance r. There would be an electrostatic force exerted between the two charges (equal and opposite). The energy stored between two point charges is given by the following relationship: E P = kq 1 q 2 r E P = electric potential energy (J) K= Coulomb's constant q= charge (C) r=distance between the two charge(m)

51 Energy is a scalar. It is important to keep track of the signs. One should notice that the relationship on the previous slide is very similar to the gravitational potential energy between two masses. The relationship can be found in a similar manner. Again, electric potential energy is defined to be zero (unbound) when the two charges are an infinite distance apart.

52 Example: What is the electric potential energy stored between an electron and proton 3.50 cm apart? Solution: E P = kq 1q 2 r = (9x109 )(1.6x10 19 )(-1.6x10 19 ) (0.035) = -6.58x10-27 J

53 Energy Conservation and Point Charges Recall energy conservation can be stated: ΔE P + ΔE K = 0 For two point charges it could also be stated by the following: E pi + E ki = E pf + E kf or kq 1 q 2 r i mv 2 i = kq 1q 2 r f mv 2 f

54 Example: An electron 15.0 cm away approaches another fixed electron with a speed of 4.5x10 3 m/s (v). How close (D, distance of approach) will the two electrons get to each other? D v e- e- e- Solution: (9x10 9 )( 1.6x10 19 )( 1.6x10 19 ) cm (9.11x10 31 )(4.5x10 3 ) 2 = (9x109 )( 1.6x10 19 )( 1.6x10 19 ) D Solve for D: D= 2.50x10-5 m

55 Example: How much energy is stored between 3 protons 5.00 cm apart (equilateral triangle) from one another? It would be the same as the total energy required to assemble these charges when they are initially an infinite distance away. p+ d d p+ d p+

56 Solution: It takes no energy to place the 1 st charge as there is no electric force to work against. p+ The energy required to bring the second proton from an infinite distance away to 5.00cm from the 1 st charge would be: p+ 5 cm E 1 = kq 1q 2 r = (9x109 )(1.6x10 19 )(1.6x10 19 ) (0.05) p+

57 Solution continued: When bringing the final 3 rd charge from an infinite distance away one must work against the electrostatic forces of both the first two charges. p+ 5 cm 5 cm p+ 5 cm p+ The energy required to do this is: E 2 = (9x109 )(1.6x10 19 )(1.6x10 19 ) (0.05) + (9x109 )(1.6x10 19 )(1.6x10 19 ) (0.05) The total energy required is E 1 +E 2 = 1.38x10-26 J *What if the distance between charges were not equal? How would this change things?

58 Electric Potential due to Point Charges Recall the electric potential energy (E P )of a charge q 1 a distance r from a charge q would be given by: E P = kq 1q r The electric potential (V), due to charge q, would be given by: V = E p q 1 V = kq r

59 Example: What is the electric potential 12 mm from an alpha particle (Helium nucleus)? Solution: V = kq r = (9x109 )(2)(1.6x10 19 ) = 2.4x10-7 V One may have to find the electric potential due to multiple charges. Remember that electric potential is a scalar so there is no direction associated with it. To find the electric potential due to multiple charges at a particular location one needs only to add numbers. These numbers could be negative or positive.

60 Example: Find the electric potential at location p (see below) due to the electron and proton. p 6 cm 90 0 p+ e- 8 cm Solution: V P = (9x109 )(1.6x10 19 ) (9x109 )( 1.6x10 19 ) 0.1 = 9.6x10-9 V

61 Example: Two protons are positioned as shown on right. A a) Calculate the electric potential at positions A and B (V A, and V B ) b) Find the change in electric potential (Voltage) where ΔV =V B -V A c) If an electron were placed at position A, what speed would it acquire when it reached position B p 10 cm 7 cm B 7 cm 10 cm p

62 Solution: a) V A = (9x109 )(1.6x10 19 )(2) 0.1 = 2.88x10-8 V V B = (9x109 )(1.6x10 19 )(2) 0.07 = 4.114x10-8 V b) V=V B -V A = 1.234x10-8 V c) ΔE k + ΔE p = 0, in more detail: v f = 1 2 mv 2 f 1 2 mv 2 i + qδv = 0 2qΔV m = 65.8 m/s

63 Equipotential Lines (Surfaces) Equipotential lines are lines of constant electric potential For a point charge the lines would be concentric circles about the charge. For parallel plates the lines would be parallel to the plates.

64 Examples of Equipotential lines

65 Question to consider: How is the electric field related to the electric potential? Examining the electric field lines and equipotential lines on the previous slide might help to answer this question.

66 D.C Electricity

67 Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple electric cell. The Electric Battery

68 A battery transforms chemical energy into electrical energy. Chemical reactions within the cell create a potential difference between the terminals by slowly dissolving them. This potential difference can be maintained even if a current is kept flowing, until one or the other terminal is completely dissolved.

