# AC vs. DC Circuits. Constant voltage circuits. The voltage from an outlet is alternating voltage

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1 Circuits

2 AC vs. DC Circuits Constant voltage circuits Typically referred to as direct current or DC Computers, logic circuits, and battery operated devices are examples of DC circuits The voltage from an outlet is alternating voltage AC circuits are much more complicated because circuit elements can introduce lag A typical analysis of AC circuits requires Fourier techniques

3 DC Circuits A voltage difference can be created using a battery This gives rise to an electric field inside the wire Charge can not redistribute itself to cancel the field generated by a battery, because the battery maintains a constant voltage difference even as charge builds up on it.

4 Electro-motive Force For historical reasons, the voltage applied to a circuit is sometimes referred to as an electro-motive force, or emf. Note that the emf is not a force.

5 Current Current is defined as the amount of charge that moves through a crosssection of a wire per unit time: I = Δq Δt

6 Current By convention, we say that current points in the direction that positive charge flows. In most materials, the electrons move. We call the charge responsible for the current the charge carrier.

7 Current Since electrons have a negative charge, the motion of the electron is opposite the direction of the current. current

8 Current Current is a vector: Current has magnitude and direction Like a car traveling down a road, current in a wire can only point in one of two directions. Therefore, current can be thought of as a onedimensional vector. Like any one-dimensional vector, we denote the direction using a plus or minus sign

9 Current For a simple circuit, current flows from high voltage to a low voltage. +

10 Current For a simple circuit, current flows from high voltage to a low voltage. +

11 Current - Example A 12 V battery is connected to a battery charger for 5 hours. The charger carries charge from the low voltage end of the battery, back to the high voltage end of the battery. The charge moves the charge at a rate of 6 A, and the voltage across the battery remains 12 V during this time. (a) How much charge is transferred to the battery during this time? (b) How much energy is transferred to the battery?

12 I = 6 A ΔV = 12 V Δt = 5 hr Δq =? ΔPE =?

13 Ohm s Law Resistance is like friction for current in a wire. Resistance is denoted with by the letter R. The symbol for a resistor is:

14 Ohm s Law The current flowing through a wire is related to the voltage and resistance according to Ohm s Law: ΔV = I R

15 Ohm s Law We can find the units of the resistance using dimensional analysis:

16 Ohm s Law The flow of current through a circuit is analogous to the flow of water through a pipe: Current flow of water ΔV Pressure difference Resistance friction and viscosity of the water

17 Ohm s Law How does the water flow depend on the crosssectional area of the pipe? How does the water flow depend on the length of the pipe?

18 Ohm s Law In a similar manner, the resistance of most wires is related to the length and cross-sectional area of the wire, according to: R = ρ L A Where ρ is called resistivity and is a measure of how well a material conducts current.

19 Ohm s Law - Example An incandescent light bulb is made with a tungsten filament. When plugged into a 120 V outlet, 1.24 A of current flows through the bulb. If the radius of the filament is m and the resistivity of tungsten is Ω m, what is the length of the filament?

20 ΔV = 120 V I = 1.24 A r = m ρ = Ω m L =?

21 When current flows through a resistor the electrostatic energy is converted into heat. Using Power Lost in Resistors P = ΔE Δt we can write a formula for the power converted into heat in a circuit.

22 Power Lost in Resistors Combining this formula with Ohm s Law gives:

23 Power Lost in Resistors The power lost in a resistor is given by: P = I V P = I 2 R P = V2 R

24 Power Lost in Resistors The old-fashioned incandescent light bulbs can be treated like a resistor. What is the resistance of a 60 W light bulb that is designed to be plugged into a 120 V power outlet?

