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Download October 1st Circuits - Chapter 28
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October 1st Circuits - Chapter 28 How to Analyze Complex Circuits ! Kirchhoff’s junction rule (or current law) – ! From conservation of charge ! Sum of currents entering a junction is equal to sum of currents leaving that junction ! Kirchhoff’s ! From loop rule (or voltage law) – conservation of energy ! Sum of changes in potential going around a complete circuit loop equals zero Circuits (Figs. 28-9a, 28-9b) ! ! ! ! What is i through the battery? Label currents. New label for every branch. Pick any arbitrary direction. i through R1 or R4 is same as for battery Can use loop rule E − i1 R1 − i2 R2 − i1 R4 = 0 + - - + + - Circuits (Figs. 28-9a, 28-9b) "− i1 R1 − i2 R2 − i1 R4 = 0 ! ! Equation has 2 unknowns so need to apply loop rule again Take the loop through R2 and R3 − i3 R3 + i2 R2 = 0 i1 = i2 + i3 − (i1 − i2 ) R3 + i2 R2 = 0 i1 R3 i2 = − ( R3 + R2 ) Circuits (Figs. 28-9a, 28-9b) Now solve for i1: "− i R − i R − i R = 0 1 1 2 2 1 4 i R R 1 3 2 "− i R − −i R = 0 1 1 (R + R ) 1 4 2 3 " i1 = R 2 R3 R1 + + R4 ( R 2 + R3 ) Circuits (Figs. 28-9a, 28-9b) ! ! What is current Voltage lost (V) in R2 ? Recall that V=iR V2 = i 2 R2 "− i1 R1 − i2 R2 − i1 R4 = 0 i1 R1 + i1 R4 − " i2 = R2 " i1 = R 2 R3 R1 + + R4 ( R 2 + R3 ) Circuits (Fig. 28-13) ! Circuits where current varies with time ! RC series circuit – a resistor and capacitor are in series with a battery and a switch ! At t =0 switch is open and capacitor is uncharged so q =0 Circuits (Fig. 28-13) ! ! ! Close the switch at point a Charge flows (current) from battery to capacitor, increasing q on plates and V across plates When VC equal Vbattery flow of charge stops (current is zero) and charge on capacitor is q = CV = CE Circuits (Fig. 28-13) ! ! Want to know how q and V of capacitor and i of the circuit change with time when charging the capacitor Apply loop rule, traversing clockwise from battery q E − iR − =0 C i Circuits (Fig. 28-13) q E − iR − = 0 C ! ! ! i Contains 2 of the variables we want i and q Remember dq i = dt Substituting gives dq q R + = E dt C Circuits (Fig. 28-14a) dq q + = E R dt C ! ! Need a function which satisfies initial condition q = 0 at t = 0 and final condition of q = C E at t=∞ For charging a capacitor ( q = CE 1 − e −t RC ) Circuits (Fig. 28-14b) ( q = CE 1 − e ! ! −t RC ) Want current as a function of time For charging a capacitor dq E −t i = = e dt R RC Circuits (Fig. 28-13) ( q = CE 1 − e ! ! −t RC ) Want V across the capacitor as function of time For charging a capacitor VC ( q −t = = E 1− e C RC ) Circuits (Fig. 28-13) ! ! Want to know how q of capacitor and i of the circuit change with time when discharging the capacitor At new time t = 0, throw switch to point b and discharge capacitor through resistor R i Circuits (Fig. 28-13) ! Apply the loop rule again but this time no battery q − iR − = 0 C ! Substituting for i again gives differential equation dq q R + =0 dt C i Circuits (Fig. 28-13) dq q R + =0 dt C ! Solution must satisfy initial condition that q0 = CV0 ! For discharging a capacitor q = q0e −t RC i Circuits (Fig. 28-13) q = q0e ! −t RC i Find i for discharging capacitor with initial condition at i0 = V0/R = q0/RC at t = 0 dq q0 −t RC i= = − e dt RC Negative sign means charge is decreasing Circuits ! Charging capacitor ( q = CE 1 − e −t RC E −t i = e R ! RC ) ! Discharging capacitor q = q0 e − t RC q0 −t RC i = − e RC Define capacitive time constant – greater τ, greater (dis)charging time τ = RC