electrical circuits (differential equation)?

Question

find the charge on the capacitor and the current in the given L-C series circuit. Assume q(0)=0 and i(0)=0

L= 1 henry, C=1/16 farad, E(t)=60 volts

Answer 1

♠ according to definition of capacitance: q=u*C,
where u is potential difference on capacitor plates, q is electric charge in capacitor; thus u=q/C,
according to definition of inductance: i=(E-u)/L, where E=60V, E-u is pot diff on inductance; thus i=(E-q/C)/L
according to definition of el current: i(t)=dq/dt;
♣ thus dq/dt =(E-q/C)/L; → LC*dq=(EC –q)*dt;
→ dq/(EC –q)=dt/(LC), hence
ln(EC-q)=-t/(LC)+K, where K is integration constant;
♦ for t=0 we have ln(EC-0)=0+K, hence K=ln(EC); thus
ln((EC-q)/(EC)) =-t/(LC); → 1-q/(EC) =exp(-t/(LC)); →
q=EC*(1- exp(-t/(LC)) is equation for q(t) on C;
♥ i(t)=dq/dt= (E/L)* exp(-t/(LC));
mind i(0)= (E/L) > 0; i(0)=0 is error!
just plug in numbers now!

Answer 2

Here are some relationships to help.

"In a capacitor the current is proportional to the rate of change of the voltage.

(Vc is the voltage across the capacitor)
(VL is the voltage across the conductor)

C*(dVc/dt) = i. (where C is the capacitance in farads)

"In an inductor, the voltage is proportional to the rate of change of the current."

L*(di/dt) = VL.