# RLC second order differential equation question

• Mar 29th 2012, 03:10 AM
Paulo1913
RLC second order differential equation question
How do I set up and solve a second order differential equation to find the voltage across a capacitor knowing R, L, C, the peak to peak voltage source, and the frequency?

As far as I can tell, it is:

L*(d^2i/dt^2)+R(di/dt)+(1/C)i=wVcoswt

But I am unsure about the RHS of the equation and then what to do from there to find the voltage across the capacitor...

Any help is much appreciated
• Mar 29th 2012, 06:30 PM
Kiwi_Dave
Re: RLC second order differential equation question
1. You could "guess" the form of the answer. e.g. i = A sin (mt) + B cos (mt)
2. Calculate the derivative and second derivative of your guessed i
3. Substitute for i and its derivatives in the differential equation
4. Choose A, B and m to make the equation work

The last bit can be done by choosing special values for t such as t= 0 and/or mt = pi/2.

Another trick is to solve it with the RHS=0 first (power supply shorted out). Then solve it with the power supply turned on. You can then add these two answers together to form another solution. The two seperate solutions are known as the natural response and the forced responce respectively (and may have different frequencies).
• Mar 29th 2012, 09:25 PM
Paulo1913
Re: RLC second order differential equation question
So if I assume the form of the answer is i = A sin (mt) + B cos (mt), with i' = mA cos (mt) - mA sin (mt) and i'' = - Am^2 sin (mt) + Bm^2 cos (mt)
and then substitute into Li'' + Ri' + (1/C)i = mV cos mt
to make:

L(- Am^2 sin (mt) + Bm^2 cos (mt)) + R(mA cos (mt) - mA sin (mt)) + (1/C)(A sin (mt) + B cos (mt)) = mV cos mt

How do I get from this point to the values of A, B and m?
• Mar 30th 2012, 01:01 PM
Kiwi_Dave
Re: RLC second order differential equation question
Take a look here

RLC circuit - Wikipedia, the free encyclopedia

Alternatively, if you post your question on the differential equations page you will probably get plenty of replies. I have added some doodles below.

assume
i = A sin (mt + phi)
i' = Am cos (mt + phi)
i'' = -Am^2 sin (mt + phi)

Then

L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

this must be true for any t. So consider t = pi/2/m, then

L(- Am^2 sin (pi/2+phi)) + R(mA cos (pi/2+phi)) + (1/C)(A sin (pi/2+phi)) = 0
simplifying:

(1/C-Lm^2)
sin (pi/2+phi) + Rm cos (pi/2+phi) = 0

but sin (pi/2 + phi) = cos (phi) and cos (pi/2 + phi) = -sin(phi) so

(1/C-Lm^2)cos (phi) = Rm sin (phi)

or tan(phi) =
(1/C-Lm^2) / R / m ONE UNKNOWN FOUND!

Now if we try t = 0 in:
L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

we get

L(- Am^2 sin (phi)) + R(mA cos (phi)) + (1/C)(A sin (phi)) = mV
or

(1/(mC)-mL) sin (phi) + R cos (phi) = V/A. Perhaps this is useful since we know tan (phi)?

Now if we try t = -phi/m in:
L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

we get

RmA= mV cos m(-phi/m) = mV cos (phi)

or

RA= V cos (phi)

So

A = V cos (phi) /R and we know tan (phi) from above