25 Power Lost in Resistors P = 60 W V = 120 V R =?

26 Internal Resistance Most batteries have some resistance between the positive and V term. r int R negative terminals. + V

27 Internal Resistance Some voltage is lost due to this internal resistance. We call the voltage difference between the ends of a battery the terminal voltage V term. + r int V

28 Internal Resistance There are two things to know about internal resistance: 1. The internal resistance is always in series with the objects connected to the battery 2. The internal resistance lowers the terminal voltage by an amount given by Ohm s Law V term = emf I r int

29 Internal Resistance 930 ma flow through a 10 Ω resistor when connected to a battery with a 9.5 V emf. What is the internal resistance of the battery?

30 Internal Resistance V term. =? r int =? R = 10 Ω V

31 Equivalent Resistance Objects in a circuit can be connected in one of two ways: Series Parallel

32 Equivalent Resistance R 1 R 2 + The current through resistor 1 is equal to the current through resistor 2, I 1 = I 2 The total voltage drop across the two resistors is equal to the sum of the voltage drops across each resistor V

33 Equivalent Resistance Want a formula for ΔV total in terms of I, R 1, and R 2. R 1 R 2 + V

34 Equivalent Resistance The two resistors in series behave like a single resistor with resistance R 1 + R 2. We call this the equivalent resistance. R eq = R 1 + R 2 (series)

35 Equivalent Resistance In general, the equivalent resistance of N resistors in series is given by: R eq = R 1 + R 2 + R 3 + R 4 +

36 Equivalent Resistance The voltage across both resistors is the same. The total current that flows through the resistors is the sum of the currents through each resistor. R 1 R 2 + V

37 Equivalent Resistance R 1 Want a formula that relates I total to V, R 1, and R 2. R 2 + V

38 Equivalent Resistance Single Resistor: Parallel Resistors: I = V R I = V 1 R R 2 The two resistors are equivalent to a single resistor with a resistance given by:

39 Equivalent Resistance In general, the equivalent resistance of N resistors connected in parallel is given by: 1 R eq = 1 R R R R 4 +

40 Equivalent Resistance When there are only two resistors connected in parallel, the equivalent resistance can be rewritten as:

41 Equivalent Resistance Notice that, the equivalent resistance for resistors in parallel is always less than the resistance of either of the two resistors. R eq = R 1 R 2 R 1 + R 2

42 Equivalent Resistance If N identical resistors are connected in parallel, the equivalent resistance is equal to:

43 Equivalent Resistance - Example A 15 Ω toaster and a 10 Ω iron are connected in parallel to a 120 V source. A 20 A breaker is connected in series with the power source. Will using both appliances at the same time throw the breaker?

44 R T = 15 Ω R Iron = 10 Ω V = 120 V I =? Breaker R T R Iron V +

45 Equivalent Resistance - Example Using the circuit diagram below, find the: (a) Power dissipated in the circuit when the switch is open (b) equivalent resistance of the circuit when the switch is closed, and (c) power dissipated in the circuit when the switch is closed. + 9V 65Ω 96Ω

46 Equivalent Resistance - Example P open =? P closed =? R eq =? + 9V 65Ω 96Ω

47 Equivalent Resistance - Example A 60 W lamp is placed in series with a resistor and a 120 V source. If the voltage lost across the lamp is 25 V, what is the resistance of the unknown resistor, R? R lamp R + 120V

48 Equivalent Resistance - Example P = 60 W ΔV = 25V R lamp R =? + 120V

49 Complex Circuits When given a complex circuit of resistors, we use divide an conquer to find R eq : Identify a small segment of the circuit for which we can easily calculate an equivalent resistance. Replacing these segments with a single equivalent resistance. Repeat the process until there is only one equivalent resistance.

50 Find the equivalent resistance between points A and B. A Complex Circuits 2Ω 6Ω 1Ω 4Ω 3Ω 2Ω B 3Ω Which portion of the circuit should we calculate the equivalent resistance for?

