1. ## Electrical Circuit problem?

A simple electrical circuit contains a condenser of capacity C farads, a coil of inductance L henrys, a resistance of R ohms, and a generator which produces an electromotive force E volts, in series. If the current intensity at time t at some point of the circuit is I amperes, the differential equation governing the current I is:

$L d^2I/dt^2 + R dI/dt + 1/C I = dE/dt$

There is also a picture attached, not sure if its important so let me know.

Find I as a function of t if:

1. R=0, 1/(LC) = w^2, E= constant
2. R=0, 1/(LC) = w^2, E= Asin(alpha*t); alpha=constant≠w
3. R=0, 1/(LC) = w^2, E= Asin(wt)
4. R=50, L=5, C=9x10^-6, E= constant

If someone can just get me started, it would be much appreciated. I have utterly no clue what to do here

1. Since R=0 there would be no 1st derivative term and since E is a constant, its derivative would just be 0, it would just be:

$L d^2I/dt^2+1/CI=0$

So would I just continue and solve like a regular 2nd order DE?

$Lr^2+ r/CI = 0$

but then where would the 1/LC come in?

2. Same thing with R=0 but the right side would have a cosine instead? or an (alpha)cosx?

3. Originally Posted by Jeffman50
A simple electrical circuit contains a condenser of capacity C farads, a coil of inductance L henrys, a resistance of R ohms, and a generator which produces an electromotive force E volts, in series. If the current intensity at time t at some point of the circuit is I amperes, the differential equation governing the current I is:

$L d^2I/dt^2 + R dI/dt + 1/C I = dE/dt$

There is also a picture attached, not sure if its important so let me know.

Find I as a function of t if:

1. R=0, 1/(LC) = w^2, E= constant
2. R=0, 1/(LC) = w^2, E= Asin(alpha*t); alpha=constant≠w
3. R=0, 1/(LC) = w^2, E= Asin(wt)
4. R=50, L=5, C=9x10^-6, E= constant

If someone can just get me started, it would be much appreciated. I have utterly no clue what to do here
I assume it's actually

$L\,\frac{d^2 I}{dt^2} + R \, \frac{dI}{dt} + \left(\frac{1}{C}\right) I = \frac{dE}{dt}$.

Please use brackets where they're needed or learn better LaTeX...

Anyway

1. $R = 0, \frac{1}{LC} = \omega ^2, E = \textrm{constant}$.

Since $E = \textrm{constant}, \frac{dE}{dt} = 0$.

The DE becomes

$L\,\frac{d^2 I}{dt^2} + \left(\frac{1}{C}\right) I =0$

$\frac{d^2 I}{dt^2} + \left(\frac{1}{LC}\right) I = 0$

$\frac{d^2 I}{dt^2} + \omega ^2 I = 0$.

Now yes, proceed as you would for a 2nd order linear constant coefficient ODE.

Characteristic equation:

$m^2 + \omega^2 = 0$

$m^2 = -\omega^2$

$m = 0 \pm i\omega$.

Since these are complex conjugates, the solution is

$I = e^{0t}[C_1 \cos{(\omega t)} + C_2 \sin{(\omega t)}]$

$= C_1\cos{(\omega t)} + C_2\sin{(\omega t)}$.

Do you have any boundary conditions?

Use similar processes for the next few questions.

4. Thanks, and yes I messed up with the original equation.. it is (1/C)I

And no boundary restrictions, but just to be curious what do u mean by that and how would that change the solution?

5. I've started 2:

R=0 so no first derivative term
d(Asin(alpha*t)/dt = A(alpha)cos(alpha t)

so i have $r^2 + w^2 = (A(alpha) cos (alpha*t))/L$

which goes to

$r= sqrt[(A(alpha) cos (alpha*t))L - w^2]$

are the two roots that whole thing + and -?

so I= e^(whole thing)pi + e^(- whole thing)pi ?

6. Originally Posted by Jeffman50
I've started 2:

R=0 so no first derivative term
d(Asin(alpha*t)/dt = A(alpha)cos(alpha t)

so i have $r^2 + w^2 = (A(alpha) cos (alpha*t))/L$

which goes to

$r= sqrt[(A(alpha) cos (alpha*t))L - w^2]$

are the two roots that whole thing + and -?

so I= e^(whole thing)pi + e^(- whole thing)pi ?
You need to set up a 2nd order linear ODE, solve for the homogenous case, then solve for the particular case.

So for 2.

$R = 0, \frac{1}{LC} = \omega ^2, E = A\sin{(\alpha t)}, \frac{dE}{dt} = \alpha A \cos{(\alpha t)}$.

So the DE becomes:

$L\,\frac{d^2 I}{dt^2} + \left(\frac{1}{C}\right)I = \alpha A \cos{(\alpha t)}$

$\frac{d^2 I}{dt^2} + \left(\frac{1}{LC}\right)I = \frac{\alpha A}{L}\cos{(\alpha t)}$.

$\frac{d^2 I}{dt^2} + \omega ^2 I = \frac{\alpha A}{L}\cos{(\alpha t)}$.

Homogeneous case Characteristic Equation:

$m^2 + \omega ^2 I = 0$

$m = 0\pm i\omega$ as seen before.

So the homogeneous solution is

$I_C = C_1\cos{(\omega t)} + C_2\sin{(\omega t)}$ as before.

Particular case - use the method of undertermined coefficients:

Assume a solution of the form:

$I_P = C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)}$.

It is important to note that $\alpha \neq \omega$. If it did we have to adjust our assumption slightly.

Anyway, if $I_P = C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)}$ then

$\frac{dI_P}{dt} = -\alpha C_3\sin{(\alpha t)} + \alpha C_4\cos{(\alpha t)}$

$\frac{d^2 I_P}{dt^2} = -\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)}$.

Our original DE is:

$\frac{d^2 I}{dt^2} + \omega ^2 I = \frac{\alpha A}{L}\cos{(\alpha t)}$.

Plugging everything into the equation gives:

$-\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)} + \omega ^2 \left[C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)}\right] = \frac{\alpha A}{L}\cos{(\alpha t)}$

$-\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)} + \omega^2 C_3\cos{(\alpha t)} + \omega ^2 C_4\sin{(\alpha t)} = \frac{\alpha A}{L} \cos{(\alpha t)}$

$(\omega^2 C_3 - \alpha ^2 C_3)\cos{(\alpha t)} + (\omega ^2 C_4 - \alpha ^2 C_4) \sin{(\alpha t)} = \frac{\alpha A}{L} \cos{(\alpha t)}$

$C_3(\omega^2 - \alpha^2) \cos{(\alpha t)} + C_4(\omega ^2 - \alpha ^2)\sin{(\alpha t)} = \frac{\alpha A}{L}\cos{(\alpha t)} + 0\sin{(\alpha t)}$.

By equating like coefficients, we find

$C_3(\omega^2 - \alpha^2) = \frac{\alpha A }{L}$ and $C_4(\omega^2 - \alpha ^2) = 0$

We can therefore see that $C_4 = 0$ and $C_3 = \frac{\alpha A}{L(\omega^2 - \alpha^2)}$.

So now we have our particular solution:

$I_P = \frac{\alpha A}{L(\omega^2 - \alpha^2)} \cos{(\alpha t)}$.

Our general solution is the sum of the homogeneous solution and the particular solution.

So $I = I_C + I_P$

$I = C_1\cos{(\omega t)} + C_2\sin{(\omega t)} + \frac{\alpha A}{L(\omega^2 - \alpha^2)} \cos{(\alpha t)}$.

It also seems that you need to revise solving second order linear ODEs.