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Transcript
Transient Analysis The Growth of current in R-L Circuit a S b i R vR + vL V L - When switch S just switches to a : t=0 , i=0 , VL =0 Then current start flowing with the rate of di/dt thus a voltage develop in inductor given by di vL L e dt a S R b i vR + vL V - Using Voltage Kirchoff’s law V vR vL di iR L dt Divided by R V L di i R R dt di I i dt Where I = V/R = steady current = L/R = time constant L 1 Rearrange the equation 1 dt di I i Integrate both sides we have Then the solution is t ln I i A 1 dt (*) A ln I Apply the initial state t=0 and i=0 t I Substitute in (*) , we have ln I i Voltage 1 di Ii t / i I 1 e Or di d IL t / t / vL L L I 1 e e dt dt Substitute = L/R vL IR e t / Ve t / vL, i V vL, V/R i 0 t Decay analysis for R-L circuit a S b R vR + V - vL L i When switch S is switched to b iL will begin to decay in a reverse direction with the rate of di/dt thus a voltage develop in inductor given by di vL L e dt Using Voltage Kirchoff’s law vR vL 0 Or Rearrange 1 1 di dt i Integrate The solution iR L di 0 dt where = L/R 1 1 di dt i t ln i A Apply initial condition t=0 , i=I Or we have t ln I ln i i Ie t / I ln i Voltage LI t / d t / LI e e dt di vL L dt Substitude = L/R and IR = V ; we have vL = - V e-t/ vL, i V/R i + 0 t - vL V Example 1 1 2 r 10 R 10V L 0.1H 15 vL For network as in figure (a)Determine the mathematical expressions for the variation of the current in the inductor following the closure of the switch at t=0 on to position 1 (b)When the switch is closed on to position 2 at t=100ms, determine the new expression for the inductor current and also for the voltage across R; (c)Plot the current waveforms for t=0 to t=200ms. (a) For the switch in position 1, the time constant is L 0.1 T1 10ms R1 10 Therefore i1 I 1 e Tt 1 t 3 t 3 10 1 e 1010 1 e 1010 10 (b) For the switch in position 2, the time constant is L 0.1 T2 4ms R2 10 15 Therefore i2 Ie t / T2 t 3 t 3 10 e 41 0 0.4e 41 0 10 15 The voltage drop vR i2 R 0.4 15e t 4103 6e t 4103 i(A) 1.0 0.5 20 40 60 80 t 100 120 140 160 180 At time t=0 till t=40ms i follows the equation i1 1 e t 10103 At time t=40ms till t=100ms i is saturated which is equaled to 1 At time t>100ms i follows the equation i2 0.4e t 41 03 Example 2 For the network in example 1, the switch is closed on to position 1 as before but it is closed on to position 2 when t=10ms. Determine the current expressions and hence plot the current waveforms When switch is switched on to position 1 , i follows i1 1 e t 10103 amperes 10103 At t=10ms the magnitude of i is i1 1 e 1010 1 e 1 0.632 A 3 When switch is switched on to position 2 , i i2 Ie t / T2 0.632e t 4103 A Example 3 For network in the Figure , the switch is closed on to position 1 when t=0 and then moved to position 2 when t=1.5ms. Determine the current in the inductor when t=2.5ms. 20 1 1A 2 10mH 10 10 20 First we simplify the supply of the circuit using Norton Theorem 10 2010 20 Re 1A 10 10 is / c The simplified circuit is 10 20 10 7.5 10 1 0.33 A 10 20 1 0.33A 2 10mH 7.5 20 Switch at position 1 7.5 Current Maximum magnitude I 0.33 0.2 A 5 7.5 At time t=1.5ms L 10 10 T1 R 5 i1 0.21 e T1 t 1 Switch at position 2 3 2.0ms 0.21 e 12.5 0.106 A L 10 103 T2 0.4ms R 5 20 t2=2.5ms- t1=2.5ms-1.5ms=1ms i2 0.106e T2 t 2 0.106e 01.4 0.009 A 9mA Energy stored To neutralize the induced e.m.f which represents the power absorbed by magnetic field a voltage source is needed. The energy produced by the voltage and