Find i(t) for the circuit below. Find i(t). 412 W 0.02 F+0 10 V HAW Mem...
1. Find v(t) fort> 0 in the circuit in below: t=0 222 w 622 w + 10V 2 F +1 50 V Assume the switch has been open for a long time and is closed at t=0.
Problem 4 = 20A Find i(t) and v(t) for t > 0 in the circuit below if i(0) 2 H 202 5Ω v(t) 1Ω Problem 5 If the switch in circuit below has been open for a long time and is closed at t = 0, find vo(t) a. Suppose that the switch has been closed for a long time and is opened at t = 0. Find vo(t). b. t = 0 2Ω WW 3 F 4Ω 18 V
For the circuit of Fig. 5, find v (t) for t > 0. 612 V1=0 + 카 412 v 6 uF 9V Fig. 5 . For the circuit of Fig. 6, if v (0 - ) = 0 V and Ic (0 - ) = 5 A find v (t) for t > 0. V fir ich 0.5.12 0.5 H 0.5F 7 Fig. 6
In the circuit given below, V = 45 V and / = 9u(t) A. Find i(t) fort > 0. 10 22 www lico i(1) V 10 mF 4092 - 4H The current equation i(t) = [E+ (A091)+(Best) A, where A = B= S1 = s2 = and E= In the circuit given below, R=40. Find for t> 0. 492 ta X R WWW HH 1=0 / F 60 V H O 19 = -7.50 + [7.50 COS(8,731 +9.556 sin(8,731 ]e-11121...
Find i(t) for t> 0 in the given circuit. Assume v;= 34 V. t=0 10 22 6022 [i(t) 1 mF Vi + 40 Ω 2.5 H O (0) = –10.88te-20+ (0) A i(t) = -27.20 te-20tu(t) A i(t) = 13.60te-20tu() A O i(t) = –17.00 te-20t4() A
In the circuit given below, V = 28 V. Find it for t> 0. 32 1 H iſt) 40u(t) A 192 V 40 mF O 10 = [8.729 sin(4.5830e-29410) A O 10 = [218.232 cos(4.5831)e-2940 A it = [218.232 sin(4.5831e-2]40) A O 10 = [8.729 COS(4.583 18-2010 A
For the circuit shown below: Vs W RG R W Vs=32 sin( 20 t) V, R=30 0, L= 1.5 H, if the firing angle is 30 Find the average output voltage in V.
Question 4 Find i(t) for t >0 for the circuit below. 4Ω 12 V 5 H 3 A
1. [10pts) Find it) and (t) for 1> 0 in the circuit below if i(0) - 10 A. iſt) 32H § 202 { v(t) 310
In the given circuit, identify (0) and i(t) for t> 0. Assume 10) = 0 V and 1(O) = 2.50 A. + 5u(t) A 222 v+0.5 F ell 1 H [3.780 e-t2cos(1.3229t - 90°)]u(1) V [5 + 2.67252 e-t2cos(1.3229t - 200.79] A [0 – 2.67252e-t2cos(1.3229t – 200.7°)] A [5 – 2.67252e-t/2 cos(1.3229t+ 200.79] A [2.673e-t2cos(1.3229t – 90°)]u(0) V [1.336e-t2 cos(1.3229t+ 90°)]/(t) v