An LC circuit with L - 0.1 h, a capacitance of 0.1 f and E(t) = 100sin ( gamma t

Question

An LC circuit with L - 0.1 h, a capacitance of 0.1 f and E(t) = 100sin ( gamma t) - Initially, there is no charge and the current is zero capacitor... Find the charge and the current at any time t. What is the steady state current?

Explanation / Answer

L= 0.1 H C = 0.1 F E(t) = 100sin(wt) d2i/dt2 +(1/LC)di/dt = E(t) => y'' + 100y' = 100sin(wt) Solve ( d^2 y(t))/( dt^2)+100 ( dy(t))/( dt) = 100 sin(a t): The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving ( d^2 y(t))/( dt^2)+100 ( dy(t))/( dt) = 0: Assume a solution will be proportional to e^(lambda t) for some constant lambda. Substitute y(t) = e^(lambda t) into the differential equation: ( d^2 )/( dt^2)(e^(lambda t))+100 ( d)/( dt)(e^(lambda t)) = 0 Substitute ( d^2 )/( dt^2)(e^(lambda t)) = lambda^2 e^(lambda t) and ( d)/( dt)(e^(lambda t)) = lambda e^(lambda t): lambda^2 e^(lambda t)+100 lambda e^(lambda t) = 0 Factor out e^(lambda t): (lambda^2+100 lambda) e^(lambda t) = 0 Since e^(lambda t) !=0 for any finite lambda, the zeros must come from the polynomial: lambda^2+100 lambda = 0 Factor: lambda (lambda+100) = 0 Solve for lambda: lambda = -100 or lambda = 0 The root lambda = -100 gives y_1(t) = c_1 e^(-100 t) as a solution, where c_1 is an arbitrary constant. The root lambda = 0 gives y_2(t) = c_2 as a solution, where c_2 is an arbitrary constant. The general solution is the sum of the above solutions: y(t) = y_1(t)+y_2(t) = c_1 e^(-100 t)+c_2 Determine the particular solution to ( d^2 y(t))/( dt^2)+100 ( dy(t))/( dt) = 100 sin(a t) by the method of undetermined coefficients: The particular solution to ( d^2 y(t))/( dt^2)+100 ( dy(t))/( dt) = 100 sin(a t) is of the form: y_p(t) = b_2 sin(a t)+b_1 cos(a t) Solve for the unknown constants b_1 and b_2: Compute ( dy_p(t))/( dt): ( dy_p(t))/( dt) = ( d)/( dt)(b_2 sin(a t)+b_1 cos(a t)) = -a b_1 sin(a t)+a b_2 cos(a t) Compute ( d^2 y_p(t))/( dt^2): ( d^2 y_p(t))/( dt^2) = ( d^2 )/( dt^2)(b_2 sin(a t)+b_1 cos(a t)) = -a^2 b_2 sin(a t)-a^2 b_1 cos(a t) Substitute the particular solution y_p(t) into the differential equation: ( d^2 y_p(t))/( dt^2)+100 ( dy_p(t))/( dt) = 100 sin(a t) (-a^2 b_2 sin(a t)-a^2 b_1 cos(a t))+100 (-a b_1 sin(a t)+a b_2 cos(a t)) = 100 sin(a t) Simplify: (-a^2 b_2-100 a b_1) sin(a t)+(100 a b_2-a^2 b_1) cos(a t) = 100 sin(a t) Equate the coefficients of cos(a t) on both sides of the equation: 100 a b_2-a^2 b_1 = 0 Equate the coefficients of sin(a t) on both sides of the equation: -a^2 b_2-100 a b_1 = 100 Solve the system: b_1 = -10000/(a (a^2+10000)) b_2 = -100/(a^2+10000) Substitute b_1 and b_2 into y_p(t) = b_2 sin(a t)+b_1 cos(a t): y_p(t) = -(10000 cos(a t))/(a (a^2+10000))-(100 sin(a t))/(a^2+10000) The general solution is: Answer: | | y(t) = y_c(t)+y_p(t) = -(100 sin(a t))/(a^2+10000)+-(10000 cos(a t))/(a (a^2+10000))+c_1 e^(-100 t)+c_2 where y(t) is the current at any time t at steady state put t = infinity hence i(t=infinity) = c2 => at t=0 i=0

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