A series RLC circuit with R = 3 ohm, L = 1H, and C = 0.5F is given to you. Suppo

Question

A series RLC circuit with R = 3 ohm, L = 1H, and C = 0.5F is given to you. Suppose the initial current in the circuit i(0) = 1 A, and the initial rate of change of current i.(0) = 4A/second. Assuming the applied voltage is zero (short circuit the voltage source) determine and sketch the current izi as a function of time. What is izi(t) as t rightarrow infinity. Now assume that the inital conditions are zero. An input sin(t) (unit volts) is applied to the circuit starting from t = 0. Determine the current izs through the circuit as a function of time. Sketch each term of this function. What is izs(t) as t rightarrow infinity.

Explanation / Answer

(a) i(0) = 1A, di(0)/dt = 4A/sec Applying KVL 0 = R i + L di/dt + 1/C i dt 0 = 3i + di/dt + 1/0.5 i dt Differentiating 0 = 3di/dt + d2i /dt2 + 2i d2i /dt2 + 3di/dt + 2i = 0 (D2 + 3D + 2) i = 0 D1 = -1 D2 = -2 The current i is given by i = C1 eAt + C2eBt i = C1 e-t + C2e-2t To find C1 and C2 At t=0, i=1A 1 = C1 + C2 ------------ (1) Differentiating i di/dt = -C1e-t -2 C2e-2t At t=0, di/dt = 4A/s 4 = - C1 - 2C2 ----------- (2) Solving (1) and (2) C2 = -5 C1 = 6 i = -5e-t + 6e-2t Current at t--- will be i () = -5e- + 6e- = 0 b) With v=sint Applying KVL R i + L di/dt + 1/C i dt = sint Differentiating Differentiating Differentiating cost = 3di/dt + d2i /dt2 +2i d2i /dt2 + 3di/dt + 2i = cost (D2 + 3D + 2) i = 0 D1 = -1 D2 = -2 The complementary solution iC = C1 eAt + C2eBt iC = C1 e-t +C2 e-2t Differentiating cost = 3di/dt + d2i /dt2 +2i d2i /dt2 + 3di/dt + 2i = cost (D2 + 3D + 2) i = 0 D1 = -1 D2 = -2 The complementary solution iC = C1 eAt + C2eBt iC = C1 e-t +C2 e-2t The particular solution iP = V /Z cos [t + ] v = sint V=1V and = 1 rad/sec Z = R2 + (1/C - L )2= 9 + 2-1 = 10 = 3.162 = tan-1 {(1/CR ) + (L/R)} ] =tan-1 {(1/1.5 ) + (1/3)} ] = tan-1 1 =45o ip = 1/ 3.16 cos [ t + 45o ] = 0.3 cos[t+45o] Complete solution is i = iC + ip i = C1 e-t + C2e-2t + 0.3 cos [t+45o] At t = 0, i(0) = 0 0 = C1 + C2 + 0.3 cos[45o] C1 + C2 = -0.212 ------------(1) Differntiating di/dt = -C1 e-t - 2C2e-2t  - 0.3 sin [t+45o] At t=0 di/dt = 0 0 = -C1 -2C2 - 0.212 C1 +2C2 = - 0.212 ------------------(2) From (1) and (2) C1 = -0.212 C2 = 0 Complete solution i = -0.212 e-t + 0.3cos [t+45o] i = C1 e-t + C2e-2t + 0.3 cos [t+45o] At t = 0, i(0) = 0 0 = C1 + C2 + 0.3 cos[45o] C1 + C2 = -0.212 ------------(1) Differntiating di/dt = -C1 e-t - 2C2e-2t  - 0.3 sin [t+45o] At t=0 di/dt = 0 0 = -C1 -2C2 - 0.212 C1 +2C2 = - 0.212 ------------------(2) From (1) and (2) C1 = -0.212 C2 = 0 Complete solution i = -0.212 e-t + 0.3cos [t+45o]

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