INDUCTANCE. RC Cicuits vs LR Circuits
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- Stuart Butler
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1 INDUTANE R cuts vs LR rcuts R rcut hargng (battery s connected): (1/ )q + (R)dq/ dt LR rcut = (R) + (L)d/ dt q = e -t/ R ) = / R(1 - e -(R/ L)t ) q ncreases from 0 to = dq/ dt decreases from / R to 0 Dschargng (battery s dsconnected): 0 = (1/ )q + (R)I ncreases from 0 to /R d/ dt decreases from / L to 0 0 = (R) + (L)d/ dt q = (e-t/ R q decreases from to 0 decreases from / R to 0 = (/ R)e-(R/ L)t decreases from / R to 0 d/ dt decreases from / L to 0 note : for a capactor, = q/ U = (1/ 2)q 2 / = (1/ 2) 2 = (1/ 2)q for an nductor, = L(dI/ dt) U L = (1/ 2)LI 2 These expressons stand alone, that s, you can t get the nductor expressons by nsertng the replacements gven above nto the capactor expressons for emf () and energy (U). Assumptons about Inductors n Multple Loops When nductors are a part of multple loops, they behave n the opposte manner of how we consdered capactors. Applyng these assumptons correctly s crucal. At t = 0, there s no current through any nductor! Therefore, the loop current wll be 0 for any loop that contans an nductor. However, f the nductor s n a branch, then current can stll flow around the permeter of the loops whch share that branch. At t = ( after a long tme ), treat the nductors as f they re smply a pece of wre wth no resstance ε L = 0. Ths wll allow you to fnd the current(s) n the loops and branches.
2 omparng & ontrastng R vs LR rcuts 1. S R omplete the table below n terms of the quanttes, R, and. quantty When swtch s closed (t = 0) After a long tme q V V R U Now, suppose an nductor wth nductance L replaces the capactor. omplete the table n terms of, R, and L. (Note the symmetry) quantty When swtch s closed (t = 0) After a long tme V R d/ dt V L U L
3 2. S R 1 o R 2 Remember that for any loop, at any tme, you can always apply = R s Be sure to nclude and/ or L on the left sde and make them negatve. The reason they re consdered emf s s because they have the ablty to supply current. Be aware of what you can assume about q and at t = 0 and at t =. (a) At t = 0, the swtch s closed. Express the followng n terms of, R 1, R 2, and. V across R 1 (b) After a long tme, express the followng n terms of, R 1, R 2, and. Now, suppose an nductor of nductance L replaces. (c) At t = 0, the swtch s closed. Express the followng n terms of, R 1, R 2, and L: L (d) After a long tme, express the followng: L
4 3. R 1 R 2 Express the answ ers to the followng questons n terms of, R 1, R 2, and. (a) At t = 0, swtch S s moved to pont A. Fnd the expresson for. o A S o B (b) After a long tme, fnd the expresson for (c) Swtch S s then moved to pont B. Fnd the expresson for at that nstant. Now suppose an nductor of nductance L replaces. Express the answers to the followng n terms of, R 1, R 2, and L. (d) At t = 0, swtch S s moved to pont A. Fnd the expresson for. (e) After a long tme, fnd the expresson for L (f) Swtch S s then moved to pont B. Fnd the expresson for L at that nstant.
5 1998E2. In the crcut shown above, the swtch S s ntally n the open poston shown, and the capactor s uncharged. A voltmeter (not shown) s used to measure the correct potental dfference across resstor R 1. a. On the crcut dagram above, draw the voltmeter wth the proper connectons for correctly measurng the potental dfference across resstor R 1. b. At tme t = 0, the swtch s moved to poston A. Determne the voltmeter readng for the tme mmedately after t = 0. c. After a long tme, a measurement of potental dfference across R 1 s agan taken. Determne for ths later tme each of the followng.. The voltmeter readng The charge on the capactor d. At a stll later tme t = T, the swtch S s moved to poston B. Determne the voltmeter readng for the tme mmedately after t = T. e. A long tme after t = T, the current n R 1 reaches a constant fnal value I f.. Determne I f. Determne the fnal energy stored n the nductor. f. Wrte, but do not solve, a dfferental equaton for the current n resstor R 1 as a functon of tme t after the swtch s moved to poston B 2008E2 In the crcut shown above, A and B are termnals to whch dfferent crcut components can be connected. (a) alculate the potental dfference across R 2 mmedately after the swtch S s closed n each of the followng cases.. A 50 Ω resstor connects A and B. A 40 mh nductor connects A and B. An ntally uncharged 0.80 μf capactor connects A and B. (b) The swtch gets closed at tme t = 0. Sketch graphs of the current n the 100 W resstor R 3 versus tme t for the three cases. Label the graphs R for the resstor, L for the nductor, and for the capactor.
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