Electricity & Magnetism Lecture 9: Electric Current

Electricity & Magnetism Lecture 9: Electric Current Today s Concept: Electric Current Electricity & Magne8sm Lecture 9, Slide 1

Battery and Bulb Which current flow model is correct? Figure 22-5: Four alternative models for current flow Model A There will be no electric current left to flow in the wire attached to the base of the battery Model B The electric current will travel in a direction toward the bulb in both wires Model C The direction of the current will be in the direction shown, but there will be less current in the return wire Model D The direction of the current will be as shown, and it will be the same in both wires

Activity 22-5 + - + + + - + - + A + + - A - A When Figure the 22-9: switch Two bulbs is connected closed, in series and with a voltmeter bulbs and are an iden8cal A)Top bulb is brighter B)BoHom bulb is brighter C)Both are equally bright

Activity 22-4 - + + - + - + + + - When the switch is closed A) bahery = bulb B) bahery < bulb C) bahery > bulb

+ + A + - + + - + - + A How dow the currents measured compare? A)I leo < I right B)I leo = I right C)I leo > I right

+ + A + - + + - + - + A How dow the oltages measured compare? A) leo < right B) leo = right C) leo > right

Your stuff so which way does DC current flow? - _- How many different things will sigma symbolize??? Since R = ρl/a, the greater the cross sec8onal area, the smaller the resistance, but the greater the length the higher the resistance. Is that why long cables have to be very thick? What if I put ammeter right between + and -? the part rela8ng to the ohm's law and current density stuff makes no sense to me.

A Note on Units Force is newtons: N = kg m/s 2 Electric Field: newton/coulomb (N/C = /m ) Electric poten;al: newton- meter/coulomb = volt! kg m2 /s 2 C = Capacitance: farad = coulomb/volt! farad is big, we usually use µf = 10 6 F pf = 10 12 F (µµf in olden days, puffs now) (nf = 10 9 not customary in N. America)

Key Concepts: 1) How resistance depends on A, L, σ, ρ 2) How to combine resistors in series and parallel 3) Understanding resistors in circuits σ is conductivity here (not surface charge density) ρ is resistivity here (not volume charge density). Today s Plan: 1) Review of resistance & preflights 2) Work out a circuit problem in detail Electricity & Magne8sm Lecture 9, Slide 4

I s A L Observables: = EL I = JA Ohm s Law: J = σ E Conduc8vity high for good conductors. I/A = σ /L I = /(L/σ A) R = Resistance ρ = 1/σ I = /R R = L σa Electricity & Magne8sm Lecture 9, Slide 5

This is just like Plumbing! I is like flow rate of water is like pressure R is how hard it is for water to flow in a pipe R = L σa To make R big, make L long or A small To make R small, make L short or A big Electricity & Magne8sm Lecture 9, Slide 6

1 CheckPoint: Two Resistors 2 Same current through both resistors Compare voltages across resistors A B C A B C Electricity & Magne8sm Lecture 9, Slide 7

CheckPoint: Current Density The SAME amount of current I passes through three different resistors. R 2 has twice the cross- sec8onal area and the same length as, and R 3 is three 8mes as long as but has the same cross- sec8onal area as. In which case is the CURRENT DENSITY through the resistor the smallest? σ 2σ σ Same Current Electricity & Magne8sm Lecture 9, Slide 8

Resistor Summary Series Parallel Wiring oltage Current Resistance R 2 Each resistor on the same wire. Different for each resistor. total = 1 + 2 oltage Divider Same for each resistor I total = I 1 = I 2 Increases R eq = + R 2 R 2 Each resistor on a different wire. Same for each resistor. total = 1 = 2 Different for each resistor I total = I 1 + I 2 Current Divider Decreases 1/R eq = 1/ + 1/R 2 Electricity & Magne8sm Lecture 9, Slide 9

