Question

Could you please show all your work step by step? thank you!

The voltage across a 40-4F capacitor is shown in the figure below. Determine the current waveform. v(O) (V) 20 0.8 1.0 1.2 0 0.2 0.4 0.6 - (ms) -20

Answer 1

Given, C= your a tross if V (+) is the voltage Capacitor, then Current through Capacitor icts is given by, dv.ct) ict) = dt v(t) = Velt) 90 V 6.2 0.4 2 t (msec) - 20V ربعة أهم is t20, for to Ve → voltage aross Capacitor cov ict C. dco) dt ict) 0 Amp for tco Velt) ii) for os t <0-2 dhe det change in Uc in time chang 0 0.2 20-0 (0.2-0) m 5ec 20 0,2x103 100 x103 due dut =105 volts/ sec due auf its c Now, 40x10 x 105 4 ict) Amps ; for ost<0.2 msec

Cor) this can also be solved using straight line equation 90 sope equation of straight line » y = mute 0.2 msec t y ( passing through mx origin Co,o) V2-V slope ta-ti (t)= mx + 20o 105 t (0.2-0 ) msec Vet) - 105 nlow, icta с duc df C. (10 st) dt noxioo x105 ict) Amp for > 0 <t<0.2 msec 10 ) for 0.2 st 50.4 msec vet) 20 V 20-20 du det change in v int 0.4-0.2 change 0. 2 0.4 temsec) 0 2 du SO due Схо i(t): C ع Amps for 0.2stco.nl

is equation of Act) or 0.2 < t <0.4 for vet): 20 due 0 i(t)= cx due Amps. dt iv) VCE 0.4 ms sts of m sec com, 20) 20 - 20 - 20 -40 0.8-0:4 0.4 msec ols Femsec) du af E-100 X10 20 cos, -20) now, ilt) = cx lоо хоз ==40x106 3 Kloo XI0 Eq" of st line i(t) -4 Amp y matc Uct) = m + tc Vet) = -100 x103 ttc dvet) -100 x103 to dt (u) for vet) 0.8 St<lmsec -20 - (-20) Himsec) dy at O J-0.8 -20V ilt): cxdvno at ilt) for ofstclmsec

vij for Imsec st<1.2ms uct) 10 1.2 du 0 - (-20) T det 20 0.2810 1.2-1.0 = 100xióv/sec -20 ilta duc с Hox106 x103 X 100 X 10 for 4 Amp icts = Ims st<1.2ms vii) for to 1.2ms VCA for t> 1-2 ms, V(t) =o 12 IN[t) dt 0 duct C Octs- at OCH for t>l2ms Amp O