Source-free 2nd order lin. circuit

  • Thread starter EvLer
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RLC second order linear network question:
So, we are given this equation which is the same for Vc(t) and iL(t) expressed as x(t):

2nd deriv of x(t) + R/L(1st deriv of x(t)) + 1/(LC)(x(t)) = 0;

And in one of the problems it asks to find both equation for the Vc(t) and iL(t) for t < 0, and now I am confused, it seems to me that they are the same, since the solution is the same for both of them:

aS^2 + bS + c = 0;
because the coefficients are the same from differential equation, so there are the same roots for Vc(t) and iL(t), and roots are w/t imaginary part, just reals.
Am I wrong?

Thanks a lot.
 
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Andrew Mason
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EvLer said:
RLC second order linear network question:
So, we are given this equation which is the same for Vc(t) and iL(t) expressed as x(t):

2nd deriv of x(t) + R/L(1st deriv of x(t)) + 1/(LC)(x(t)) = 0;
I am not sure I understand what you are asking exactly, but perhaps this will help.

An LC circuit with resistance is simply a damped harmonic oscillator. The differential equation:

[tex]L\ddot x(t) + R\dot x(t) + \frac{1}{C}x(t) = 0[/tex]

has the general solution:

[tex]x = A_0e^{-\gamma t}sin(\omega t+\phi)[/tex]

where [itex]\omega^2 = \omega_0^2 - \gamma^2[/itex] and
[itex]\omega_0^2 = 1/LC[/itex] and
[itex]\gamma = R/2L[/itex]

The relationship between V and I in an RLC circuit is:

[tex]V = IZ[/tex] where is the impedance: [itex]Z^2 = R^2 + (\omega L - 1/\omega C)^2[/itex]

AM
 
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