First Order RL Circuits Switch S is closed for t<0, and open for t ≥ 0 When the switch S is closed for t<0, the inductor behaves as a short circuit to dc. The voltage across the inductor v=0, Hence the voltage across the R is also zero zero. + i v - The current component i (t ) V Rg Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits When the switch S is open at t=0s and remains open for t>0. Since the current through the inductor does not change instaneously. V i (0) Rg For t≥0, applying KVL, we get di L Ri 0 dt di R i0 dt L di R i d dt L ................((2) Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits Equation (2) can be solved as follows, Divide both side of (2) by i and integrate both side w.r.t t , we get 1 di R i dt dt L dt 1 R di i L dt R ln i t K L take exp onential both side , we get i (t ) e R tK L e R t L e K , at t 0, i (0) e K Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits Substituting i (0) e K in i (t ) e i (t ) i (0) e R t L R t L e K , we get for t 0 s Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits Example : In the following circuit at t<0 , the switch is closed and at t≥0, the switch is open. Find current through inductor iL at tt<0 0 , tt>0 0 and at tt=1ms 1ms Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits When t<0 switch is closed and voltage across the inductor is zero and it acts like a short circuit. circuit The current iL=2A 2A Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits For t ≥ 0, switch is open and the current through the inductor iL (t>0) does not change instantaneously from iL (t<0) At t=1ms We know that , i (t ) i (0) e i (t ) 2 e R t L for t 0 s 20 x 1 x 10 3 4 2 x 0.995 1.99 A Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RL Circuits Example : Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits For t<0 switch is closed and capacitor behaves as a open circuit By voltage division we get , v V R R Rg for t 0 Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits At t=0 and t>0 switch is open Since the voltage across the capacitors does not change instantaneously, voltage v iR remains same at t=0s V R v(0) for t 0 R Rg iC Applying KCL at RC circuit , we get, iC iR 0 dv v C 0 dt R dv 1 v (2) dt RC Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits Equation (2) can be solved as follows, Divide both side of (2) by v and integrate both side w.r.t t , we get 1 dv 1 v dt dt RC dt 1 1 d dv v RC dt 1 ln v(t ) tK RC take exp onential both side , we get v(t ) e t K RC e t RC e K , at t 0, v(0) e K Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits Substituting v(0) e K in i v(t ) e v(t ) v(0) e t RC t RC e K , we gett for t 0 s Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits Given following circuit, Find vc and v0 for I I. t<0 II. t>0 III. t=1.3ms Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits For t < 0 Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits 120 x 1250 vC 100 V 1250 250 Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits v ? What is 0 Current through the 400 Ω resistance = 0.096A g across the 400 Ω resistance i. e v0 = 38.4 V Voltage Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits For t > 0 , vc , does not change instantaneously so vc =100V Let us redraw the circuit for t ≥ 0 + 100V - Find voltage between these two points using voltage division rule Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits + + 100V v’’ - - Once you get this voltage say v’ , then get the find vo y applying voltage division v’=32V and v0 at t=0 is 25.6 V Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus First Order RC Circuits At t=1.3ms, we know that vc (t ) vc (0) e t RC for t 0s vc (t 1.3ms ) 59.5 V vc (0) 100 V and d C 4 H Req R ? Req=R= 625Ω Req=R= Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus