Question

Considering an RCL circuit where an inductor L = 1/2 H, a resistance R = 6Ω, a capacitor C = 1/50 F and a source of e(t) = 24 sin (10t) volts are connected in series. At t = 0, the switch is closed and the current starts to flow. Considering that the capacitor has an initial voltage of 5 volts, thus,V_c (0) = 5 V and V^'_c (0) = 0.

While using the general form of the equation of an RCL circuit:

$LC(\frac{d^2v_c{}}{dt^2}) + RC (\frac{dv_{c}}{dt}) + v_{c} = e(t)$

We obtain the solution;

$v_{c}(t) = e^{-6t} (\frac{75}{4}sin(8t) + 25cos(8t))-20cos(10t)$

We observe that the solution has two distinct parts.

Part 1: $e^{-6t} (\frac{75}{4}sin(8t) + 25cos(8t))$

Part 2:

What does each part specifically represent and how are they referred to?

Answer 1

homogeneouy eqn. Soln RCL so Clotait R=64 L= 0.5H ;e= If elt) = 24 sin (lot) Ve(o) = 5V V'(o)=0 - tv = e(t) RC + general en » LC (VE) - dved) vitve +(6) 5 1/2 x 1/2 24 sin (lot) 50 D. 2400 sin (lot) - Vol 4 124't loove T d?+ 127. t 100 d=-618 Vh ést (Aco 8t + B sinet) c cos tot + sintot Particular solution Up - loc sinlot tlod ws lot Yol Up " = -4 100'Ws lo tij - 100 D sin lot

Now 2400 sin lot Ip" +12 Up' + loo yp — Тоос as lot -lood sinlot - 120 C sin lot +120D coslot = 2you sinfot + louc as lot t1oo D sin lot D = ca 20 3 so 2o cos Up (1• t) Velt) - Vpt. Vh Velt) E-6 ( Aws et +B sin et +B sinet) – 20 w lot = A=25 A - 20 Velo) 5 -67 BA sinet telcset) - best (Acoottesinet) vi' (t) = e + Yoo sintot BB - 6A =d B = 75 4 Vet lo = ) Now 20 cos Bt e is sinet + 25 coset 4 Vc (t) = An