Question

4.- (25) Find it) two ways: a) in the time domain the Laplace Transform 6) using t=0 sec + 10 cost Volts 12 ww lovolts IH

Answer 1

The circuit given lo cost 14 10V to find Apply super position the dem response Consider 10V de source ilts 10=Riti dilat TH 10V ilta 10 [nette Ta HR = Hizi, 1 Hla lo [i-e'] alone 0 Consider lo cont source acting 11 11 transient Tocont 11 Steady ronnient ilt= steady + atoto inteady slobo = Vm conwt-o) vind 10 cost-45°) XL=WL =12 coo (t-4.50 -450] : $tan lehte - Tañ'(11=450 Z 10

i trannient K ett zket i(Hz No Cos (6-45°) + Ké o) + két? to find 'K' value consider i(t) at tao i Col=0 COM45° + Ke к (t = cos (f-45°) & set The total rexponse ilte ichs + c "Ct) ilt) [icēt) a 5e + ice cos(+-459) ilta 10-15é + lo conlt-45°)

6 Apply Laplace trantam 10 coat effor wat] = 1 tout ict) + di locost +10 L[k] = kls dt IOS ICS) [1+5] S 1+s? IOS + ICS) S(+1) (1+5)(1+52) A B DSTE + S st + S A: 10, B=-10 (1+52) ctcst DS+E+DSTES (its) (145) C+E=0 (+D=0 0-E, D + Echo C-5 D = 5, E-5 5 + - 5hts 5s + 5/14 SS (+32) ICS) 15 s

ilt) = 10-15ét + 5(cost+ Bint) iltl - | 0 – 15 - + Tocos (t-45°) (cost+ sint) 1 = 2 Te I cost.c0348°+ Bfnt.sings] Te [coj[t-45)