Resistor symbol (ANSI)! N. America, Japan, China(?) Symbols 4.7 k = 4700 ohm 1.8 Ω = 1.8 ohm Alternate resistor symbol (DIN)! Europe, Middle East, Aus/NZ, Africa(?) 4k7 =4700 ohm 1R8= 1.8 ohm oltage Source Electrochemical Cell ( bahery ) some;mes used for voltage source

CheckPoint: Resistor Network 1 Three resistors are connected to a bahery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. Compare the current through R 2 with the current through R 3 : A. I 2 > I 3 B. I 2 = I 3 C. I 2 < I 3 R 2 in series with R 3 Current through R 2 and R 3 is the same Electricity & Magne8sm Lecture 9, Slide 10

= R 2 = R 3 = R CheckPoint 2 CheckPoint 3 CheckPoint 4 Compare the current through with the current through R 2 I 1 I 2 Compare the voltage across R 2 with the voltage across R 3 2 3 Compare the voltage across with the voltage across R 2 1 2 Electricity & Magne8sm Lecture 9, Slide 11

CheckPoint: Resistor Network 2 Three resistors are connected to a bahery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. I 23 I 1 Compare the current through with the current through R 2 : A. I 1 /I 23 =1/2 B. I 1 /I 23 =1/3 C. I 1 = I 23 D. I 1 /I 23 =2 E. I 1 /I 23 =3 We know: Similarly: Electricity & Magne8sm Lecture 9, Slide 12

CheckPoint: Resistor Network 3 Three resistors are connected to a bahery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. Compare the voltage across R 2 with the voltage across R 3 : 2 3 23 A. 2 > 3 B. 2 = 3 = C. 2 = 3 < D. 2 < 3 Consider loop Electricity & Magne8sm Lecture 9, Slide 13

CheckPoint: Resistor Network 4 Three resistors are connected to a bahery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. 1 23 Compare the voltage across with the voltage across R 2. A. 1 = 2 = B. 1 = 1/2 2 = C. 1 = 2 2 = D. 1 =1/2 2 =1/5 E. 1 =1/2 2 = 1/2 in parallel with series combina8on of R 2 and R 3 Electricity & Magne8sm Lecture 9, Slide 14

Calculation R 2 In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. R 3 R 4 What is 2, the voltage across R 2? Conceptual Analysis: Ohm s Law: when current I flows through resistance R, the poten8al drop is given by: = IR. Resistances are combined in series and parallel combina8ons R series = R a + R b (1/R parallel ) = (1/R a ) + (1/R b ) Strategic Analysis: Combine resistances to form equivalent resistances Evaluate voltages or currents from Ohm s Law Expand circuit back using knowledge of voltages and currents Electricity & Magne8sm Lecture 9, Slide 15

Calculation R 2 R 3 R 4 In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. What is 2, the voltage across R 2? Combine Resistances: and R 2 are connected: A) in series B) in parallel C) neither in series nor in parallel Parallel Combina8on R a Series Combina8on R a R b R b Parallel: Can make a loop that contains only those two resistors Series : Every loop with resistor 1 also has resistor 2. Electricity & Magne8sm Lecture 9, Slide 16

Calculation R 2 In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. R 3 R 4 What is 2, the voltage across R 2? We first will combine resistances R 2 + R 3 + R 4 : Which of the following is true? A) R 2, R 3 and R 4 are connected in series B) R 2, R 3, and R 4 are connected in parallel C) R 3 and R 4 are connected in series (R 34 ) which is connected in parallel with R 2 D) R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3 E) R 2 and R 4 are connected in parallel (R 24 ) which is connected in parallel with R 3 Electricity & Magne8sm Lecture 9, Slide 17

Calculation R 2 R 3 R 4 In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. What is 2, the voltage across R 2? R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3 Redraw the circuit using the equivalent resistor R 24 = series combina8on of R 2 and R 4. R 3 R 3 R 3 R 24 R 24 R 24 (A) (B) (C) Electricity & Magne8sm Lecture 9, Slide 18

