Circuits and Ohm’s Law Physics 102: Lecture 05 PRACTICE EXAMS AVAILABLE-CLICK ON WEB PAGE Conflict Exam sign up available in grade book Homework, keep lots of digits! Exam I Feb. 18

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 5

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 7

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 –Double potential difference  double current –I = (VA)/ (  L) A L 10

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% The bigger the resistor, the greater the resistance. The charge has to travel the same distance in each resistor The greater the area, the easier it is for current to flow through R =  L /A

Comparison: Capacitors vs. Resistors Capacitors store energy as separated charge: U=1/2QV –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!

Visualization Practice… –Calculate I when  =24 Volts and R = 8  –Ohm’s Law: V =IR Simple Circuit R  I 15 I I = V/R = 3 Amps

Resistors in Series One wire: –Effectively adding lengths: R eq =  (L 1 +L 2 )/A –Since R  L add resistance: Visualization 16 RR = 2R R eq = R 1 + R 2

Resistors in Series 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 20 R eq = V eq = V 1 + V 2 = R 1 + R 2 I eq I eq

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 % 67% 22% V 1 = I 1 R 1 = I x 1 V 10 = I 10 R 10 = I x 10 Since they are connected in series, the current is the same for every resistor. If charge goes through one resistor, it must go through other. Note: I is the same everywhere in this circuit! ACT: Series Circuit

Practice: Resistors in Series Calculate the voltage across each resistor if the battery has potential V 0 = 22 volts. 28 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 

Resistors in Parallel Two wires: –Effectively adding the Area –Since R  1/A add 1/R: 29 R R = R/2 1/R eq = 1/R 1 + 1/R 2 Visualization

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 32

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 36% 26% 38% 35 I battery = I 2 + I 3 ACT: Parallel Circuit

Practice: Resistors in Parallel Determine the current through the battery. Let E = 60 Volts, R 2 = 20  and R 3 =30 . 35 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

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

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 38

Preflight 5.6, 5.7 Which configuration has the smallest resistance? % 5% 69% “The one has only one resistor so it has the least resistance” “In parallel, the resistors add like this: [(1/R) + (1/R)]^-1 which is the same as R/2. This means the resistance is halved compared to a single R. In series, the total resistance is just R + R or 2R, twice the resistance of A.” Which configuration has the largest resistance? 2 80% Nice Work! R 2R R/2

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!!

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   48 : I 123 = 44 V/22  A Power delivered by battery? P=IV = 2  44 = 88W

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 48

See you next Monday! Read 18.5, 18.7 Extra Problems from Ch 18: –Conceptual 9 –Multiple Choice 4 –Problems 24, 25, 40, 41, 42