51 Complex Circuits A 2Ω 6Ω 1Ω 4Ω 3Ω 2Ω B 3Ω

52 Complex Circuits A 2Ω 6Ω 4Ω 3Ω 6Ω B

53 Complex Circuits A 2Ω 6Ω 4Ω 2Ω B

54 Complex Circuits A 2Ω 4Ω 8Ω B

55 Complex Circuits A 2Ω 2.67Ω B

56 Complex Circuits 1Ω 2Ω A 7Ω 6Ω B 3Ω

57 Complex Circuits 1Ω 2Ω A 7Ω 6Ω B 3Ω

58 Complex Circuits 3Ω A 7Ω 6Ω B 3Ω

59 Complex Circuits 3Ω A 7Ω 6Ω B 3Ω

60 Complex Circuits A 7Ω 2Ω B 3Ω

61 Complex Circuits A 7Ω 2Ω B 3Ω

62 Complex Circuits A 12Ω B

63 Kirchoff s Laws While equivalent resistance is useful, it does not allow us to find the voltage drop and current flowing through each individual resistor in a circuit. For this we will need an additional set of equations.

64 Kirchoff s Laws Junction Rule Charge can not be created or destroyed. Therefore, the current flowing into the junction must equal the current flowing out of the junction.

65 Kirchoff s Laws Junction Rule For example: 5 A 3 A I =?

66 Kirchoff s Laws Loop Rule When an electron, starting at point A, goes around a loop in the circuit, and returns to point A, it must have the same voltage that it started with. R V A

67 Kirchoff s Laws Loop Rule Because voltage decreases when current flows through a resistor, the voltage is higher on the side that the current flows in and lower on the side the current flows out. A I B V A > V B

68 Kirchoff s Laws Together these two rules are called Kirchoff s Laws: The current flowing into a junction must equal the current flowing out of the junction The total voltage change around any closed loop must always equal zero volts.

69 Kirchoff s Laws - Example Find I 1, I 2, I 3, and V AB in the circuit below. I 1 A 10 A 10 Ω 14 Ω 8 Ω B I 3 I 2

70 Kirchoff s Laws - Example Use Kirchoff s Laws to solve for all the currents in the circuit below: I 1 5 Ω 12 V I 2 I 3 5 Ω 10 Ω

71 Kirchoff s Laws - Example Use Kirchoff s Laws to solve for all the currents in the circuit below: I 1 4 Ω 2 V 4 Ω 10 V I 2 I 3 3 Ω 2 Ω

72 Making Measurements Measurements in circuits are commonly made using a multi-meter. Which is capable of measuring current, voltage, and resistance.

73 V Making Measurements Ammeters measure current An ammeter must be connected in series, so that the current flows through the ammeter. A + R +

74 V Making Measurements To prevent the ammeter from effecting the circuit, the ammeter has almost no resistance through it. A + R +

75 Making Measurements Voltmeters must be connected in parallel, so that it can measure the voltage difference between two points in the circuit. + V +

76 Making Measurements The voltage between two points can be measured by combining an ammeter and Ohm s Law. By placing a known resistance inside an ammeter, and measuring the current, an ammeter can measure voltage.

77 Capacitors A capacitor is made by placing two metal plates close to each other. When connected to a battery, the plates become oppositely charged.

78 Capacitors Using Gauss s Law, we can calculate the electric field between two oppositely charged parallel plates

79 Capacitors We can split the region up into three sections. Between the two plates the electric fields from the plates add. Outside the plates the electric fields cancel I II III

80 Capacitors Suppose the plates are infinitely long, and have charge per unit area, σ. Consider the following Gaussian Surface:

81 Capacitors The only flux comes through one end of the cylindrical Gaussian Surface. Therefore: Φ = q enc E A = ε 0 σ A ε E E = σ ε 0 A + + +

82 Voltage Across a Capacitor To find the voltage across the capacitor we use: ΔV = E Δx,E

83 Voltage Across a Capacitor The voltage across a capacitor is given by: ΔV = q d A ε 0

84 Voltage Across a Capacitor Notice that the voltage across the capacitor is proportional to the charge on it: ΔV = q d A ε 0 d Also, notice that describes the design of the A ε 0 two plates, and has nothing to do with voltage or charge.

