Question

For the circuit shown, the switch has been open for a long time and it closes a t=0. The initial energy stored in capacitor is 250 . 102 w 3V 05v, 522 1c 20 F Question 6: What is the initial condition of the voltage of the capacitor? (A) 0 V (C) 1V (E) None of the above (B) 5 V (D) 3V (E) Question 7: What is the value of v, at t=0*? (A) -3V (C) 2V (E) None of the above (B) -2V (D) 3V Question 8: What is the time constant of the circuit at > 0? (A) 16 us C) 25 us None of the above (B) 10 us (D) 100 us Question 9: What is the final condition of the voltage of the capacitor? (A) -3 V (C) 1V (E) None of the above (B) 2.25V (D) 3V Question 10: What is the expression of the current of the capacitor at t> 0? (C) 2.75 +2.25 -40000 A (E) None of the above (D) -2.2 e 40000 A (A) -5 e62500 (B) -3 -10000 A A

Answer 1

is Qcm): Y UC00) : 9.25 V 23. Therefore, the final condition of the voltage of the capacitor in Q.25 V. Hence , option -B is correct option for Question 9. Q10. 24 • The voltage across the capacitor, Welt for trois calculated using formula The ų (t) = 6 (60) + (6.(0) - ()) e tro We know that ų 0) = 5 V Uc (0) = 2.250 and = 100 usec C = 100x1068 -4 2 10'8 25. Substituting Occo), O. (W) and r in Oelt), we get rt 2.25 + (5-2.25).e 10+ VCH : V 720 -100005 Oc(t) = 2.25 +2.75 e V; to 26. But current through the capacitor icct) for to is calculated using d ( (t): c Ult) Butz C= 20 uF -100006 d to . ... lelt) = 20x10*** (2.25+ 275.2

-6 • i (t) = 20X10 -10000t d (275xe * d (a+b) = X (67:25) dt dt da do dt dt But d at (2-25) = 0 ict) : 90xtoʻx[0+ 46-75% e lot) ] (2.75 x = 10000t) 2 375 de ce asook) d -100005 1 & 75x (-lo00o) x 2 since (d eat a eat at d dt -10000t -100005 ( 2:35 Xe ) = -215 00 Xe d at 27. Substituting (2.75 x 2160oot) in ielt), we ger -10000+) A iclH : 20x10 x (-2500 xe -10000t :ic(t) = -0.55 xe A tro 28. Therefore, expoession of the current of capal itor for t>o is -1000ot is (t) =-0.55x e A for to Hence, option -E is correct option for Question 10.