Physics 150 Electric current and circuits Chapter 18
Electric current Let s imagine a conductor with a potendal difference between its ends - - - - NegaDve (lower) potendal V a E! + + + + PosiDve (higher) potendal V b PosiDve charges will start moving to the lem (away from posidve potendal and toward the negadve potendal) NegaDve charges will move to the right (away from negadve and toward the posidve potendal) Neutral atoms will just get polarized 2
Electric current Electric current is the modon of the charged pardcles due to the potendal difference (voltage). I = Δq Δt NotaDon for current is: I Unit of current: Ampere [A] How big is Ampere? Household BaWery Powered Lightning Bolt To hurt you To kill you ~ 1-10 A ~ 100 ma ~ 100 A to 100 ka ~ 40 ma ~ 0.1 A to 0.2 A 3
Exercise: Electric current If 5x10 20 electrons flow past a given cross- secdon of the metal wire in 10.0 minutes, what is the current in the wire? What do we know: N e = 5x10 20 electrons Δt = 10 min I = Δq Δt I = Δq Δt First, find the total charge in the wire è Equal to number of electrons Dmes the charge of one electron Δq = Ne Δq = 5 10 20 1.6 10 19 C = 80C Then, convert minutes to seconds: I = 80C 600 = 0.13A t =10min 60s 1min = 600s 4
DirecDon of the current By convendon the direcdon of the current is equal to the direcdon of modon of posidve charges. (In reality, a flow of negadve electrons determines the current). Current will always flow from high potendal to lower potendal - along the direcdon of the electric field. 5
Wait, according to Chapter 16 But we stated that an electric field inside of a conductor must be zero (E=0) and a conductor is an equipoten6al surface (V ab =0). è electrons will move undl the internal field was canceled - - - - Lower potendal V a + + + + Higher potendal V b How can we achieve the non zero potendal difference and constant movement of charged pardcles? 6
ElectromoDve force We need an object which is capable of creadng the potendal difference and constantly providing electrons to one end of the conductor and removing them from the other. ε Such an object is a source of electromo0ve force (EMF). The EMF does not create charges, but increases the energy of charges passing through it. 7
ElectromoDve force Examples of sources of EMF are: Ba>eries Generators Capacitors ε - + At high potendal At low potendal EMF is NOT a force it is a potendal difference (voltage). The circuit symbol for a bawery (EMF source) The work done by an ideal bawery in pumping a charge q is W = qε. The amount of charge entering the conductor per unit Dme is equal to the amount of charge leaving the conductor per unit Dme. 8
Microscopic view of current No electric field random modon (but very fast) v ~10 6 m/s Electric field sdll random modon with an average velocity or dri4 velocity v d ~10-4 m/s 9
Microscopic view of current 10
DriM velocity Let s, calculate the number of charges (N e ) that pass through the shaded region in a Dme Δt. Number of charges per unit volume is n = N e V = N e A Δx N e = n(aδx) N e = na(v d Δt) The current in the wire is: Δq Δt en Δt e I = = = neav d 11
Exercise: DriM velocity I = neav d A copper wire of cross- secdonal area 1.00 mm 2 has a constant current of 2.0 A flowing along its length. What is the drim speed of the conducdon electrons? Copper has 1.10 10 29 electrons/m 3. What do we know: A = 1 mm 2 I = 2 A n = 1.10 10 29 electrons/m 3 I = neav d v d = I nea Convert 1 mm 2 to m 2 : 1 mm 2 = 1x10-6 m 2 v d = 2.0 A ( 1.10 10 29 m 3 ) 1.60 10 19 C ( )( 1.00 10 6 m 2 ) =1.1 10 4 m/sec = 0.11 mm/sec 12
Resistance and Ohm s law What determines how much current we get for a given applied electric field or potendal difference? properdes of the conductor determine the resistance R Ohm s law ΔV = IR Units of resistance: Ohm = Volt per ampere [Ω] 13
