Question

) In the cirant below the inductor is initialy uncharged, a) find i(t). b) plot this function using Mathal's ezplot function, what is the maximum value of i(t)?() at what time you achieve this? do 2.1 w 24v 36H, io=0 2) Given the ckt below Hilt +12V 10.1 12.1 AM zor Sulit Show Por su preved = PT, ABORDE - SV

Answer 1

son 22 is given 64 2h 0.00 9. wybiere wamec Wolteplecy Ac cind tos DC WD = D. the resistence of inductor XL. hence for lo XL=0 le voltage drop due to sell inductance is a E WL 0 Sapth - l de mot or di or integrating both side in - Straight line (6 2 here Ilt) ait hence graph will be Max. Cursent through inductor will be Imax 24 = 12A Ildim Cover is opposite to the apphed Crosent so this conibe mox only when I (5) 20 and this is possible only (@hen: (f.eo) as besar Iltif- then Ilt) = and Current in the branch will be maximum.

12 V (2) 1st replacingl all Voltage Sourcea Source by Simple It's Sloe fle Cav 40 N 12 Intl Is In 6 Il Is Here three variables so three ecuation required im ABEA T:+ 260,+1) +3 (Tatry) 29 21; +512 +373 29 im B C DEB 2 Uit Is) – 12(5-1) + 10.9,* = 12 24. I +2 I 2 - 12 I3 12 I, + I2 - 6Iz for HE DEH -3 (In+Ts) – 12 (13-1) +5 = 0 121, +372 +15 T2 = 5 © Solving get I = 0.UA I 5.22 Iz 6.7A Now if you caculate power doop across each resistance and Power supply by each Source then it will same o с we
​​​​​​power supply by source will be voltage ×cuurent supply

And power absorbed at each resistance is equals to v square divided bye resistance

Your answer for 2nd will be proved

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