Calculation In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. R 3 R 24 What is 2, the voltage across R 2? Combine Resistances: What is the value of R 234? R 2 and R 4 are connected in series = R 24 R 3 and R 24 are connected in parallel = R 234 A) R 234 = 1 Ω B) R 234 = 2 Ω C) R 234 = 4 Ω D) R 234 = 6 Ω R 2 and R 4 in series R 24 = R 2 + R 4 = 2Ω + 4Ω = 6Ω (1/R parallel ) = (1/R a ) + (1/R b ) 1/R 234 = (1/3) + (1/6) = (3/6) Ω -1 R 234 = 2 Ω Electricity & Magne8sm Lecture 9, Slide 19

Calculation In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. I 1 = I 234 R 234 234 R 24 = 6Ω R 234 = 2Ω What is 2, the voltage across R 2? and R 234 are in series. 234 = 1 + 2 = 3 Ω Our next task is to calculate the total current in the circuit Ohm s Law tells us: I 1234 = /234 = I 1234 234 = 18 / 3 = 6 Amps Electricity & Magne8sm Lecture 9, Slide 20

Calculation = I 1234 234 In the circuit shown: = 18, = 1Ω, R 2 = 2Ω, R 3 = 3Ω, and R 4 = 4Ω. R 24 = 6Ω R 234 = 2Ω I 1234 = 6 A What is 2, the voltage across R 2? a I 234 = I 1234 Since in series with R 234 I 1 = I 234 R 234 234 234 = I 234 R 234 = 6 x 2 b = 12 olts What is ab, the voltage across R 234? A) ab = 1 B) ab = 2 C) ab = 9 D) ab = 12 E) ab = 16 Electricity & Magne8sm Lecture 9, Slide 21

Calculation = 18 = 1Ω R 234 R 3 R 24 R 2 = 2Ω R 3 = 3Ω R 4 = 4Ω R 24 = 6Ω R 234 = 2Ω Which of the following are true? I 1234 = 6 Amps A) I 234 = 6 Amps 234 = 24 B) I 234 = I 24 C) Both A+B D) None 234 = 12 R 3 and R 24 were combined in parallel to get R 234 oltages are same! What is 2? Ohm s Law I 24 = 24 / R 24 = 12 / 6 = 2 Amps Electricity & Magne8sm Lecture 9, Slide 22

Calculation I 1234 R 2 I 24 = 18 R 3 R 24 R 3 = 1Ω R 2 = 2Ω R 3 = 3Ω Which of the following are true? A) 24 = 2 B) I 24 = I 2 C) Both A+B D) None R 2 and R 4 where combined in series to get R 24 The Problem Can Now Be Solved! Ohm s Law 2 = I 2 R 2 = 2 x 2 = 4 olts! R 4 Currents are same! R 4 = 4Ω. R 24 = 6Ω R 234 = 2Ω I 1234 = 6 Amps I 234 = 6 Amps 234 = 12 24 = 12 I 24 = 2 Amps What is 2? Electricity & Magne8sm Lecture 9, Slide 23

Quick Follow-Ups I 1 I 2 R 2 I 3 R 3 = R 4 What is I 3? A) I 3 = 2 A B) I 3 = 3 A C) I 3 = 4 A 3 = 234 = 12 I 3 = 3 /R 3 = 12/3Ω = 4A What is I 1? We know I 1 = I 1234 = 6 A a b R 234 = 18 = 1Ω R 2 = 2Ω R 3 = 3Ω R 4 = 4Ω R 24 = 6Ω R 234 = 2Ω 234 = 12 2 = 4 I 1234 = 6 Amps NOTE: I 2 = 2 /R 2 = 4/2 = 2 A I 1 = I 2 + I 3 Make Sense? Electricity & Magne8sm Lecture 9, Slide 24