69 Several cells connected together make a battery, although now we refer to a single cell as a battery as well.

70 Electric Current Electric current is the rate of flow of charge through a conductor: Unit of electric current: the ampere(amp), A. 1 A = 1 C/s. Charge is not used up (destroyed) in a resistor. The charge that enters must equal the charge that leave. Conservation of Charge.

71 Example: 4.00x10-3 C of charge flowed through a light bulb in a time of 0.25 s. a) What was the average current through the bulb in this time? b) How many electrons passed through the bulb per second on average? Solution: a) = 4.00x10-3 /0.25= 1.6x10-2 A b) # electrons = 1.6x10-2 /1.6x10-19 = 1x10 17 electrons

72 A complete circuit is one where current can flow all the way around. Note that the schematic drawing doesn t look much like the physical circuit!

73 In order for current to flow, there must be a path from one battery terminal, through the circuit, and back to the other battery terminal. Only one of these circuits will work:

74 By convention, current is defined as flowing from + to -. Electrons (the mobile charge carriers) actually flow in the opposite direction, but not all currents consist of electrons (e.g. proton beam in particle accelerator).

75 An ammeter is a device used to measure current in circuit. The entire current must pass through the device so it is connected in series. V I R

76 Voltage Voltage is an electric potential difference. If there is a voltage there an electric field which causes a force on the charges. This potential difference forces electrons through the circuit. The voltage between positions A and B is the energy(or work) per unit charge to move the charge from A to B. Units of voltage: J/C or V(volt) Voltage is always measured between two locations. A voltmeter is therefore connected across something to measure it s voltage

77 For example to measure the voltage across a resistor (in parallel): V I R The voltage (energy per unit charge) V provided by the battery is equal to the energy per unit charge lost in the form of heat (and radiation) in the resistor (R).

78 Example: If J of energy is dissipated in a resistor when 1.25x10 20 electrons pass through it, what must be the voltage across the resistor? Solution: The total charge that passed through the resistor is: Q= (1.6x10-19 C)(1.25x10 20 )= 20 C The energy per unit charge (or voltage) is: V=E/Q= (100 J)/(20 C) = 5.00V

79 Resistivity and Resistance The resistance in an electric circuit is the opposition to the movement of charge through the conductor. The unit of resistance is the ohm(ω) The resistance of a wire is directly proportional to its length and inversely proportional to its crosssectional area: The constant ρ, the resistivity, is characteristic of the material. The lower the resistivity the better the conductor

80 Standard resistors are manufactured for use in electric circuits; they are color-coded to indicate their value and precision.

81

82 Materials that are better conductors have a lower resistivity, and vice versa for insulators.

83 Ohm s Law: Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

84 The ratio of voltage to current is the resistance: 1Ω=1J s/c 2 Ohm s Law The voltage is across a resistor is equal to the product of the resistance and the current through it. If one increases the voltage across a resistor the current should increase in a linear manner.

85 Example: If there is a voltage of 20.0 V across a 5.00 Ω resistor, what current flows through the resistor? Solution: I = V R = 20.0/5.00= 4.00 A

86 In many conductors, the resistance is independent of the voltage; this relationship is called Ohm s law. Materials that do not follow Ohm s law are called nonohmic. Unit of resistance: the ohm, Ω. 1 Ω = 1 V/A.

87 For any given material, the resistivity increases with temperature. When doing problems one generally assume the resistances remain constant. Semiconductors are complex materials, and may have resistivities that decrease with temperature.

88 Electric Power Power, as in kinematics, is the energy transformed by a device per unit time:

89 The unit of power is the watt, W. For ohmic devices, we can make the substitutions:

90 Example: A 20.0Ω resistor has a current of 0.50 A going through it. How much energy is dissipated in the resistor in a time of 5.00 minutes? Solution: P = I 2 R = (0.50) 2 (20)= 5.0 W E = Pt = (5.0)[(5)(60)]= 1500 J

91 What you pay for on your electric bill is not power, but energy the power consumption multiplied by the time. We have been measuring energy in joules, but the electric company measures it in kilowatthours, kwh.

92 Power in Household Circuits The wires used in homes to carry electricity have very low resistance. However, if the current is high enough, the power will increase and the wires can become hot enough to start a fire. To avoid this, we use fuses or circuit breakers, which disconnect when the current goes above a predetermined value.

93 Fuses are one-use items if they blow, the fuse is destroyed and must be replaced.

94 Circuit breakers, which are now much more common in homes than they once were, are switches that will open if the current is too high; they can then be reset.

95 Superconductivity In general, resistivity decreases as temperature decreases. Some materials, however, have resistivity that falls abruptly to zero at a very low temperature, called the critical temperature, T C.