85 Voltage Across a Capacitor Capacitance is a quantity which describes the plates ability to hold charge (think capacity). We define capacitance according to the formula: q = V C

86 Voltage Across a Capacitor If we rearrange: ΔV = q d A ε 0 to solve for q, and compare this with: q = V C We can write a formula for the capacitance of two parallel plates.

87 Voltage Across a Capacitor Two parallel plates of area A and separated by a distance d, have a capacitance of: C = ε 0 A d This is known as a parallel plate capacitor.

88 Voltage Across a Capacitor We can find the units of capacitance using dimensional analysis We call the quantity C a Farad in honor of Michael V Faraday who discovered elecro-magnetic induction.

89 Voltage Across a Capacitor A parallel plate capacitor is connected to a 9V battery. After the capacitor is charged, you remove the capacitor from the battery without removing the charge from the plates of the capacitor. You pull on the plates of the parallel plate capacitor so that the distance between the plates is doubled. What is the new voltage across the capacitor?

90 V 0 = 9 V q f = q 0 d f = 2 d 0 V f =?

91 Energy Stored In a Capacitor When a capacitor is connected to a battery, the battery moves electrons from one plate of the capacitor to the other plate. This process continues until the voltage across the capacitor equals the voltage supplied by the battery. To calculate the energy stored in the capacitor consider the work done by the battery.

92 Energy Stored In a Capacitor As charge is moved from one plate to the other, it becomes more difficult to move the charges.

93 Energy Stored In a Capacitor The total energy stored in the capacitor is the sum of the ΔPE for each charge moved from one plate to the other. The energy required to move a charge Δq across a capacitor with a charge q is: ΔPE = Δq V = q C Δq

94 Energy Stored In a Capacitor Taking the sum of these potential energies PE total = PE In the limit that Δq 0: = Δq q C U = 0 Q q C dq = Q2 2 C

95 Energy Stored In a Capacitor Using V = q C voltage: we can rewrite this in terms of U = q2 2 C = V C 2 2 C = 1 2 C V2

96 Energy Stored In a Capacitor We can also get rid of the capacitance and write it in terms of voltage and charge U = 1 2 = 1 2 C V2 q V V2 = 1 2 q V

97 Energy Stored In a Capacitor We have three formulas for the energy stored inside a capacitor: U = q2 2 C U = 1 2 C V2 U = 1 2 q V

98 Equivalent Capacitance Recall that: A capacitor can be created by placing two conducting surfaces close to each other When a voltage is applied to the plates of a capacitor, charge from one plate moves to the other plate The charge on the capacitor is equal to: q = VC

99 Equivalent Capacitance Just like resistors, capacitors can be connected in series or in parallel Once again, we want to find the equivalent capacitance in each of these scenarios

100 Equivalent Capacitance The charge on capacitors in series is always the same: This is because the charge on the capacitor plates comes from moving charge from another plate.

101 Equivalent Capacitance Since the total voltage across the two plates is equal to the sum of the voltages across each plate: V = V 1 + V 2 = q C 1 + q C 2 = q 1 C C 2 = q C eq Therefore, 1 C eq = 1 C C 2

102 Equivalent Capacitance In general, the equivalent capacitance of many capacitors in series is: 1 C eq = 1 C n = 1 C C C 3 +

103 Equivalent Capacitance Like resistors, the voltage across capacitors in parallel is the same. The total charge on the capacitors is equal to the sum of the charges on each. q = q 1 + q 2 = VC 1 + VC 2 = V(C 1 + C 2 ) = VC eq

104 Equivalent Capacitance Therefore, the equivalent capacitance for capacitors in parallel is: C eq = C n = C 1 + C 2 + C 3 +

105 Equivalent Capacitance - Example Suppose two capacitors are in parallel. Show that the sum of the energy stored in the capacitors is equal to the energy that would be sored in an equivalent capacitor. In other words, demonstrate that: U 1 + U 2 = U eq.

