# Lecture05bd - Physics 102: Lecture 05 Circuits and Ohm's...

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Unformatted text preview: Physics 102: Lecture 05 Circuits and Ohm's Law Physics 102: Lecture 5, Slide 1 Summary of Last Time Capacitors Physical Series Parallel Energy C = ke0A/d C=Q/V 1/Ceq = 1/C1 + 1/C2 Ceq = C1 + C2 U = 1/2 QV Summary of Today Resistors Physical Series Parallel Power R = r L/A V=IR Req = R1 + R2 1/Req = 1/R1 + 1/R2 P = IV Physics 102: Lecture 5, Slide 2 Electric Terminology Current: Moving Charges Symbol: I 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 = Volt Coulomb/sec P = VI Physics 102: Lecture 5, Slide 3 60 W= 60 J/s Physical Resistor Resistance: Traveling through a resistor, electrons bump into things which slows them down. Units: Ohms W R = r L /A r: Resistivity: Density of scatterers L: Length of resistor A: Cross sectional area of resistor I A - Ohms Law I = V/R Cause and effect (sort of like a=F/m) Double potential difference double current I = (VA)/ (r L) Physics 102: Lecture 5, Slide 4 L potential difference cause current to flow resistance regulate the amount of flow 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. 61% 1. R1 > R2 1 2 4% 2. 35% 3. R1 = R2 R1 < R2 R = r L /A 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 = ke0A/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... e I R Calculate I when e=24 Volts and R = 8 W Ohm's Law: V =IR I = V/R = 3 Amps Physics 102: Lecture 5, Slide 7 Resistors in Series One wire: Effectively adding lengths: Req=r(L1+L2)/A Since R L add resistance: R Req = R1 + R2 R = 2R 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. I1 = I2 = Ieq Both have voltage drops: V1+V2 V1 + V2 = Veq V1 V2 R1 R2 I Req I I Physics 102: Lecture 5, Slide 9 CheckPoint 2.1 Compare I1 the current through R1, with I10 the current through R10. 10% 1. 57% 2. 33% 3. I1<I10 I1=I10 I1>I10 "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." R1=1W e 0 R10=10W Note: I is the same everywhere in this circuit! Physics 102: Lecture 5, Slide 10 ACT: Series Circuit R1=1W Compare V1 the voltage across R1, with V10 the voltage across R10. 1. V1>V10 2. V1=V10 3. V1<V10 e 0 R10=10W V1 = I1 R1 = I x 1 V10 = I10 R10 = I x 10 Physics 102: Lecture 5, Slide 11 Practice: Resistors in Series Calculate the voltage across each resistor if the battery has potential 0= 22 volts. Simplify (R1 and R2 in series): R12 = R1 + R2 V12 = V1 + V2 I12 = I1 = I2 = 11 W = 0 = 22 Volts R1=1W e0 R2=10W e0 R12 = V12/R12 = 2 Amps Expand: V1 = I1R1 V2 = I2R2 = 2 x 1 = 2 Volts = 2 x 10 = 20 Volts R1=1W Check: V1 + V2 = V12 ? Physics 102: Lecture 5, Slide 12 e0 R2=10W Resistors in Parallel Two wires: Effectively adding the Area Since R 1/A add 1/R: 1/Req = 1/R1 + 1/R2 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: I1 + I2 = Ieq Voltage is same across each: V1 = V2 = Veq I1+I2 I1 I2 V R1 I1+I2 R2 V Req I1+I2 V Physics 102: Lecture 5, Slide 14 CheckPoint 3.1 What happens to the current through R2 when the switch is closed? 18% Increases 37% Remains Same 45% 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 Ibattery = I2 + I3 Practice: Resistors in Parallel e R2 R3 Determine the current through the battery. Let = 60 Volts, R2 = 20 W and R3=30 W. Simplify: R2 and R3 are in parallel e 1/R23 = 1/R2 + 1/R3 R23 = 12 W V23 = V2 = V3 = 60 Volts I23 = I2 + I3 = V23 /R23 = 5 Amps Physics 102: Lecture 5, Slide 17 R23 ACT / CheckPoint 4.1,4.2 1 2 3 Which configuration has the smallest resistance? 30% A. 1 B. 2 5% C. 3 65% Which configuration has the largest resistance? B. 2 Physics 102: Lecture 5, Slide 18 R 2R R/2 76% Try it! R1 Calculate current through each resistor. R1 = 10 W, R2 = 20 W, R3 = 30 W, e = 44 V Simplify: R2 and R3 are in parallel 1/R23 = 1/R2 + 1/R3 V23 = V2 = V3 I23 = I2 + I3 : R23 = 12 W e R2 R3 R1 e R23 Simplify: R1 and R23 are in series R123 = R1 + R23 V123 = V1 + V23= e I123 = I1 = I23 = Ibattery : I123 = 44 V/22 W = 2 A : R123 = 22 W e R123 Power delivered by battery? P=IV = 244 = 88W Physics 102: Lecture 5, Slide 19 Try it! (cont.) Calculate current through each resistor. R1 = 10 W, R2 = 20 W, R3 = 30 W, e = 44 V Expand: R1 and R23 are in series R123 = R1 + R23 V123 = V1 + V23= e I123 = I1 = I23 = Ibattery : I23 = 2 A : V23 = I23 R23 e R1 R123 e = 24 V R1 R2 R23 Expand: R2 and R3 are in parallel 1/R23 = 1/R2 + 1/R3 V23 = V2 = V3 I2 = V2/R2 =24/20=1.2A I23 = I2 + I3 I3 = V3/R3 =24/30=0.8A e R3 Physics 102: Lecture 5, Slide 20 Summary Series R1 R2 R2 Each resistor on a different wire. Same for each resistor. Vtotal = V1 = V2 Different for each resistor Itotal = I1 + I2 Decreases 1/Req = 1/R1 + 1/R2 Parallel R1 Wiring Voltage Current Resistance Each resistor on the same wire. Different for each resistor. Vtotal = V1 + V2 Same for each resistor Itotal = I1 = I2 Increases Req = R1 + R2 Physics 102: Lecture 5, Slide 21 ...
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## This note was uploaded on 02/04/2012 for the course PHYSICS 102 taught by Professor Demarco during the Spring '11 term at University of Illinois at Urbana–Champaign.

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