Question

on Given LRC series circuit with L=0.05h R = 1 e, c = 0.04 f, Elt) = 0 V, q10) = 70, the the charge and i(o)=0 A. Find and the current, Capacitater

Answer 1

The differential equation associated with RLC Series Circuit with input voltage Elt) is given as dan R dq I c q = // Elt). L at + t dt? on Where qut) is the Charge Capacitor. Substitute the given values L -0.05 H, R=1, C=0.04 F, Elt) = OV. d²q - dq Ooor dt + q dt²t 0.05x004 deq + 20 de at + 500g -0. dte Where Let us solve this Second Order differential operation D2q + zod q + 500 q = 0 differential operator (D2+ 200 + 50019 Ausiliary Equation m²+zom1500 =0 -20 IN 12072_4x500 d dt. m 2 m The roots -10 1201 are complex m=azit Como fugate of the form

RE -lot The general solution of the DE is qft) C, cos (pt) + C₂ sin (bt)) → qlt) - e hot (a.cos(20k) + G sin(2ot) To find the arbitary Constants , and le given initial conditions q(0 = 7.6 and ů (o) 7 dqlo) let us use the - OA dt q10) - è lolo) (, cos (o) + sinco) 7 Ci to C 7 da TE 2-10 e lºt (, cos (204) + Sin (2001) + etot (-200, Sin (20+} + 20€ Cos (2001). -(106, to) + (+20 (2) dt tro न = -106, +20C2 CG 호 -Ž (7) 9=7 ₂ = 3.5

.. The Charge on the Capacitor is qCEJ 7 cos(20t) + 3.5 sinf 5 Sin(Zot] -rot (60) C. dt The cussent in the Circuit is given as dq(t) ict) -10 e lot [7 cos(20+) +3.5 Sin (204] + e 10€ [-140 SPS(204) + 70 COSC2sty] i(t)= ē lot [670 +70) Cos(206) + (-35-140) Sin (203) (-175 Sin(20t)) -lot 7 ilti. The current ilt/ -JOE -175 e Sin (20+)