To determine The solution of the initial value problem for the RLC -circuit when R = 8 Ω , L = 0.2 H , C = 12.5 × 10 − 3 F , and E ( t ) = 100 sin 10 t V and sketch the solution. Explanation 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 particular solution of (1) is given by, I p = a cos ω t + b sin ω t 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 = 8 Ω , L = 0.2 H , C = 12.5 × 10 − 3 F , and E ( t ) = 100 sin 10 t V . First compute the derivative of E ( t ) = 100 sin 10 t with respect to t as follows. E ′ ( t ) = d d t ( 100 sin 10 t ) E 0 ω cos ω t = 100 ( 10 cos 10 t ) From the above equation E 0 ω cos ω t = 100 ( 10 cos 10 t ) , the values of E 0 and ω are obtained as E 0 = 1 00 and ω = 10 . Now substitute R = 8 , L = 0.2 , C = 0.0125 , and E 0 ω cos ω t = 100 ( 10 cos 10 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 0.2 I ″ + 8 I ′ + 1 0.0125 I = 100 ( 10 cos 10 t ) 0.2 I ″ + 8 I ′ + 80 I = 100 ( 10 cos 10 t ) Hence the current in the given RLC circuit is the solution of second order ordinary differential equation 0.2 I ″ + 8 I ′ + 80 I = 100 ( 10 cos 10 t ) . The auxiliary equation of 0.2 I ″ + 8 I ′ + 80 I is 0.2 m 2 + 8 m + 80 = 0 . Obtain the roots of 0.2 m 2 + 8 m + 80 = 0 as shown below. 0.2 m 2 + 8 m + 80 = 0 m 2 + 40 m + 400 = 0 ( m + 20 ) 2 = 0 m = − 20 ( double root ) Here the root is a double real root and hence the solution y h ( t ) is y h ( t ) = ( c 1 + c 2 t ) e − 20 t . Now compute the particular solution of 0.5 I ″ + 4 I ′ + 10 I = 500 ( 2 cos 2 t ) as shown below. First obtain the value of S as follows. S = ω L − 1 ω C = 10 ( 0.2 ) − 1 10 ( 0.0125 ) = 2 − 1 0.125 = 2 − 8 = − 6 Using the value of S , obtain the value of a as follows. a = − E 0 S R 2 + S 2 = − 100 ( − 6 ) 8 2 + ( − 6 ) 2 = 600 64 + 36 = 600 100 = 6 Now obtain the value of b as shown below

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ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.9, Problem 16P

To determine

The solution of the initial value problem for the RLC-circuit when R=8Ω, L=0.2H, C=12.5×10−3F, and E(t)=100sin10tV and sketch the solution.

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Chapter 2.9, Problem 15P

Chapter 2.9, Problem 17P

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