Question

A 50 uF capacitor is subjected to a voltage pulse having a duration of 4 s. The pulse is described by the following equations: - Spv. Osis 2 s; 0.10) = 3-5(t - 4) V. 28 si s 4s; elsewhere. a) Derive the expressions for ic fort 0 b) Find the power and energy Answer

Answer 1

; ; 0<t< 21 sec) 2<t <41 sec) dt Given (563 v (t) = -5 (t-4) 3 3 (in ) 0 c=50 (MF) ; elsewhere Therefore, i c(t) = cave(t) Sexo (5+3) ; Os + 4 21 sec) Cina) cx (-517-4)3;2 st 24(sey Lex d (0) ; elsewhere p 50x10-6x 1572; osts 26sed ² lc (t) = 50 x 10-6x1-15 (t-47²); (inA 23t< 4(sec) lo ; elsewhere 7.5x10-4ť ; 03+<2(sec) -7.5x104(4-4)}; 2 t<4 (sec) 2 (in A) o; elsewhere | by Ans. ( Therefore, (W) power, þct) = Volt) x iect)

bet) = W 1 5+3 x 7.5*10-4x t²; 0<t< 2 (sec) ) -5 (t-4)3x - 7.5 x 10-4x (+-4) 2; 2 <t <4 (sec) lo; elsewhere ost2 2 (sec) My b(+) = (in W) 3-75x10-3 +5 ; 05 tk 2 (sec) 3.75 X 10-3[t-4)5; 2 st 54(sec) lo; elsewhere to Anso Therefore, Energy (t) = ( pct at t 2 ( 3.45x10-345dt ; osts 2(sec) p 3-75x103 (+) Edt + $ 3.75x103 (+-4) ist ; 25t<4(sec) Energy ( 4 sec) ; tz4( sec ) 0.625x10–3 +6 ; 05t5 21 sec) 0.625 X 10–34672 + 0.625X10-964-4677 ; 25t_4(ec) 0.625X103[+61 +0425*10**{O COM i tz4(sec) 6-2007

- Energy (t) = + (mJ) 0.625+", ost3215ec) 40+0.625f4t-416–62)67 ; 25t: 41 sec) 40+0.625x26; tz4(sec) - Energy (t) = 50.62576 ; 05t521 sec) 40+0.625(+-4)6_40; 25t<4(sec) L 40-40 ; t Z4 (sec) (MI) 0.62576 ; 03t< 21 sec) => Energy (+) = (mJ) 0.6254-4); 2t<4(see 0; tz 4 (SEC) Anso
If you need more information related to this problem, then please let me know in the comments. Also, please let me know in the comments if the image is not clear. I will upload it again.