Answer to Question #85563 in Electric Circuits for Rinku Rai

Question

In a series LCR circuit with L = 5/3 H, R = 10 Ω and C = 1/30 F and a source of emf
E = 600 V, determine the charge on the capacitor as a function of time. It is given that
the initial charge on the capacitor is zero and the initial current in the circuit is 9A.

Answer 1

For a series LCR circuit

U=U_C+U_L+U_(R ) → U=q/C+L (d^2 q)/(dt^2 )+R dq/dt → (d^2 q)/(dt^2 )+R/L dq/dt+q/LC=U/L →

(d^2 q)/(dt^2 )+10/(5/3) dq/dt+q/(5/3∙1/30)=600/(5/3) → (d^2 q)/(dt^2 )+6 dq/dt+18q=360

(d^2 q)/(dt^2 )+6 dq/dt+18q=360

The solution of the equation are

q(t)=(-17 sin⁡〖(3t)-20 cos⁡(3t)e^(-3t) ) 〗+20

If t→∞ then q→20 C.