Conceptual quesdon Ohm s law Q1 If the potendal difference (voltage) across a resistor is doubled: 1. Only the resistance is doubled 2. Only the current is doubled 3. Only the resistance is halved 4. Both current and resistance are doubled ΔV = IR Resistance is the property of the object and it does not depend on voltage and current. 14
Resistance and resisdvity Resistance (R) is a property of an object increases with increasing length of wire L (charge has further to push through ) decreases with increasing cross- secdonal area of wire A (charge has more possible paths through the wire) depends upon the material the wire is made of ρ R = L ρ A Symbol for resistance: ρ (Greek lewer rho) is the resisdvity which is a property of the material at a given temperature. Resistance does not depend on the shape of the area A! 15
Conceptual quesdon Resistance Q2 A cylindrical copper rod has resistance R. It is reformed to twice its original length with no change of area. Its new resistance is: A. R B. 2R C. R/2 D. R stays the same R = L ρ A What about resisdvity? ResisDvity is the property of the material it does not depend on the length and area of the resistor. 16
Temperature dependence The resisdvity of a material depends on its temperature: ( + ( T )) ρ = ρ 1 α T 0 0 where ρ 0 is the resisdvity at the temperature T 0, and α is the temperature coefficient of resisdvity. R = R ( 0 1+α ( T T )) 0 Some materials, when very cold, have a resisdvity which abruptly drops to zero. Such materials are then superconductors. 17
Exercise: Resistance of a conductor A common flashlight bulb is rated at 0.3 A and 2.7 V. If the resistance of the bulb s tungsten filament at room temperature (20 0 C) is 1 Ω, esdmate the temperature of the filament when the bulb is turned on. R = R ( 0 1+α ( T T )) 0 1+α ( T T 0 ) = R R T T 0 = 0 Resistance at room temperature T 0 = 20 0 C =20+273 K = 293 K is R 0 = 1 Ω Resistance at unknown temperature T can be found from Ohm s law R R 0 1 α ΔV = IR R = ΔV I α (tungsten) = 4.50 10 3 C 1 T = 20 C + R = 2.7V 0.3A = 9Ω 9Ω 1Ω 1 = 20 C+1777.8 C=1797.8 C 4.50 10 3 1 C What do we know: I = 0.3 A ΔV = 2.7 V R 0 = 1 Ω T 0 = 20 0 C α = 4.5x10-3 0 C - 1 18
A simple loop circuit A current will only flow around a closed loop. I = ε R Where ε is electromodve force (EMF) The potendal difference between the terminals of an ideal EMF device is equal to the EMF of the device 19
A single- loop rules Loop rule: The algebraic sum of all changes in potendal encountered in a complete traversal of any loop of a circuit must be zero (Kirhhoff s loop rule). Resistance Rule: For a move through a resistance in the direcdon of the current, the change in potendal is - ir; in the opposite direcdon +ir. EMF Rule: For a move through an ideal emf device in the direcdon of the emf arrow, the change on the potendal is +ε; in the opposite direcdon is - ε 20
PotenDal in a circuit A liwle like water flowing downhill The emf is like the pump that gets the water to the top of the hill potendal rises Across every resistor potendal drops ε Ir IR = 0 21
Ideal baweries vs. real baweries Ideal BaWery Keeps potendal difference between its terminals. Real BaWeries have an internal resistance, r, which is small but non- zero. Resul6ng (terminal) voltage supplied by the bawery ΔV = V b V a = ε Ir By applying Ohm s law ΔV = IR = ε Ir I = ε R + r 22
Exercise: A dim flashlight As a flashlight bawery ages its emf stays approximately constant, but its internal resistance increases. A fresh bawery has an emf of 1.5V and negligible internal resistance. When the bawery needs replacing its emf is sdll 1.5V but its internal resistance has increased to 1000 V. If this old bawery is supplying 0.5 ma, what is its terminal voltage? ΔV = ε Ir What do we know ε =1.5 V r i = 1000 Ω = 0.5 ma ΔV =1.5V 0.5 10 3 A 1000Ω ΔV =1.5V 0.5V =1V 23
