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Lcr Capacitance
L-C-R Circuit Help.... 2nd order differential equation..?

Suppose in a LCR circuit, we have a battery of pd V, a Inductor with self inductance 'L' , a resistor with resistance 'R' and a capacitor with capacitance 'C' , all circuit elements are connected in series, Find the current 'i' and charge 'q' in the circuit.....

My approach-

By Kirchoff's law-
V - iR - q/C - L(di/dt) =0

V - R(dq/dt) - q/C - L(d^2d/dt^2) = 0
and I'm Unable to solve second order differential equation..... Please help...

RATING ASSURED!!!!

So you have:

V - R(dq/dt) - q/C - L(d^2q/dt^2) = 0

q'' + (R/L)*q' + q/(LC) = V/L

(Note that this is formally equivalent to the differential equation for a damped harmonic oscillator subjected to a constant forcing.)

This is a second-order, linear, non-homogeneous ODE with constant coefficients. Start by solving the homogeneous equation"

q'' + (R/L)*q' + q/(LC) = 0

As usual for this class of equations, we seek a solution of the form:

q = exp(k*t), so q' = k*exp(*t) q'' = (k^2)*exp(k*t)

Plugging these into the homogeneous equation and cancelling the common factor of exp(k*t) yields the characteristic equation:

k^2 + (R/L)*k + 1/(L*C) = 0

k = [-(R/L) ± sqrt((R/L)^2 - 4/(L*C))]/2 = [-R/(2L) ± sqrt((R/(2L))^2 - 1/(L*C))]

The character of the solutions differs depending on whether the quantity, (R/L)^2 - 4/(L*C)) is greater than zero, less than zero, or equal to zero. These correspond to the cases of an overdamped, underdamped, or critically damped oscillator. The first and third of these cases lead to a solution that is simply a decaying exponential. The more interesting case is the second, and I'll go through the derivation of the most convenient form of the solution for this case.

Let ω = sqrt(1/(L*C) - (R/(2L))^2), then:

k = -R/L ± i*ω

where i = sqrt(-1)

The solution to the homogeneous equation is then:

q_h(t) = a*exp((-R/L + i*ω)*t) + b*exp((-R/L - i*ω)*t)

where a and b are constants of integration.

Remember that exp( y+z) = exp(y)*exp(z), so the homogeneous solution can be written as:

q_h(t) = exp(-R*t/L) * [a*exp(i*ω*t) + b*exp(-i*ω*t)]

The pesky imaginary exponentials can be eliminated using Euler's relationship:

exp(w*i*t) = cos(w*t) + i*sin(w*t).

Remembering that cos(-x) = cos(x) and sin(-x) = -sin(x), Euler's relationship simplifies the solution to:

q_h(t) = exp(-R*t/L) * [ A*cos(ω*t) + B*sin(ω*t)]

where A = (a+b) and B = i*(a-b) are just different ways of writing the integration constants.

Now we need to look for the particular solution for the original nonhomogeneous equation. Assuming the V is constant, the method of undetermined coefficients suggests that we look for a solution of the form:

q_p = d, where d is a constant. Then, q_p' = q_p' = 0. Plugging these into the original ODE gives:

d/(LC) = V/L

d = V*C

So the particular solution is:

q_p(t) = V*C

The complete solution is the sum of the homogeneous and particular solutions, so:

q(t) = exp(-R*t/L) * [ A*cos(ω*t) + B*sin(ω*t)] + V*C