Question

anything would be greatly appreciated help me understand how to solve this :(

6.) Find the energy stored in the inductor/ 2 seconds and place answer in box. 3.2 10 2018.01 ANNN Question 7 and 8: Find the voltage Vard) and the current loud) for the circuit shown in Figure 4 with (1)-0 for 1 0 and (i)(1 +) V for 120. The capacitor is uncharged for <0. 7.) Find an expression for our) with 120 and place your answer in the box. Vinle olf t tout(t) Figure 4 - Questions 748 8.) Find an expression for loadi) with 120 and place your answer in the box. low Kiw) Ausle) 20H 3 10H Question 9 and 10: Find the voltage v1) and the current (1) for the circuit shown in Figure 5 with 600 for 0 and 10) (r + 1) A for 120. The inductors are not magnetized for 1 <0. 9.) Find an expression for (i) with 120 and place your answer in the box. Figures Questions 9 & 10 10.) Find an expression for i() with 120 and place your answer in the box.

Answer 1

I vout (t) Vin(t) Given circuit is a noninverting amplifier as the input is connected to non investing node. Fora non inverting op Amp Vors= (1+ RF In the given circuit C- series connection) R = 22 +8-2 Ri= 102 RF = 10113 RE = 10x3 = (parallel connection) i Vout = It 30/3] Vim CS Scanned with CamScanner CS scanned from the momentin (t) = (+7t), t20

Voit (D) = (1+3) (t+t) ; to Vout (t) = 16 (t+E); to 1. Vout (t) = 1.23 (t+t) ; t o 1 (8) For on op Amp according to vis. Venu = Vnoninu Venu = m(t) . By KCL at Vont junction Vin (E) - Vent) + icut (t) = ic (t) For a capacitor ic (t) = ( dve. dt here are je, dvot) :ic(t) = C. Vo Ct = 0.1 d (1-23)(++} ic (t) = 0.123 (2t+)) ; to CS scanned with CamScanner

Vint) - Vout(t) t + iout (t) = 0.123 (2+1) 3x10 3 to 33. (ert)– 1.23 (tab) - iet b= onero) [txt=1232= 12 3 4 ) a joue ) = ogue #70103 is fout (t) = 0.2u1t40123 0234-0.25t] ut (t) = 0.2u6t to 123 1-2 t tt-1-23 t = 0-246 € + 0-123 + " It+t]X0-23 = 0-2467 + 0-123 +0.1 (tzt) ; t o = o.lt + 0.346t+ 0.123 cijout (t) = 0,4 +0-3467 +0.123 jt »,0 CS Scanned with CamScanner