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InterviewSolution
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Obtain the differential equation for a LC circuit. |
Answer» Solution :L-C circuit is shown in figure. In their circuit capacitor (C )and inductor ( L ) are connected. Let at time t=0, capacitor is charged and charge on it is `q_(m)`.  The moment the circuit is comleted the charge on the capacitor starts decreasing giving rise to currentin the circuit. Let q and I be the chrage and current in the circuit at time t. Since`( dI)/( dt) ` is positive the induced emf in L will have polarity as shown in figure, Means `V_(a) GT V_(b)`. As q decreases I increases, ` :. I = - (dq)/(dt)` Induced emf in inductor at any isntant, `V = epsilon = - L (dI)/(dt)` and p.d. across capacitor `= ( q)/( C )` According to Kirchhoff.s loop rule, `- L (dI)/(dt) + ( q)/( C ) = 0` but `I = - ( dq)/( dt)` `:. (dI)/(dt) = - ( d^(2)q)/( dt^(2))` `:. + L ( d^(2) q)/( dt^(2)) + ( q)/( C ) = 0 ` `:. (d^(2) q)/( dt^(2)) + ( q)/( LC ) = 0 ` `:. (d^(2)q)/( dt^(2)) + omega_(0)^(2) q= 0 ` `[ :. (1)/(LC) = omega_(0)^(2) ]` which is second order DIFFERENTIAL equation of linear charged. This equation has the form of general formula, for a simple harmonic oscillator.The charge therefore, OSCILLATES with a natural frequency. `omega_(0) = (1)/( SQRT(LC))` `:. 2 pi f_(0) = (1)/( sqrt(LC))` `:.` Its frequency is `f_(0) = (1)/(2pi ) sqrt((1)/( LC))` |
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