Question

In the circuit shown in (Figure 1), S, has been closed for a long enough time so that the current reads a steady 3.50 A. Suddenly, S2 is closed and S, is opened at the same instant Figure 1 of 1 S, 2.0 mH 5.0 MF Lun Prov

Answer 1

Y 3.5 A | Ans: When the switch s, is closed and ₂ is opens the circuit is like this. A E L= 2mH R According to data given - Cursent through winductos I = 3:58 Inductance of Inductes L-2MH Now Energy stored in the Inductor U24112 1:1 X2X103 x (3.5)2 U: 12.25 310-3J. Now Switch Sy is closed and switch s open and the circuit becomes- 2 Is 1=2mH C = 5ME at time When the capacitas is Initially unchanged the charge on the capacitor any t is given by Q10 = Rmax Sin (wt) where domex = Maximum charge on capacitors w = rad/s. √ LC And Current in the circuit is given by I(t) = {max w Coswt).

U = Almaxf2c Continued We see that Ql) = Qmax, when sinut = 1 [:| I's the maximum value of simwt) ] OR wt = goº Also - Putting values in above expressiom- 19.25 X10-3: Qmax 2X5 X106 Omax 19.25 X 10-8 Omax = 19.95 X10-8 3.58104c Hence- Maximum charge on Capacitos = 3.5X10-4C Ans Post: By the formula IlÐ Amax w Coslwt) and wt = 90° It) = Amax W cos(90°) Ilt) = 0 [i cos(90%) = 01 When charge on capacitor is maximum = OA the current Ans.