Question

Solve the RCL circuit ?

Answer 1

See below.

Explanation:

We have

`e(t)=R_1 i + L_1 (di)/(dt)+1/C_1 int i dt`

introducing now `q = int i dt` we have equivalently

`e(t)=R_1 dot q + L_1 ddot q + 1/C_1 q`

The solution to this differential equation can be composed as

`q = q_h + q_p` where

`R_1 dot q_h + L_1 ddot q_h + 1/C_1 q_h = 0`

and

`q_h = e^(lambda t)` is the general solution to this type of equation so

`(R_1 lambda+L_1 lambda^2+1/C_1)e^(lambda t)=0`

so

`lambda = (-C_1R_1 pm sqrt((C_1R_1^2-4L_1)C_1))/(2C_1L_1) = (lambda_1, lambda_2)`

and

`R_1 dot q_p + L_1 ddot q_p + 1/C_1 q_p = e_0 cos(omega t)`

assuming `q_p = a sin (omega t)+b cos(omega t)` and substituting

we have two conditions

`{(b - C_1 e_0 - b C_1 L_1 omega^2 + a C_1 omega R_1=0),(a - a C_1 L_1 omega^2 - b C_1 omega R_1=0):}`

so

#{(a=(e_0C_1^2omega R_1)/((C_1 L_1 omega^2-1)^2 + C_1^2 omega^2 R_1^2)),(b=(e_0C_1(1 - C_1 L_1 omega^2))/((C_1 L_1 omega^2-1)^2 + C_1^2 omega^2 R_1^2)):}#

and finally

`q(t) = K_1e^(lambda_1t)+K_2e^(lambda_2t)+ a sin (omega t)+b cos(omega t)`

Here `K_1,K_2` are constants to determine according to the initial conditions.

To obtain `i` is necessary to do `i = (dq)/(dt)`

Numerical results are left to the reader.

NOTE:

Attached a plot showing the permanent voltage drops associated to `q_p` for

`((V, "black"),(R_1,"red"),(L_1,"blue"),(C_1,"green"))`

and also the transient or `q_h = int_0^t i_h dt` in blue

Answer 2

I am a bit rusty so take it carefully...! :-)

Explanation:

Have a look:

Although the various voltages drops do not seem to match the original `10V` remember that we need to compare them as:
`V=sqrt(V_R^2+(V_L-V_C)^2)` to have our `10V`!
This is due to the phase change introduced by our components in the circuit (have a look at Phasors to have an idea).

Hope I didn't confuse you even more!