# electrical circuit problem

• August 7th 2012, 04:54 PM
math254
electrical circuit problem
Consider an electrical circuit comprising an inductor and a capacitor, with no external applied voltage, and no electrical resistance. The current, i, through the inductor is related to the charge, q, on the capacitor by the differential equation: where the inductance, L, and capacitance C, are positive constants.

L(di/dt)+q/c=0

(a)Given that the charge and the current are related through i=dq/dt derive a second order differential equation for q(t).

i know that the second order differential is

L(d2q/dt2)+q/c=0

b)assuming that the initial charge q(0)=q0 and the initial current i(0)=i0, show that

q(t)=q0cos(wt)+(i0/w)sin(wt)

where w^2=1/LC

c) hence show that q(t) can be written as

q(t)=Acos(wt+fi)

where A and fi are to be determined

thanks for the help
• August 7th 2012, 07:55 PM
GJA
Re: electrical circuit problem
Hi, math254.

Your DE is a second order linear homogeneous equation. One route you could take is the following:

1. Set up what's called the "characteristic equation." This is done by replacing your DE with the polynomial

$L\lambda^{2}+\frac{1}{C}=0$

(if you haven't seen this before and need more details let me know).

2. Solve the above for $\lambda.$ Note: The two solutions for $\lambda$ will be a complex numbers.

3. The general solution to your DE is then

$q(t)=e^{\alpha t}(c_{1}\cos(\omega t)+c_{2}\sin(\omega t))$

where $\alpha$ is the real part of $\lambda$ and $\omega$ is the imaginary part of $\lambda$. See Equation 11 of http://www.stewartcalculus.com/data/...arEqns_Stu.pdf

4. Now use the initial conditions to determine $c_{1}$ and $c_{2}$.

5. To prove part c) I would suggest visiting http://www.mathcentre.ac.uk/resource...a-alphaetc.pdf. There is a nice outline showing how to express the sum of sine and cosine as one cosine function.

Does this help sort out the confusion?

Good luck!
• August 8th 2012, 08:38 AM
math254
Re: electrical circuit problem
thank you :)