Question

#1. Find the current in and voltage Vc in the circuit below. Also find the energy stored in the Capacitor and the inductor il 3 H m + 10 A 692 6 Ω VC 4F 12. Find Equivalent Capacitance at terminals a-b a 1 MF u 2uF 2 F 2uF 3 uF 3 uF 3 uF 3 uF b

Answer 1

1) given, 3H Zorro 4F IOA (1 Let us consider the Dc Steady state condition. In DC steadystate copaciter will be open cirevited and inductor will be short circuited. is + for Ve WW ban IOA (T rule il can be By current division found as 11:10 6 Get 6 -- 5A

Voltage Ve=6% Vc 6(5) Vc = 30 V Ener stored in a capacitor Ecal teat / CV² ET() (2017 Ec = 1800 Energy stored in the inductor 622112 EL (3) (5) 2 E = 37.5J

2) IMF MF 2MF T 2MF 2MF 34F 3.MF BMF Тэр when n' capacitors of capacitance c Fored ove in parallel then their equelent C+ C+ C+ on can be given as capacitance Ceq = nc circuit two similarly in the above Capacitances of Du IMF are in parallel They can be replaced by 2(14f) capacitance

Three 2 MF Copccitor sore porollel, their equivalent capacitance is 3 (245)=GXf Four 3 up capacitors can be replaced by 4 (34F) = 12 rf Q 2(1MF) =2rR 3(2.MF) = GMF 4 (345) = 1245 1 b. all three Cepecitenees 2 rf, buf, 12 ur are in series. Their equivalent capacitance, Cab of It I 2rr 12 MF GMF Ette + 6 1 / ₂ Cob lo 12

12t사 42 Cab 2 - 이 R4 ㅗ G [] gb 10 he Cab 24 15 18 Ceb = 1.333 ×10% Ccb=1·333A