Lecture 2

# Lecture 2 - ΔV = E − IR Fluid Circuits Electric...

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Unformatted text preview: ΔV = E − IR Fluid Circuits Electric Circuits Object Flowing Fluid Electron (charge) Flows Through Pipe Metal Current m3/s C/s Energy Density J/m3 J/C Driven By Pressure Diﬀerence Voltage Diﬀerence Energy Added By Pump Battery Energy Dissipated By Resistance Resistance Fluid Circuits Object Flowing Electric Circuits Fluid Electron (charge) Fluid Electric Circuits Circuits Energy J/m3 Density J/C Fluid Circuits Flows Through Electric Circuits Pipe Metal Fluid Circuits Current Electric Circuits m3/s C/s Fluid Circuits Electric Circuits Driven By Pressure Diﬀerence Voltage Diﬀerence Energy Added By Pump Battery Fluid Circuits Energy Dissipated By Resistance Electric Circuits Resistance RPipe € 8ηL =4 πr L 1L R=ρ = A kA € Do Clicker Question 2,3 ΔP = E Pump − IR € ΔV = E − IR ΔV = E − IR From the last two lines, we see that:  ­ΔVDA = ΔVAD 0=E ­IR € E I= R ΔVAB =  ­IR ΔVBC = 0 ΔVCD = E ΔVDA = 0 ΔVAD = E ­IR Do Clicker Question 4 ΔV = E − IR VA=0 VB=0 VC=0 VD=0 VE=0 VA 0 6 16 16 0 VB  ­6 0 10 10  ­6 VC  ­16  ­10 0 0  ­16 VD  ­16  ­10 0 0  ­16 VE 0 6 16 16 0 Do Clicker Question 5 RS = R1 + R2 + R3 + ... ΔVS = ΔV1 + ΔV2 + ΔV3 + ... € € L 1L R=ρ = A kA Do Clicker Question 6 1 11 1 =+ + + ... RP R1 R2 R3 ΔVP = ΔV1 = ΔV2 = ΔV3 IP = I1 + I2 + I3 + ... € € € L 1L R=ρ = A kA Do Clicker Question 7 • A+B? • A+C? • A||C? • (A+B)||C? • D+E? • D||E? • E||F? • (E||F)+D? P = IΔV PBattery = IE 2 PR = −I R ΔV PR = − R 2 ΔV = 0 = E − IR1 − IR2 € ΔV = 0 = E1 − I1R1 − E 2 − ( I1 − I3 ) R2 ΔV = 0 = E 2 − I3 R3 − ( I3 − I1 ) R2 ΔV I P Battery 16V 2A 32W R3  ­6V 2A  ­12W R5  ­10V 2A  ­20W ΔV I P Battery 15V 8A 120W R3  ­15V 5A  ­75W R5  ­15V 3A  ­45W RT = RA + 1 1 1 + RB RC + RD 3E IT = 5R IT = IA 2 IB = IT 3 1 IC = ID = IT 3 5R = 3 ...
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## This note was uploaded on 08/21/2011 for the course PHYS 7B taught by Professor Mcraven during the Spring '10 term at UC Davis.

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