Question

For the circuit below, determine: (a) The maximum energy stored in the inductor. (b) What percentage of the maximum energy remains in the inductor at t = 0.4s? 222 240 =° Soe is 30.34 Son

Answer 1

Solutions Switching occurs al to 7 fos t o the cicuit is in strady stale Now steady state > Fonductor Shost ckt + The resultant circuit is mit 704 24 6 or Xi(o) 6. - 22 34 che at t= 0 (steady State) €007-2(8) = 16 Ampere 278 This since inductor current cannot change instantaneously + 1€(0) = 1607)= {(0) = 16 Ampere 7 --- 0 The Manimum Energy stored in the Inductor Elmari Llib = 4x033(76)?

- 0.384 Joules mm at te 6 • Now off after switching, once again , the che reaches steady state Steady stałe > Fonductor Short biscuit The resultant ckt is ze un tot Lekt at + 0 (Steady State) = TE (0) = 0+1 -- ~ Now the general equation of current through inductor for txo is given by į 4) = < (8) + (60-i2())e Ampere where Tz 4 Reg is the time constant of the - Circuit Reg --Equivalent resistance seen by Inductor

after switching 22 um ckt to Find > Req . . Req= 6//2 = 12 = 1.52 776 4Req= 115 sec -- ® using eq0, 2 and 6, 9 becomes (t) = 0+ (16-0) e Amperes oz /126 = 16e amperes . -5t -57 tro Now at t= 0.4 seconds 5 (104) CAE) = 16€ 5(0:4) = 0.2765 Ampere Energy at t-045 Bon į Liebe 204 = 3X03x (0-210572 = 70031 mToules

Now 1. Energy at t=0:4 sec Compared to Elmore (Maximum lénergy J is ML (0:4). 7. 03/X100 - 1-83% 0-384 (man)