Question

Consider an LC circuit with L = 1H, C = 1F. Suppose the circuit devices are connected to a voltage source f given by:

t si 0 <t < 6 f(t) = 6 si t 26

If the capacitor is initially discharged and no current is flowing through the circuit, determine the charge on the capacitor at any time in t. Can someone please  solve it STEP BY STEP without skipping any step please? It would be really helpful. I´m lost ):

Answer 1

The diffrential equation for the LC circuit is

$L\frac{\mathrm{d}^2 q }{\mathrm{d} t^2}+q/C = f(t) \\$

Since initially capacitor is discharged and no current is flowing, the initial conditions are, q(0)=0 and dq/dt(0) =0

For L=1 and C=1, we get

$\frac{\mathrm{d}^2 q }{\mathrm{d} t^2}+q = f(t)$

For  $0 \leqslant t < 6, ~ f(t) = t$

and $t \geq 6, ~ f(t) = 6$

The solution (complementary) of homogeneous differential equation $\frac{\mathrm{d}^2 q }{\mathrm{d} t^2}+q = 0$ is

q_c(t) = A sin[t]+B cos[t]

The particular solution (it is the solution which satisfy the nonhomogeneous differential equation) is

$q_p(t)= \left\{\begin{matrix} t &for~ 0 \leqslant t < 6 \\& 6 ~for~ t \geq 6 & \end{matrix}\right, \\$

Now q(t) = q_c(t) +q_p(t)

Using initial conditions (for q_p=t)

q(0)=0 => Asin[0]+B cos[0]+0 = 0 => B=0

and dq/dt(0)=0 => A cos[0]-Bsin[0]+1=0 => A=-1

So q= t-sin[t]

Now using initial conditions (for q_p=6)

q(0)=0 => Asin[0]+B cos[0]+6 = 0 => B=-6

and dq/dt(0)=0 => A cos[0]-Bsin[0]=0 => A=0

So q= 6-6cos[t]

We can solve it using Mathematica as well. The results are same.

In[8]= DSolve[{"[t] +9[t] =t, q[0] == 0, '[0] == 0), q[t], t] DSolve[{q"[t] + [t] = 6, 9[0] == 0, '[0] == 0), q[t], t] Out[8] = {{q[t] →t - Sin[t]}} Out[9]= (( [t] →-6 (-1 + Cos[t])))

So at any time the charge is given by

$q(t)= \left\{\begin{matrix} t-sin[t] & ~for~ 0 \leqslant t < 6\\ 6(1-cos[t])~for~ t \geq 6 & \end{matrix}\right, \\$