Physics 102: Lecture 5, Slide 1 Circuits and Ohm’s Law Physics 102: Lecture 05

Physics 102: Lecture 5, Slide 2 Summary of Last Time Capacitors –PhysicalC =  0 A/d C=Q/V –Series1/C eq = 1/C 1 + 1/C 2 –ParallelC eq = C 1 + C 2 –EnergyU = 1/2 QV Resistors –PhysicalR =  L/AV=IR –SeriesR eq = R 1 + R 2 –Parallel1/R eq = 1/R 1 + 1/R 2 –PowerP = IV Summary of Today

Physics 102: Lecture 5, Slide 3 Electric Terminology Current: Moving Charges –Symbol: I –Unit: Amp  Coulomb/second –Count number of charges which pass point/sec –Direction of current is direction that + flows Power: Energy/Time –Symbol: P –Unit: Watt  Joule/second = Volt Coulomb/sec –P = VI

Physics 102: Lecture 5, Slide 4

Physics 102: Lecture 5, Slide 5 Physical Resistor Resistance: Traveling through a resistor, electrons bump into things which slows them down. R =  L /A –  : Resistivity: Density of bumps – L: Length of resistor – A: Cross sectional area of resistor Ohms Law I = V/R –Cause and effect (sort of like a=F/m) potential difference cause current to flow resistance regulate the amount of flow –Double potential difference  double current –I = (VA)/ (  L) A L

Physics 102: Lecture 5, Slide 6 Preflight 5.1 Two cylindrical resistors are made from the same material. They are of equal length but one has twice the diameter of the other. 1.R 1 > R 2 2.R 1 = R 2 3.R 1 < R % 5% 24%

Physics 102: Lecture 5, Slide 7 Comparison: Capacitors vs. Resistors Capacitors store energy as separated charge: U=QV/2 –Capacitance: ability to store separated charge: C =  0 A/d –Voltage drop determines charge: V=Q/C Resistors dissipate energy as power: P=VI –Resistance: how difficult it is for charges to get through: R =  L /A –Voltage drop determines current: V=IR Don’t mix capacitor and resistor equations!

Physics 102: Lecture 5, Slide 8

Physics 102: Lecture 5, Slide 9 Visualization Practice… –Calculate I when  =24 Volts and R = 8  –Ohm’s Law: V =IR Simple Circuit R  I I I = V/R = 3 Amps

Physics 102: Lecture 5, Slide 10 Resistors in Series One wire: –Effectively adding lengths: R eq =  (L 1 +L 2 )/A –Since R  L add resistance: RR = 2R R eq = R 1 + R 2

Physics 102: Lecture 5, Slide 11 Resistors in Series: “Proof” that R eq =R 1 +R 2 Resistors connected end-to-end: –If charge goes through one resistor, it must go through other. I 1 = I 2 = I eq –Both have voltage drops: V 1 + V 2 = V eq R1R1 R2R2 R eq R eq = V eq = V 1 + V 2 = R 1 + R 2 I eq I eq

Physics 102: Lecture 5, Slide 12

Physics 102: Lecture 5, Slide 13 Preflight 5.3 Compare I 1 the current through R 1, with I 10 the current through R I 1 <I 10 2.I 1 =I 10 3.I 1 >I 10 R 1 =1  00 R 10 =10  Compare V 1 the voltage across R 1, with V 10 the voltage across R V 1 >V V 1 =V V 1 <V 10 V 1 = I 1 R 1 = I x 1 V 10 = I 10 R 10 = I x 10 Note: I is the same everywhere in this circuit! ACT: Series Circuit

Physics 102: Lecture 5, Slide 14 Practice: Resistors in Series Calculate the voltage across each resistor if the battery has potential V 0 = 22 volts. R 12 = R 1 + R 2 V 12 = V 1 + V 2 I 12 = I 1 = I 2 = 11  R 12 00 = V 0 = 22 Volts = V 12 /R 12 = 2 Amps Expand: V 1 = I 1 R 1 V 2 = I 2 R 2 = 2 x 1 = 2 Volts = 2 x 10 = 20 Volts R 1 =1  00 R 2 =10  Check: V 1 + V 2 = V 12 ? Simplify (R 1 and R 2 in series): R 1 =1  00 R 2 =10 