96 Experiments have shown that currents, once started, can flow through these materials for years without decreasing even without a potential difference. Critical temperatures are low; for many years no material was found to be superconducting above 23 K. More recently, novel materials have been found to be superconducting below 90 K, and work on higher temperature superconductors is continuing. They now have materials that are superconducting at 133 K.

97 DC Circuits

98 Kirchhoff's circuit laws The Current Law (Junction Rule) The current law arises from the basic principle that charge is conserved. The law states that the total current which enters a junction, must equal the total current which leaves the junction. In the circuit element above there are currents (conventional) entering and leaving the junction. Applying the law, one would obtain the relationship: I 1 +I 2 =I 3 If this condition were not true there would either be more less charge entering the junction compared to the charge leaving the junction. This would suggest that it is not in steady state.

99 The Voltage Law This law arises from the basic principle of conservation of energy. The law states that the sum of all the voltage (potential difference) increases and decreases around a complete path must add to zero. Since you have returned to the same position the change in electric potential energy (or change in electric potential) must equal zero.

100 Resistors in Series and in Parallel A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.

101 According to Kirchhoff s rules the current through each resistor is the same, and the sum of the voltage drops across the resistors equals the battery voltage.

102 From this we get the equivalent (or total) resistance (that single resistance that gives the same current in the circuit). The equivalent resistance in this case is a resistor that could replace the group of resistors and draw the same current from the battery with the same voltage across it.

103 Example: Series Circuit a) What is the total resistance of the circuit below? b) Label the direction of conventional current. c) What current leaves the voltage source? d) Determine the current through each resistor, and the voltage across each resistor. 28 V 2.00 Ω 1.00Ω 4.00 Ω

104 Solution: a) Req= 2+4+1= 7.0 Ω 28 V I 2.00 Ω b) See diagram Using the equivalent circuit, one can calculate the current leaving the battery. c) I = V R = 28/7= 4.00 A 1.00Ω 4.00 Ω 28 V I If one assumes the wires have no resistance, the voltage of the source must equal the voltage across the total resistance (Req). d) Req The current I ( 4A) is the same current through each resistance since the resistors are in series. Using ohm s law on can calculate the voltage across resistor. For the 2.00 Ω, 4.00 Ω, and 1.00 Ω resistors the voltages are 8.0V, 16 V, and 4.0 V respectively.

105 In the movie below a light bulb is connected to a battery (switch open). The voltage across the light bulb is initially 9.00 V. Assume the wires have near zero resistance, and the battery always produces a constant voltage. What will happen to the current and voltage when the switch is closed and resistance of the wires in increased?

106 Explanation of previous slide: With the switch open no current flows through the circuit and the voltage across the bulb must be zero. The current would start to flow when the switch is closed. The current drops and the voltage across the bulb must decrease if the resistance in the wires increases. The total voltage across the entire circuit would still remain at 9.00 V since the the battery is assumed to be a constant voltage source. Batteries actually have an internal resistance, so in reality the voltage across the circuit (terminal voltage) is not a constant as it depends on the current. We will examine this later.

107 A parallel connection splits the current; the voltage across each resistor is the same:

108 The total current is the sum of the currents across each resistor: This gives the reciprocal of the equivalent resistance:

109 Example: Parallel Circuit a) What is total resistance of the circuit? b) What is the voltage across each resistor and the current through each resistor? c) What current leaves the battery? 6.00 V 2.00Ω 6.00Ω

110 Solution: a) 1 R Eq = 1 R R 2 =1/2 +1/6= 4/6 Req = 1.50 Ω 6.00 V I b) According to Kirchhoff s voltage law, the voltage across each resistor must be equal ( V= 6.00 V). This is assuming zero resistance in the wires connecting the resistors. Using Ohm s Law one can calculate the current through each resistor ( I = V ). The currents through the 2.00 Ω and 6.00 Ω R resistors are 3.00 A, and 1.00 A respectively. c) One could determine the current leaving the battery by using ohm s law [ I = V ] for the equivalent circuit R above, or by using the junction (current) rule [ I = I 1 + I 2 ] for the original circuit. I= 3 +1 = 4.00 A I I 1 I Ω 2.00Ω 6.00Ω 6.00 V

111 Example: Combination Circuit a) In the circuit shown calculate the equivalent (total) resistance of the entire circuit. Reduce the circuit in steps. b) Calculate the current that leaves the battery. c) Find the voltage across each resistor and the current through each resistor d) What was the power dissipated in the 5Ω resistor? e) How much energy does the entire circuit use in 10 V= 24 V minutes? 4Ω 3Ω 1Ω 5Ω