Resistors in series = R eq ΔV 1 ΔV 2 ΔV 3 The current through each resistor is the same. Resistors following one amer the other We add them together to get the total resistance. R eq = R 1 + R 2 + R 3 +... PotenDal drops across every resistor ΔV 1 = IR 1 ΔV 2 = IR 2 ΔV 3 = IR 3 ΔV = ΔV 1 + ΔV 2 + ΔV 3 = IR 1 + IR 2 + IR 3 24
Resistors in parallel ΔV = R eq 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 + The potendal difference (voltage) across each path is the same. ΔV 1 = ΔV 2 = ΔV 3 = ΔV Current is branching (like a river) è I = I 1 + I 2 +I 3 25
Resistors in series and parallel (a reference set) ε = ΔV 1 + ΔV 2 + ΔV 3 I = I 1 = I 2 = I 3 R = R 1 + R 2 + R 3 ε = ΔV 1 = ΔV 2 = ΔV 3 I = I 1 + I 2 + I 3 1 R = 1 + 1 + 1 R 1 R 2 R 3 26
Christmas tree Old- Dme Christmas tree lights had the property that, when bulb burned out, all the lights were out. How were these lights connected? In series è when one bulb burns out it breaks the circuit è no current è all bulbs are out How could you rewire them to prevent all the lights from going out when one of them burned out? Connect them in parallel connecdon è when one bulb burns out it breaks the circuit only across its own path 27
CombinaDon circuits What to do when we have a mix of series and parallel combinadons? 1. Select a group of resistors connected in either series or parallel. In this case the two resistors on the right connected in parallel (current is branching) 2. Calculate the equivalent resistance for the group. 3. Go to the first step - Select a new group of resistors connected in either series or parallel in the new loop 4. Keep going Dll you get what you want 28
Simple analysis: all resistors are the same (a) The two verdcal resistors are in parallel with one another, hence they can be replaced with their equivalent resistance, R/2. (b) Now, the circuit consists of three resistors in series. The equivalent resistance of these three resistors is 2.5 R. (c) The original circuit reduced to a single equivalent resistance. 29
Exercise: A resistor network Three idendcal resistors with resistance of 6 Ω are connected to a bawery with an emf of 18 V a) Find the equivalent resistance of the resistor network b) Find the current through the bawery and in each resistor 1. First, resistors R 2 and R 3 are connected in parallel. Their combined resistance is 1 = 1 + 1 1 = 1 R 23 R 2 R 3 R 23 6 + 1 6 = 1 3Ω R 23 = 3Ω 2. Second, resistors R 1 and R 23 are now connected in series. Their combined resistance is: R eq = R 1 + R 23 = 6 Ω + 3 Ω = 9 Ω 30
Part (a) Part (b) Current through the bawery can be found using Ohm s law I = ε R eq = 18V 9Ω = 2A By looking at the simplified graphs we can say that the same current flows through R 1 and R 23 => I 1 =I 23 = I =2 A This means that the voltage drop across R 1 is ΔV 1 = I 1 R 1 = 2 A x 6 Ω = 12 V and across R 23 is ΔV 23 = I 23 R 23 = 2 A x 3 Ω = 6 V which adds up to the 18 V provided by the bawery Since R 2 and R 3 are connected in parallel => ΔV 2 = ΔV 3 = ΔV 23 =6 V => Current across R 2 is I 2 = ΔV 2 /R 2 = 6 V/6Ω = 1 A => Current across R 3 is I 3 = ΔV 3 /R 3 = 6 V/6Ω = 1 A 31
AddiDonal problems (a) (b) (c) Which resistors in this circuit are in parallel with each other? What is the equivalent resistance of the circuit? What are the size and direcdon of current i 1 in the figure, if R 1 = 9.0 ohms and R 2 = 18 ohms? 32
Circuit Analysis SoluDon (a) Resistors R 2 are in parallel 1 1 1 1 Equivalent resistance = + + = R e R R R 2 2 2 2 3 R 2 R e = 2 R 3 2 (b) Resistors are in connected in series with R2 1 R2 Equivalent resistance of the circuit Re = Re 2 + R1 = + R1 3 18Ω R e = + 9Ω = 15Ω 3 R (c) Current I through R e : I = ε R e = 12V 15Ω = 0.8A Voltage drop across R1 : ΔV 1 = IR 1 = 0.8A 9Ω = 7.2V Voltage drop across all R2 : ΔV 2 = ε ΔV 1 =12V 7.2V = 4.8V Current through : R 2 I 1 I 1 = ΔV 2 R 2 = 4.8V 18Ω = 0.267A 33