current developed is stored in the form of magnetic field. This is given by di vL L p ivL and dt di p iL dt Average energy absorbed Total energy absorbed di w pdt iL dt iLdi dt 1 2 W iLdi Li 0 2 I L 0 1 2 LI 2 The Growth of current in R-C Circuit a S b + R i vR vC V C - When switch S just switches to a : t=0 , q=0 , VC =0 Then voltage start develops across the capacitor with the rate of dVC/dt thus a current flowing in the circuit is given by dv C iC C dt a S b + R i vR vC V C - Using Voltage Kirchoff’s law Or V = vR + vC V – vC = vR = iR Substitute i we have Or dvC V vC RC dt dvC V vC dt Where =RC Rearrange we have dvC dt V vC dv C dt V vC Integrate t ln( V vC ) A Then we have At t=0 we have vC=0 thus A=ln V Substitute A we have Or t V ln V vC vC V 1 e t / Substitute VC in the current formulation d CV t t / i CV (1 e ) e dt V t / i e R dvC iC dt vC, i V vC V R i t Voltage development and current decay in serial RC circuit Decay analysis for R-C circuit a S b + V R i vR vC C - When the switch is connected to b, vC = V. The accumulated charges in C now start to discharge via R. The initial current iI is given by: V iI R In the process of discharging, voltage across C, vC is decaying. When charges are full discharged, vC = 0 and the final current iF = 0. Using voltage‘s Kirchoff law vC = -vR i Therefore But i C vR vC R R dvC dt Rearrange Substitute we have dt dvC RC vC Then we have C dvC v R dt R And substitute RC = dt dvC vC Integrate We get dt dvC vC t ln vC A If t=0 ; vC=V , then we have Substitute A we have We get A = ln V t V ln v C ln V ln vC V et / vC or vC Ve t / For current we have vC V t i e R R vC , i V vC 0 -V/R t i Decaying of Voltage and currrent in RC circuit Example 4 A capacitor is made of two pieces of aluminium foil separated by a piece of paper of 0.2mm thick and having a relative permittivity of r = 5. Determine the area of aluminium required to produce a capacitance of 1200 pF? Assume that the permittivity of the free space is o ,8.854 x 10-12 F/m. C = 1200 pF = 1200 x 10-12 F d = 0.2 mm = 2 x 10-4 m r = 5; dan o = 8.854 x 10-12 F/m A = ? C r o A d 1200 1012 2 104 2 A 54 . 2 cm r 0 5 8.854 1012 Cd Example 5 Two capacitors, C1 = 1200 pF and C2 = 2200 pF are connected in parallel. The combination is connected to an a.c voltage source VS = 155 V having a series resistor RS. After the voltage across the capacitor reach a steady state, the connection of capacitor to the battery is disconnected and discharged via a resistor RP. 1. Draw the circuit. 2. Calculate Rs so that the energy stored in capacitor reaches 25 mJ in 250ms during capacitor charging 3. Calculate Rp , after 1ms discharging, the voltage drop is 20% from its steady state. 4. Calculate the energy stored in capacitor after 1.5ms discharging. (1) RS VS 155 V RP C1 1200 pF C2 2200 pF (2) Total capacitance CT C1 C2 1200 2200 3400 pF W CvC 25 106 J Given 2 1 2 Voltage across cap. During charging rewrite e then t / RS C 2W 2 25 106 vC 121 V 12 CT 3400 10 vC VS 1 e t / R S C vC VS vC 1 VS VS t RS C ln vC 1 et / R C VS or or S e t / R SC VS VS vC VS 155 ln 1.517 VS v C 155 121 t 250 106 RS 48.5k 12 1.517 C 1.517 3400 10 (3) vC VSe t / R During discharging P CT Given that vc drop 20% from its steady voltage , vC= 0.8VS 0.8VS VSet / R et / R t R PC P P CT CT or 0.8 e t / R P CT 1.25 ln 1.25 0.223 t 1103 RP 1.32M 12 0.223 CT 0.223 3400 10 (4) After 1.5ms discharge 1.5 103 0.334 6 12 RP C 1.32 10 3400 10 t v C 1.55e t / RP CT 1.55e 0.334 111V W CvC 12 3400 1012 1112 21mJ 1 2 2