In a series lcr circuit with l = 5/3 h, r = 10 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
Answers
Answered by
0
Answer:The RLC circuit is descrided by the initial value problem
L\frac{d^2 q}{dt^2}+R\frac{d q}{dt}+\frac{1}{C}q=EL
dt
2
d
2
q
+R
dt
dq
+
C
1
q=E
q(0)=q_0,\quad I(0)=q'(0)=I_0q(0)=q
0
,I(0)=q
′
(0)=I
0
where q(t)q(t) is a charge on the capacitor.
In our case we have
\frac{5}{3}\frac{d^2 q}{dt^2}+10\frac{d q}{dt}+30q=600
3
5
dt
2
d
2
q
+10
dt
dq
+30q=600
or
\frac{d^2 q}{dt^2}+6\frac{d q}{dt}+18q=360
dt
2
d
2
q
+6
dt
dq
+18q=360
and
q(0)=0,\quad I(0)=q'(0)=9q(0)=0,I(0)=q
′
(0)=9
Solution of the IVP
q(t)=e^{-3t}(20e^{3t}-20\cos(3t)-17\sin(3t))q(t)=e
−3t
(20e
3t
−20cos(3t)−17sin(3t))
Explanation:
Similar questions