Power and energy in circuits If a charge is moving through potendal difference, the change in potendal energy is ΔU = qδv A change in potendal energy in Dme is equal to power P = ΔU Δt = q Δt ΔV = IΔV The energy/power added to the circuit by the EMF must be removed by the resistor. For an EMF source: P = Iε For a resistor: P 2 2 ΔV = IΔV = I R = The resistor dissipates this R energy as heat. 34
Conceptual quesdon 40 vs 100 WaWs Q3 Which has a greater resistance, R, a 100 W or a 40 W incandescent light bulb? (The radng of a light bulb tells you how much power it dissipates when connected to a 110 V circuit. It does not tell you how much light it emits.) A. A 100 W bulb has more resistance B. A 40 W bulb has more resistance C. Both the same D. Need more info Brightness of the bulb depends on how much power it dissipates è how much current it draws from the power grid è resistance of the bulb 35
Conceptual quesdon Which bulb is brighter Q4 Assume that the four light bulbs and two baweries are idendcal. Which circuit gives off more light? A. The circuit with the two bulbs in parallel B. The circuit with the two bulbs in series C. Both the same D. Need more info Series: R eq = 2R => Current I = ε/2r => Power dissipated P = I 2 R = ε 2 /4R Parallel: Current through every resistor I = ε/r => Power dissipated P = I 2 R = ε 2 /R 36
Conceptual quesdon Add a wire Charge flows through a light bulb. Suppose a wire with negligible resistance is connected across the bulb as shown. When the wire is connected, r wire ε Q5 A. All the charge condnues to flow through the bulb. B. All the charge flows through the wire. C. Half the charge flows through the wire, the other half condnues through the bulb. D. None of the above + Charge/current always follows the smallest resistance path. 37
Exercise: Power that radio A student kept his 9.0 V, 7.0 W radio turned on at full volume from 5:00 p.m. undl 2:00 a.m. How much charge went through it? What do we know: ΔV = 9.0 V P = 7.0 W Δt from 5:00 pm to 2:00 am I = Δq Δt Δq = IΔt Time: t = 9 h = 9 3600 s = 32400 s We can find the current from power dissipated by the radio P = IΔV I = P ΔV I = 7W 9V = 0.78A Δq = IΔt = 0.78A 32400s = 25272C = 2.53 10 4 C 38
Kirchhoff s rules JuncDon rule The sum of the current entering any juncdon must be equal to the sum of the currents leaving that juncdon I in = I out A juncdon is a place where two or more wires (or other components) meet. 39
Kirchhoff s rules loop rule The algebraic sum of the potendal difference in any loop must equal zero. ΔV = 0 For any circuit we must calculate this rule for the minimum number of loops such that each circuit element is crossed at least once. The direcdon of the loop is your choice. 40
Kirchhoff s rules loop rule Sign convendons 41
Exercise: Circuit analysis I Calculate the current through each ideal bawery in the figure. I R1 R1 + ε I 1 R 3 + 2 I II ε 1 ε 2 I I + R 1 I a b I 2 I 2 R 1 I 2 1. Since the direcdons of the currents are not given assume it yourself JuncDon rule: I = I 1 + I 2 2. Draw two loops and choose the direcdons of both Loop I: ε 1 IR 1 I 1 R 2 ε 2 IR 1 = 0 2 I 2I 1 3 I = 0 R 1 =1Ω R 2 = 2Ω ε 1 = 2.0V ε 2 = ε 3 = 3.0V Loop II: ε 2 + I 1 R 2 I 2 R 1 ε 3 I 2 R 1 = 0 3+ 2I 1 I 2 3 I 2 = 0 42
Exercise: Circuit analysis I I = I 1 + I 2 2 I 2I 1 3 I = 0 3+ 2I 1 I 2 3 I 2 = 0 2I 1 = 2I 2 I 1 = I 2 SubsDtute this into the first equadon I = I 1 + I 2 = 2I 1 And place this into the second equadon 2 2I 1 2I 1 3 2I 1 = 0 6I 1 = 1 I 1 = 0.167A I 2 = I 1 = 0.167A I = 2I 1 = 0.333A Current is a scalar it has to be a posidve number. Minus signs in the current values mean that all the currents actually flow in the direcdons opposite to the ones we assumed. 43
Exercise: Circuit analysis II Loop I Loop II a) Apply the juncdon rule at each juncdon and write the equadons that result. b) Choose two loops and write the equadons resuldng from the loop rule. c) Find i, r and E JuncDon rule: I in = I out i =1A + 2A = 3A Loop II: ε +1A 1Ω 2A 3Ω = 0 ε = 1V + 6V = 5V Loop I: 12V ir 1A 1Ω ε = 0 12V 3A r 1V 5V = 0 6V = 3A r r = 2Ω 44