112 Solution: a) 6Ω b) Using Ohm s Law: I= 24/6= 4.00 A c) 4Ω Voltage across the 4 Ω resistor: V 4 =IR= (4)(4) = 16 V One can now use Kirchhoff s voltage law to find the voltage across the 3 Ω resistor: V 3 =0, V 3 = 8V Now use Ohm s law to find the current through the 3 Ω resistor : I 1 =V/R= 2.67 A One can now use the current law to find the current through the 1 Ω and 5 Ω resistors. 4=2.67+I 2, I 2 = 1.33 A And finally one can use Ohm s law to find the voltages across the 1 Ω and 5 Ω resistors. V=IR, V 1 = 1.33 V, V 5 = 6.67 V I I 1 1Ω V= 24 V 3Ω I 2 5Ω d) P=VI= (6.66 )( 1.33) = 8.89 W e) P=VI, E=Pt= VIt = (24)(4)[(10)(60)] = 57.6 kj

113 Example: Assume all the light bulbs (A,B,C, and D) have the same resistance and the brightness of a bulb is proportional to the power dissipated in the bulb. Describe what happens to the brightness of bulb before, and after the the switch S is closed. Also make sure you make relative comparisons of brightness between the bulbs. A B S V C D

114 Solution: Initially the same current flows through bulbs A,B, and C so all should have the same brightness. There is no current going through bulb D. When the switch is closed the total resistance of the circuit is decreased. This would increase the current leaving the voltage source which means bulbs A and B would become brighter (equal) than they were initially. Since C and D are in parallel they have the same voltage (also same current since equal resistances) and would have the same brightness. Bulbs C and D would be less bright than bulbs A and B as they have half the current. A tougher part to the question is how does the brightness of bulbs C and D compare to the brightness of bulb C (A and B) before the switch was closed? It turns out that the brightness of bulbs C and D after the switch is closed would be less bright than bulbs C, A and B before the switch was closed. To answer the final question examine the currents (in terms of V and R, where R is the resistance of one bulb) through each bulb, before and after the switch is closed, then compare. The relative values of the currents would be directly related to the relative brightness. If you are having difficulty with this, see me for help.

115 EMF and Terminal Voltage Electric circuit needs battery or generator to produce current these are called sources of emf (electro motive force, voltage source). Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf: This resistance behaves as though it were in series with the emf source.

116 Discharging Battery- (conventional current leaves + terminal) In this case a battery is connected to an external resistance (R). Using a kirchhoff voltage loop one can state: ε = IR + Ir ε Ir = IR = V This is the terminal T r! I r = ε = voltage (V) Internal resistance(ω) emf (V) R = I = Current (A) External resistance (Ω) R V T = ε Ir Q: What happens to the current, and terminal voltage when the external resistance is increased?

117 Q: What happens to the current, and terminal voltage when the external resistance is increased? Answer: When the external resistance is increased the total resistance of the circuit is increased which would decrease the current. A lower current means a smaller voltage drop across the internal resistance which would mean a larger terminal voltage. Supporting math: I = ε R + r V T = ε Ir

118 It is actually possible to have a terminal voltage greater than the emf of the battery. This occurs in the case when a battery is charging. This would mean that conventional current is entering the positive terminal of the battery. Remember what is actually occurring, is that electrons are entering the negative terminal. One can charge a battery by connecting another battery with a greater emf in parallel as shown.! 1 r In this case ε 1 > ε 2. The emf of 1 I cell 1 overpowers the emf of cell 2 and drives conventional current into cell 2.! 2 r 2 In this case the terminal voltage of cell 2 s: V T = ε 2 + Ir 2 In general one can use the equation: V T = ε ± Ir

119 Example: A 6.00 Ω (R) resistor was connected to a battery with an emf of 8.00 V(ε) and caused a current of 1.00 A to flow through the circuit. a) What is the terminal voltage of the battery? b) What is the internal resistance of the battery? r! I R

120 Solution: a) V T = IR = ( 1)(6)= 6.0 V b) r = ε V T I = ( )/1= 2.00 Ω

121 Example: When a 3.00 Ω resistor was connected to a battery, it drew a current of 3.00 A. When this external resistance was replaced with a 5.00 Ω resistor, a current of 2.00 A was drawn. What is the emf (ε) and internal resistance (r) of this battery? Hint: You will need to solve a system of equations. Solution: Using Kirchhoff voltage loops (or terminal voltage equation) one can state: r! I = 3.00 A r! I = 2.00 A 3.00 Ω 5.00 Ω ε 3r = (3)(3) = 9 ε 2r = (2)(5) =10 Solving the system gives: ε= 12 V, and r= 1.00 Ω

122 Electric Hazards Even very small currents 10 to 100 ma can be dangerous, disrupting the nervous system. Larger currents may also cause burns. Household voltage can be lethal if you are wet and in good contact with the ground. Be careful!

123 A person receiving a shock has become part of a complete circuit.

124 Faulty wiring and improper grounding can be hazardous. Make sure electrical work is done by a professional.

125 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.