undefined Solution The steady-state current in the RLC -circuit when R = 12 Ω , L = 1.2 H , C = 20 3 ⋅ 10 − 3 F , and E ( t ) = 12000 sin 25 t V . The steady-state current in the given RLC -circuit is I = e − 5 t ( A cos 10 t + B sin 10 t ) − 400 cos 25 t + 200 sin 25 t _ . Formula used: The current in an RLC -circuit is the solution of nonhomogeneous second-order ordinary differential equation, L I ″ + R I ′ + 1 C I = E ′ ( t ) = E 0 ω cos ω t ⋯ ⋯ ( 1 ) The steady-state current in the RLC -circuit is the particular solution I p = a cos ω t + b sin ω t of (1) where a = − E 0 S R 2 + S 2 , b = E 0 R R 2 + S 2 and S = ω L − 1 ω C . Calculation: It is given that, R = 12 Ω , L = 1.2 H , C = 20 3 ⋅ 10 − 3 F , and E ( t ) = 12000 sin 25 t V . First compute the derivative of E ( t ) = 12000 sin 25 t with respect to t as follows. E ′ ( t ) = d d t ( 12000 sin 25 t ) E 0 ω cos ω t = 12000 ( 25 cos 25 t ) From the above equation E 0 ω cos ω t = 12000 ( 25 cos 25 t ) , the values of E 0 and ω are obtained as E 0 = 12000 and ω = 25 . Now substitute R = 12 , L = 1.2 , C = 20 3 ⋅ 10 − 3 , and E 0 ω cos ω t = 12000 ( 25 cos 25 t ) in equation (1) and obtain a second order nonhomogeneous ordinary differential equation as shown below. L I ″ + R I ′ + 1 C I = E 0 ω cos ω t 1.2 I ″ + 12 I ′ + 1 ( 20 3 ⋅ 10 − 3 ) I = 12000 ( 25 cos 25 t ) 1.2 I ″ + 12 I ′ + 150 I = 12000 ( 25 cos 25 t ) Hence the current in the given RLC circuit is the solution of second order ordinary differential equation 1.2 I ″ + 12 I ′ + 150 I = 12000 ( 25 cos 25 t ) . First compute the solution of the homogeneous part as shown below. The auxiliary equation of 1.2 I ″ + 12 I ′ + 150 I = 0 is 1.2 m 2 + 12 m + 150 = 0 and the roots of auxiliary equation are obtained as follows. 1.2 m 2 + 12 m + 150 = 0 m 2 + 10 m + 125 = 0 m = − 10 ± 10 2 − 4 ( 1 ) ( 125 ) 2 = − 10 ± − 400 2 On further simplification, m = − 10 ± 20 i 2 = − 5 ± 10 i Here the roots are complex conjugate. Therefore, the solution of the homogeneous part is I h = e − 5 t ( A cos 10 t + B sin 10 t ) . Now compute the particular solution of 1.2 I ″ + 12 I ′ + 150 I = 12000 ( 25 cos 25 t ) as shown below. By the above formula, the particular solution is I p = a cos ω t + b sin ω t of (1) where a = − E 0 S R 2 + S 2 , b = E 0 R R 2 + S 2 and S = ω L − 1 ω C . Now compute the value of S as shown below. S = ω L − 1 ω C = 25 ( 1.2 ) − 150 25 = 30 − 6 = 24 Using the value of S , obtain the value of a as follows. a = − E 0 S R 2 + S 2 = − 12000 ( 24 ) 12 2 + ( 24 ) 2 = − 288000 144 + 576 = − 288000 720 = − 400 Now obtain the value of b as shown below. b = E 0 R R 2 + S 2 = 12000 ( 12 ) 12 2 + ( 24 ) 2 = 144000 144 + 576 = 144000 720 = 200 Obtain the particular solution by substituting a = − 400 , b = 200 and ω = 25 in I p = a cos ω t + b sin ω t as shown below. I p = a cos ω t + b sin ω t = − 400 cos 25 t + 200 sin 25 t The current in the given RLC -circuit is obtained as follows. I = I h + I p = e − 5 t ( A cos 10 t + B sin 10 t ) − 400 cos 25 t + 200 sin 25 t Therefore, the steady-state current in the given RLC -circuit is I = e − 5 t ( A cos 10 t + B sin 10 t ) − 400 cos 25 t + 200 sin 25 t _ .

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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