Question

2. The switch in the circuit shown in Figure 1 has been in position 'a' for a long time. At t=0, the switch moves instantaneously to position b. a) Find the numerical expression for io(t) when t > 0. b) Find the numerical expression for volt) fort > 0 1 = 0 b! 51 602 45 A 2150 vo 20 10 mH 240 V 1

Answer 1

(1) - tlz lolt) = 1 L I = 9 -> Colt) is current through in duchor L = 10m H. We will use following equahn to find current Lolt) through the inductor L. [1260+)- i_lo)] e + izbo) where 4160+) Current through inductor at time t= 0+ 11600) Current through induchor at Home t= a : Time Constant of RL arcuit - (11 Req Req: Equivalent Reoistor across L. > Inductor for the DC supply act as a short circuit in a steady stele Condehon. Yence circuit was connecled to point a for a long before t = 0 , The arcuit as time t-o- just befor to will be like shown in dig 1. At t=0- Now Resistors R and Re R3 600 are in parallel and equivalent Resistance of two resister is given as Fig 1 Athime t= '0- Pulting Rj and R2 Faith you to choose 4_12 6192 time R,=52 310-) T I 1 R₂ 2012 I lost s.c. +407 7 Req= RiX Re Res I ww R1 + R2 value of Scakegd Tith 5720 CamScanner R3 Rea + Reg 240V CS

now circuit in dig ! can be ne draw as in Hg 2, with Req = 42 now Current in the Circuit can be calculated as 240 240 Voltage I = Total Resistance 60+4 R3+ Req 240 74 2 I- A I --> I ¥12 ER2 & Re Current Division Rule If Current I flows ble too Resistors R, and R2 which are in parallel, then Current Jy and I are given as • R2 fig 3 X I I,= (Ritha) 12 - ' Ri I R1+R2 1- (3 X I Using concept of eqn . In dg 1 I is known, then i Ra i_(0-) R+R2 Pulling all the values 10-) 20 240 64 - 3 A X 4110-) = 3A since Current in Inducher does not change ab rukelly so í (0-) from ean ® _10+) = 1/(0-) = 3A 3A os ill0+) 15 Scanned with CamScanner

(3 Now Arcuil at t>O : When switch go to position b shown in fig4 -uu Ri Clearly Ra and Ra are 33 RA R2 = 20-22 (1) 454 Hillt) - 152 figa to 2 in parallel and can find equivalent Resistor using en Rax R4 Rd+ R4 20x15 do +15 6 Re 300 35 = 60 7 2 = R1 = 5.2 2 Re Gore 2 DI-454 Reg L'a Floma figs " (0) I! RE Corcuit in dig 4 can be Redrawn as circuit in dig 5 Now at t=0, Cercuit in sig 5 goes to sleady stole con dihon and inductor L becomes short Grcuiled as shown in Hg 6. Now Current iclo con be Calculated usong Current 9 ) I' division Rule as Rep2 (60) figo. At tao =160435 ) * 45 = 28.42 www Reg S.C X I ' RI+ Req² 6011 60 Scanned with CamScanner 199+5

( 1 (0) 28.42 1 6 - Rj+ Req? How from fig 6, el is very clear that eqivelent Resister Req across L is Req = R1 + Req? .. 5+ 60 7 13.57 22 Req Puting Rey and L inlo egr I A C I= 10X10-3 13.57 13.57 10x10-3 1357 +: 1/ = 1357 1 Putting Gu the values inlo egn -(13574) sed t) = [(3 – 28.42)] 28.42 + -(1357 +) 25.42 e Al Ans 8 lolt) = 28.42 6 voldoge Qeydiss Inductor Le given by V = L di at so Here ľ ) lot) - (1957)+. - (359€ Ver 1 (08.42 - 218.42 – 25.42 € on diferen Hating V = L [ [o -1357)+ 25.42 e x-1357 - 1357 + 25.44 x 1351 x @ L x Pulding value of L and Calculating |px|o3 x 85. 48 x 1351 x -(13577) Scanned with CamScanner CS

V2 -1357 + = 3.5 X10 e -1857+ x109) V co VZ? (3.5x@-13577 Lelt) VR Min T Now Drawing Leash loop only Writing KVL egn Volt) = VR + VL = ielt x R + VL + M eelso VL Volt) and (9 from ean @ (1357)+7 -1357 t... 3 + 3.5 e VdA) [ 28.42 -25.42 e -1857+ 3.5 x103 e -1357 + = 148.1 -187-1 e -13577 148 - (187.1+ 3.5x10%) e 148 + 3.37 x103 e - 1357 + V Ans. CS Volt) Scanned with CamScanner