EE201 Lecture 17 P 1 Capacitors and Inductors cont Inductors in Series voltage division see lect 15 P 9 Ln vin t vn t Leq Example Find Leq and v1 t v1 t 4mH 16e 10t V 16mH 12mH Leq Step 1 Find Leq Leq 4 mH 16 mH 12 mH 32 mH EE201 Lecture 17 P 2 Step 2 Find v1 t v1 t 4 mH 32 mH 16e 10t V 2e 10t V Inductors in Parallel current division see lect 15 P 13 1 Lj iin t iLj 1 n L n Example Find Leq and i3 t i3 t 100e mA t Leq 16mH 32mH 32mH EE201 Lecture 17 P 3 Step 1 Leq computation 1 Leq 1 1 1 4 16mH 32mH 32mH 32mH Leq 8mH Step 2 Current division 1 32 mH i3 1 8 mH 1 is i3 25e t mA 4 100e t mA EE201 Lecture 17 P 4 Example Find Leq and iout t v1 t 6H iout 3H 6H 9sin 2 t A Leq 6H 3H 6H Leq 6H 2H 8H 1 6 iout 1 6 1 3 9sin 2 t 3sin 2 t A EE201 Lecture 17 P 5 Capacitors in Series voltage division see lect 16 P 9 1 Cn vC vCn 1 Ceq Example Find Ceq and v2 t in the circuit 16 F 16 F v2 16 cos t V 30 F 16 F 10 F Step 1 Ceq 1 1 16 F 1 16 F 1 56 F Ceq 7 F EE201 Lecture 17 P 6 Step 2 Find v2 t using voltage division v2 t 1 16 v 7 16cos t s 1 7 16 v2 t 7 cos t V Example Find Ceq and vout 18 cos t V Step 1 Ceq 8mF 12mF 24mF 60mF 8 24 32 60 12 mF 72 mF Ceq 6 mF 10mF 16 mF vout EE201 Lecture 17 P 7 Step 2 1 60 vout 18cos t 18cos t 1 6 1 60 1 12 vout 3 cos t V Capacitors in Parallel current division see lect 16 P 11 Cn iC iCn Ceq Example Find i2 i2 16 cos t A 8mF 12mF 24mF 60mF Branch 1 2 EE201 Lecture 17 P 8 Step 1 From example on P 6 equivalent capacitances for branch 1 and 2 are known C1 6 mF C2 10mF Ceq 16 mF Step 2 Calculate current through branch 2 i2 i2 C2 Ceq 10mF 16mF iC 16 cos t A i2 10 cos t A EE201 Lecture 17 P 9 Introduction to First Order Circuits RL and RC Circuits Inductor resistor RL Circuit Combine an inductor with a resistor to examine how initial inductor current decays with time R iL vL L diL v L RiL dt diL L dt RiL 0 First order differential equation describing how inductor current behaves with time diL dt R iL 0 L EE201 Lecture 17 P 10 Capacitor resistor RC Circuit Combine a capacitor with a resistor to examine how initial capacitor voltage decays with time R vC ic C vC dvC ic C dt R dvC dt 1 vC 0 RC Solutions for Differential Equations General Differential Eqn dx dt x iL or vC time constant L R RL RC RC 1 x 0 EE201 Lecture 17 P 11 Solution form for general differential equation x t e t to x t0 17 1 x t e K t to x t0 where K frequency Note 1 K 1 In RL circuit driven by an independent source for a long time the inductor behaves as a short 2 In RC circuit driven by an independent source for a long time the capacitor behaves as an open General strategy reduce complicated circuit attached to inductor or capacitor to Thevenin equivalent resistor attach inductor or capacitor to resistor and use Eqn 17 1 EE201 Lecture 17 Example Find vL t for t 0 1 0 1 t 0 P 12 iL t 11 V 9 0 5 H vL t Step 1 Find initial condition iL 0 For all times t 0 circuit is configured as below 0 1 t 0 1 iL t 11 V 9 0 5 H vL t EE201 Lecture 17 P 13 Assume long time operation because voltage source has been on for long time Therefore inductor behaves as a short 11 V 10A i 0 L 1 1 iL 0 iL 0 because of the continuity property of inductor current iL 0 Was assumption of long time operation valid Calculate time constant for circuit when t 0 s L RTH 0 5 Vs A 1 10 0 45 s A long time is about 3 5 time constants Therefore 2 3 sec after the voltage source turned on in this case at t the inductor behaves as a short EE201 Lecture 17 P 14 Step 2 Find iL t t 0 s Configuration of circuit for t 0 s 1 0 1 t 0 11 V 9 0 5 H RTH 10 iL t exp RTH t i 0 L iL t 10 e 20t A L Step 3 Calculate vL t t 0 vL t L diL dt 100e 20t V iL t vL t EE201 Lecture 17 P 15 Example Find vC t t 0 1 t 0 t 1 9 10V 72 4 5 0 1F vC t Step 1 Find vC 0 vC 0 For all t 0 capacitor is open vC 0 9V vC 0 Step 2 Find vC t 0 t 1 RTH 8 vC t vC 0 e t to RC v C t RTHC 0 8 s 1 RC 1 25 s 1 EE201 Lecture 17 P 16 Therefore vC t 9 exp 1 25t V Step 3 Find vC t 1 t RTH 3 0 1F v C t vC 1 vC 1 9e 1 25 vC 1 vC 1 2 58 V vC t vC 1 e t to RC Step 4 Plot vC t vs t vC t 9 0 8 s vC t 2 58 e t 1 0 3 V 2 58 0 3 s 1 2 t
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