Question

An LRC series circuit has L = 20 henries, R = 180 ohms, C = 1/280 farad, and an applied voltage E(t) = 10 sin t. Assuming no initial charge on the capacitor, but an initial current of 1 Ampere at t = 0 when the voltage is first applied, find the steady-state charge q(t) on the capacitor and the steady-state circuit current i(t).

Answer 1

$LQ''+RQ'+Q/C=E(t)$

$20Q''+180Q'+280Q=10sint$

$k^2+9k+14=0$

$Q=\frac{-9\pm \sqrt{81-56}}{2}$

$k_1=-7,\ k_2=-2$

Homogeneous Solution:

$Q=c_1e^{-7t}+c_2e^{-2t}$

Particular Solution:

$Q=Asint+Bcost$

$Q'=Acost-Bsint$

$Q''=-Asint-Bcost$

$-Asint-Bcost+9(Acost-Bsint)+14(Asint+Bcost)=10sint$

$(-A-9B+14A)sint+(-B+9A+14B)cost=10sint$

$9A+13B=0$

$13A-9B=10$

$A=-13B/9$

$-169B-81B=90$

$B=-9/25$

$A=13/25$

$Q=\frac{13}{25}sint-\frac{9}{25}cost$

General Solution:

$Q=c_1e^{-7t}+c_2e^{-2t}+\frac{13}{25}sint-\frac{9}{25}cost$

$I=\frac{dQ}{dt}=-7c_1e^{-7t}-2c_2e^{-2t}+\frac{13}{25}cost+\frac{9}{25}sint$