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Solve the differential equation of L-C circuit and obtain the expression of current. |
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Answer» Solution :The differential equation of L-C circuit is, `(d^(2)Q)/( dt^(2))+(1)/(LC) q = 0` The solution of the equation is, `q= q_(m) COS ( omega_(0) t+ phi)`…(1) where `q_(m)` is the maximum VALUE OFQ and `phi` is phase constant. At t= 0, `q= q_(m)` `:. q_(m) = q_(m) cos phi` `:. 1 = cos phi` `:. phi = 0^(@)` `:.` From equation (1), `q = q_(m) cos ( omega_(0) t)` Differentiating w.r.t. time t, `(DQ)/(dt) = q_(m) omega_(0)( - sin omega_(0)t)` but `(dq)/(dt)= -1` `:. -I = - q_(m) omega_(0) sin omega_(0) t ` `:. I = I_(m) sin omega_(0) t ` where,` q_(m) omega _(0) =I_(m) ` `:. I = I_(m) sin omega_(0) t` |
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