Physics 102: Lecture 5, Slide 15 Resistors in Parallel Two wires: –Effectively adding the Area –Since R  1/A add 1/R: R R = R/2 1/R eq = 1/R 1 + 1/R 2

Physics 102: Lecture 5, Slide 16

Physics 102: Lecture 5, Slide 17 Resistors in Parallel Both ends of resistor are connected: –Current is split between two wires: I 1 + I 2 = I eq –Voltage is same across each: V 1 = V 2 = V eq R eq R2R2 R1R1

Physics 102: Lecture 5, Slide 18 Preflight 5.5 What happens to the current through R 2 when the switch is closed? Increases Remains Same Decreases What happens to the current through the battery? (1)Increases (2)Remains Same (3)Decreases ACT: Parallel Circuit

Physics 102: Lecture 5, Slide 19 Practice: Resistors in Parallel Determine the current through the battery. Let E = 60 Volts, R 2 = 20  and R 3 =30 . 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 R2R2 R3R3  R 23  R 23 = 12  = 60 Volts = V 23 /R 23 = 5 Amps Simplify: R 2 and R 3 are in parallel

Physics 102: Lecture 5, Slide 20

Physics 102: Lecture 5, Slide 21 Johnny “Danger” Powells uses one power strip to plug in his microwave, coffee pot, space heater, toaster, and guitar amplifier all into one outlet. 1. The resistance of the kitchen circuit is too high. 2. The voltage across the kitchen circuit is too high. 3. The current in the kitchen circuit is too high. Toaster Coffee PotMicrowave 10 A5 A10 A 25 A This is dangerous because… (By the way, power strips are wired in parallel.) ACT: Your Kitchen

Physics 102: Lecture 5, Slide 22 Voltage Current Resistance Series Parallel Summary Different for each resistor. V total = V 1 + V 2 Increases R eq = R 1 + R 2 Same for each resistor I total = I 1 = I 2 Same for each resistor. V total = V 1 = V 2 Decreases 1/R eq = 1/R 1 + 1/R 2 Wiring Each resistor on the same wire. Each resistor on a different wire. Different for each resistor I total = I 1 + I 2 R1R1 R2R2 R1R1 R2R2

Physics 102: Lecture 5, Slide 23 ACT/Preflight 5.6, 5.7 Which configuration has the smallest resistance? Which configuration has the largest resistance? R 2R R/2

Physics 102: Lecture 5, Slide 24

Physics 102: Lecture 5, Slide 25 Parallel + Series Tests Resistors R 1 and R 2 are in series if and only if every loop that contains R 1 also contains R 2 Resistors R 1 and R 2 are in parallel if and only if you can make a loop that has ONLY R 1 and R 2 Same rules apply to capacitors!!

Physics 102: Lecture 5, Slide 26 Try it! R1R1 R2R2 R3R3 Calculate current through each resistor. R 1 = 10 , R 2 = 20  R 3 = 30    V Simplify: R 2 and R 3 are in parallel 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 Simplify: R 1 and R 23 are in series R 123 = R 1 + R 23 V 123 = V 1 + V 23 =  I 123 = I 1 = I 23 = I battery : R 23 = 12  : R 123 = 22  R 123  R1R1 R 23   : I 123 = 44 V/22  A Power delivered by battery? P=IV = 2  44 = 88W

Physics 102: Lecture 5, Slide 27 Try it! (cont.) R1R1 R2R2 R3R3 Calculate current through each resistor. R 1 = 10 , R 2 = 20  R 3 = 30    V Expand: R 2 and R 3 are in parallel 1/R 23 = 1/R 2 + 1/R 3 V 23 = V 2 = V 3 I 23 = I 2 + I 3 Expand: R 1 and R 23 are in series R 123 = R 1 + R 23 V 123 = V 1 + V 23 =  I 123 = I 1 = I 23 = I battery R 123  R1R1 R 23   : I 23 = 2 A : V 23 = I 23 R 23 = 24 V I 2 = V 2 /R 2 =24/20=1.2A I 3 = V 3 /R 3 =24/30=0.8A

Physics 102: Lecture 5, Slide 28