Question

part 2 and 3

48-0 → IR VIL 14 ut + & Ic +04 24 from the circuit find ... ה dil dt dir It=ot dt tot dic at It=ot 2 LVL at It=ot dvr I dt It=ot dva dt 31 Ve(t) for all t

Answer 1

Solution IR y yor m 18 144(0) 7 12UA for & Co Yon I, (o) (9 Uclo) 24A Llot) 11(6)2.44 = (eco) and 1152 V 10t)] t=ot) 48r IR UR 1400( 1 (un (1) 1152 (1) 24 At toot) 17 de to dt at a = of

V lot) E dielot) de dillot de Velot) Now, Velot) = 7 At tot :. By Nodal Analysis -14 + 24 + IR = 0 'IR = -10 Amp and UR = ( R) VR = (-10x48) 6480 VOIT) UR E-480 By RUL -ų + VR + 1152=0 ų = ( k + 1152) - 4480 +1152) ų = 672 Volt Put VL 107) - 672 V in ea d_1109) dt (672) 7 qo Affec dillot) = 96A1sec dt And wer girlot) : dt Answer o Alpec Ålot) Since, constanta

and iekot) .cx duclot) dt diuclot) le lot) -) dt С. and - + C – 2430 +24) iclot) = Cl cloth to +24) é cot) = 14 Amp 14 d Vc cof) = dt (te) 14 18 ulsec dURI dt Ul8ec lat tiot due dt o vsec p=07 and lc Cot ) = 14 Armo Answer die von dr • Amplee. t=of