What if capacitors occur muldple Dmes in a Capacitors in parallel circuit? ε C 1 C 2 C 3 C 4 The potendal difference (voltage) across capacitors in parallel must by the same The charges on the plates add up ΔV 1 = ΔV 2 = ΔV 3 = ε Q eq = Q 1 +Q 2 +... ε C eq = ε C 1 +ε C 2 +... C eq = C 1 + C 2 + + C n = C i. n i= 1 45
Capacitors in series C 1 C 2 C 3 C 4 +Q - Q +Q - Q +Q - Q +Q - Q ε For capacitors in series the charge on the plates is the same. ε = ΔV 1 + ΔV 2 + ΔV 3 +... Q C eq = Q C 1 + Q C 2 +.. 1 = 1 + 1 + + 1 n 1 =. C eq C 1 C 2 C n i=1 C i 46
Exercise: Equivalent capacitance Find the equivalent capacitance of the following circuit: 12.0 nf 4.00 nf 6.00 nf 3.00 nf 8.00 nf 2.00 nf 3. 4.0 nf, 6.0 nf and 3.0 nf capacitors are connected in series 1 = 1 C 3 4nF + 1 6nF + 1 3nF = 9 12nF 3.00 nf 1. 2.0 nf and 3.0 nf capacitors are connected in parallel C 1 = 2nF + 3nF = 5nF 2. These two are in series with 8.0 nf capacitor 1 = 1 C 2 8nF + 1 5nF = 13 40nF C 2 = 40 nf = 3.08nF 13 C 3 = 12 9 nf =1.33nF 47
Exercise: Equivalent capacitance Find the equivalent capacitance of the following circuit: 1.33 nf 4. C 2 = 3.08 nf and C 3 = 1.33 nf capacitors are connected in parallel 12.0 nf C 4 = 3.08nF +1.33nF = 4.41nF 3.08 nf 5. This is in series with 12.0 nf capacitor 1 C eq = 1 4.41nF + 1 12nF = 0.31 1 nf C eq = 3.22nF 48
RC circuits Suppose we put a capacitor and a resistor in series with a bawery Switch R - + C When we close the switch Current will flow Charge will start to build up on the capacitor + - ε As more charge builds up on the capacitor, it gets harder for current to flow (repelled by the charges already on the capacitor plate) stored charge goes up amount of current flow goes down 49
Charging a capacitor V C ( t) = ε ( 1 e t τ ) I ( t) = I 0 e t τ Q C ( t) = Cε ( 1 e t τ ) I 0 = ε R and τ = RC τ=rc is the Dme constant 50
Exercise: Charging a capacitor An inidally uncharged 70- µf capacitor is hooked up to a 560- kω resistor. What is the charge on the capacitor 8 s amer making the connecdon with an ideal 9.00- V bawery? Q C ( t) = Cε ( 1 e t τ ) Q C $ ( t) = 70 10 6 F 9V 1 e 8s 560 10 ' 3 Ω 70 10 6 F & ) % ( Q C ( t) = 630 10 6 C( 1 e 8s 39.2s ) Q C ( t) = 630 10 6 C 0.18 =116.310 6 C 51
Exercise: Charging a capacitor A capacitor, C = 3.1 µf, and a resistor, R = 5.3 kω, are connected to a bawery, ε = 15 V, as shown. At t = 0 the switch is closed. At what Dme is the current across the resistor equal to 25% of its maximum original value? What do we know: C = 3.1 µf R = 5.3 kω ε = 15 V I ( t) = I 0 e t τ When current is at 25% of its original value => I = 0.25I 0 0.25I 0 = I 0 e t τ 0.25 = e t τ Now we have to deal with a logarithmic funcdon, but A = e B B = ln A t τ = ln0.25 Time constant τ = RC ε Natural logarithm built in your calculator R S C t = τ ln0.25 t = 3.1 10 6 F 5.3 10 3 Ω ln0.25 = 22.78 10 3 s 52
Measuring voltage and current Ideal voltmeter draws no current connected in parallel (infinite resistance) Ideal ammeter causes no change in potendal change connected in series (zero resistance) V A 53
Review Electric current I = Δq Δt The current in the wire is: Resistors in series R eq = R 1 + R 2 + R 3 +... ΔV = ΔV 1 + ΔV 2 + ΔV 3 = IR 1 + IR 2 + IR 3 Δq Δt en Δt e I = = = Resistance R = L ρ A Ohm s law ΔV = IR neav ρ R = R0 d Temperature dependence ( + ( T )) = ρ 1 α T 0 0 ( 1+α ( T T0 )) Resistors in parallel 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 +... ΔV 1 = ΔV 2 = ΔV 3 = ΔV Current is branching (like a river) è I = I 1 + I 2 +I 3 54
Power for an EMF For a resistor: P = IΔV = I R = 2 Δ P = Iε V R 2 Review Kirchhoff s rules JuncDon rule Loop rule I in = I out ΔV = 0 Capacitors in series 1 C eq = 1 C 1 + 1 C 2 + + 1 C n = n 1. i=1 C i Charging a capacitor V C ( t) = ε ( 1 e t τ ) Q C ( t) = Cε ( 1 e t τ ) Capacitors in parallel C eq = C 1 + C 2 + + C n = C i. n i= 1 I ( t) = I 0 e t τ I 0 = ε R and τ = RC 55