Physics 102: Lecture 05 Circuits and Ohm s Law Physics 102: Lecture 5, Slide 1
Summary of Last Time Capacitors Physical C = ke 0 A/d C=Q/V Series 1/C eq = 1/C 1 + 1/C 2 Parallel C eq = C 1 + C 2 Energy U = 1/2 QV Summary of Today Resistors Physical R = r L/A V=IR Series R eq = R 1 + R 2 Parallel 1/R eq = 1/R 1 + 1/R 2 Power P = IV Physics 102: Lecture 5, Slide 2
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 + charge flows Power: Energy/Time Symbol: P Unit: Watt Joule/second = Physics 102: Lecture 5, Slide 3 Volt Coulomb/sec P = VI 60 W= 60 J/s I
Physical Resistor Resistance: Traveling through a resistor, electrons bump into things which slows them down. R = r L /A r: 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)/ (r L) L A - I Physics 102: Lecture 5, Slide 4
CheckPoint 1.1 Two cylindrical resistors are made from the same material. They are of equal length but one has twice the diameter of the other. 1 2 1. R 1 > R 2 2. R 1 = R 2 3. R 1 < R 2 Physics 102: Lecture 5, Slide 5
Comparison: Capacitors vs. Resistors Capacitors store energy as separated charge: U=QV/2 Capacitance: ability to store separated charge: C = ke 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 = r L /A Voltage drop determines current: V=IR Don t mix capacitor and resistor equations! + + - Physics 102: Lecture 5, Slide 6
Simple Circuit I Phet Visualization Practice Calculate I when e=24 Volts and R = 8 W Ohm s Law: V =IR e I R I = V/R = 3 Amps Physics 102: Lecture 5, Slide 7
Resistors in Series One wire: Effectively adding lengths: R eq =r(l 1 +L 2 )/A Since R L add resistance: R R eq = R 1 + R 2 = 2R R Physics 102: Lecture 5, Slide 8
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: I R eq V 1 + V 2 = V eq V 1 R 1 V 1 +V 2 I V 2 R 2 I Physics 102: Lecture 5, Slide 9
CheckPoint 2.1 Compare I 1 the current through R 1, with I 10 the current through R 10. R 1 =1W e 0 R 10 =10W 1. I 1 <I 10 2. I 1 =I 10 3. I 1 >I 10 Physics 102: Lecture 5, Slide 10
ACT: Series Circuit R 1 =1W Compare V 1 the voltage across R 1, with V 10 the voltage across R 10. e 0 R 10 =10W 1. V 1 >V 10 2. V 1 =V 10 3. V 1 <V 10 Physics 102: Lecture 5, Slide 11
Practice: Resistors in Series R 1 =1W e 0 R 2 =10W Calculate the voltage across each resistor if the battery has potential ε 0 = 22 volts. Simplify (R 1 and R 2 in series): R 12 = R 1 + R 2 V 12 = V 1 + V 2 I 12 = I 1 = I 2 = 11 W = ε 0 = 22 Volts = V 12 /R 12 = 2 Amps e 0 R 12 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 =1W Check: V 1 + V 2 = V 12? e 0 R 2 =10W Physics 102: Lecture 5, Slide 12
Resistors in Parallel Two wires: Effectively adding the Area Since R 1/A add 1/R: 1/R eq = 1/R 1 + 1/R 2 Used in your house! R R = R/2 Physics 102: Lecture 5, Slide 13
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 I 1 +I 2 I 1 +I 2 I 1 I 2 V R 1 R 2 V R eq V Physics 102: Lecture 5, Slide 14 I 1 +I 2
CheckPoint 3.1 What happens to the current through R 2 when the switch is closed? Increases Remains Same Decreases Physics 102: Lecture 5, Slide 15
ACT: Parallel Circuit What happens to the current through the battery when the switch is closed? (A) Increases (B) Remains Same (C) Decreases Physics 102: Lecture 5, Slide 16
Practice: Resistors in Parallel e R 2 R 3 Determine the current through the battery. Let ε = 60 Volts, R 2 = 20 W and R 3 =30 W. Simplify: R 2 and R 3 are in parallel 1/R 23 = 1/R 2 + 1/R 3 R 23 = 12 W V 23 = V 2 = V 3 = 60 Volts I 23 = I 2 + I 3 = V 23 /R 23 = 5 Amps e R 23 Physics 102: Lecture 5, Slide 17
ACT / CheckPoint 4.1,4.2 1 2 3 Which configuration has the smallest resistance? A. 1 B. 2 C. 3 Which configuration has the largest resistance? Physics 102: Lecture 5, Slide 18
Try it! Calculate current through each resistor. R 1 = 10 W, R 2 = 20 W, R 3 = 30 W, e = 44 V e R 1 R 2 R 3 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 : R 23 = 12 W e R 1 R 23 Simplify: R 1 and R 23 are in series R 123 = R 1 + R 23 V 123 = V 1 + V 23 = e I 123 = I 1 = I 23 = I battery : R 123 = 22 W : I 123 = 44 V/22 W = 2 A e R 123 Power delivered by battery? P=IV = 2 44 = 88W Physics 102: Lecture 5, Slide 19
Try it! (cont.) Calculate current through each resistor. R 1 = 10 W, R 2 = 20 W, R 3 = 30 W, e = 44 V e R 123 Expand: R 1 and R 23 are in series R 1 R 123 = R 1 + R 23 V 123 = V 1 + V 23 = e I 123 = I 1 = I 23 = I battery : I 23 = 2 A : V 23 = I 23 R 23 = 24 V e R 23 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 I 2 = V 2 /R 2 =24/20=1.2A I 3 = V 3 /R 3 =24/30=0.8A e R 1 R 2 R 3 Physics 102: Lecture 5, Slide 20
Summary Series Parallel R 1 Wiring Voltage Current Resistance R 1 R 2 Each resistor on the same wire. Different for each resistor. V total = V 1 + V 2 Same for each resistor I total = I 1 = I 2 Increases R eq = R 1 + R 2 R 2 Each resistor on a different wire. Same for each resistor. V total = V 1 = V 2 Different for each resistor I total = I 1 + I 2 Decreases 1/R eq = 1/R 1 + 1/R 2 Physics 102: Lecture 5, Slide 21