106 Equivalent Capacitance - Example We want to relate U 1 + U 2 to U eq. The energy stored in a capacitor is given by: Therefore, U = q2 2C U 1 + U 2 = q q 2 2 2C 1 2C 2 However, this is not very useful because q 1 q 2.

107 Equivalent Capacitance - Example Using q = VC, we can rewrite this formula in terms of V U 1 + U 2 = q q 2 2 2C 1 2C 2 What quantity is the same for capacitors connected in parallel? U 1 + U 2 = V C V C 2 2C 1 2C 2 = V2 2 C 1 + C 2 = V2 C eq 2 2

108 Equivalent Capacitance - Example A 7 μf and 3 μf capacitor are connected in series across a 24 V battery. What voltage is required to store the same amount of energy is the capacitors were connected in parallel?

109 Equivalent Capacitance - Example C 1 = 7 μf V s = 24 V C 2 = 3 μf To determine the voltage required for a parallel connection, we first need to find the energy stored in a series connection. Just like in the previous example, we can substitute C eq into the energy formula.

110 Equivalent Capacitance - Example C 1 = 7 μf V s = 24 V For capacitors in series: 1 C 2 = 3 μf C eq = 1 C C 2 or C eq = C 1C 2 C 1 +C 2 Therefore, C eq = (7 μf)(3 μf) (7 μf + 3 μf) = 2.1 μf

111 Equivalent Capacitance - Example C 1 = 7 μf V S = 24 V For capacitors in parallel: C 2 = 3 μf C series = 2.1 μf C eq = C 1 + C 2 Therefore, C eq = 3 μf + 7 μf = 10 μf

112 Equivalent Capacitance - Example C 1 = 7 μf V s = 24 V C 2 = 3 μf Since we are not given the charge on the capacitors, we must combine: U = q2 2C To get: and q = VC U = 1 2 C V2

113 Equivalent Capacitance - Example C s = 2.1 μf V s = 24 V Setting the energies equal to each other, and solving for V parallel gives: U series = U parallel 1 2 C s V s 2 = 1 2 C p V p 2 V P 2 = V P C s C p V s 2 = 11 V C p = 10 μf

114 RC Circuits When a capacitor is connected to battery, the battery causes charge to move from one plate of a capacitor to the other. However, as charge builds up on the plates, the repulsive electrostatic force caused by this charge makes it more and more difficult to add additional charge to the capacitor. As a result, the rate that charge builds up on the capacitor slows.

115 RC Circuits Consider applying the loop rule to the following circuit: We get: V = IR + q C

116 The current through the resistor is equal to the rate that the charge on the capacitors changes: I = dq dt RC Circuits Plugging this into the loop rule gives: V = dq dt R + q C

117 RC Circuits V = dq dt R + q C This is called a differential equation. We want to find q t that satisfies this equation. First, lets rewrite this slightly: dq dt = V R q R C

118 RC Circuits dq dt = V R q R C Recall that d dx ex = e x, Therefore, the solution to: dq dt = q RC Is: q = e t RC

119 RC Circuits dq dt = V R q R C Taking the V R term into account gives: q t = q f 1 e t RC

120 RC Circuits A graph of q t = q f 1 e t RC looks like:

121 RC Circuits Notice that time is divided by RC: q t = q f 1 e t RC This means that the product RC tells us how fast the capacitor in a circuit will charge We call this number the time-constant: τ = RC

122 RC Circuit - Example Find the time constant of the circuit shown in the diagram below:

123 RC Circuit - Example The time constant is simply τ = R eq C eq R eq = = R 1 R 2 R 1 + R 2 4 kω 2 kω 4 kω + 2 kω = 1.3 kω

124 RC Circuit - Example The time constant is simply τ = R eq C eq C eq = C 1 + C 2 = 3 μf + 6 μf = 9 μf

125 RC Circuit - Example The time constant is simply τ = R eq C eq τ = R eq C eq = s

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