Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 1 I(s) 1/s + − 1/s s I(s) = 1s 1 1 = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2 i( t ) = 2 3 e - t 2 sin 2 t 3 i( t ) = e sin ( ) A Chapter 16, Solution 2. s + Vx − 2 4 8/s 4 s + − .Vx − 4 s + Vx − 0 + Vx − 0 = 0 8 s. | Chapter 16 Solution 1. Consider the s-domain form of the circuit which is shown below. 1 Vs 1 I S ---L----- --- - r- 1 s 1 s s2 s 1 s 1 2 2 G 3 2 2 i t -U V3 sin 2 2 A J i t - st sin A Chapter 16 Solution 2. V - Vx s V - 0 V - 0 n --- - - --- 0 s 2 8 s 2 4 - s Vx 4s 8 - 16s 32 2s2 4s Vx s2Vx 0 s Vx 3s2 8s 8 16s 32 s A Vx -16 s 2 s 3s2 8s 8 16 - - -------- r -------- r 4 .V8 4 .V8 s - J s -- j 3 3 3 3 s k vx -4 2e - 1 3333 J0-9428 t 2e - L3333-J0-9428 t u t V vx 4u t - e - 4t 3 cos 6 -4t 3 f2 2 e sin -------- V2 k 3 f 1 k 1 t V Chapter 16 Solution 3. Current division leads to f 1 1 V 15 2 5 5 0 8 s 1 1 s 10 16s 16 s k2 8 vo t 1 - e-0 625t u t V Chapter 16 Solution 4. The s-domain form of the circuit is shown below. s 1 s 1 6 VA 10 s Vo s Using voltage division 10 s f1 V s s 6 10 s I s 1 10 f1 s2 6s 101 s 1J _10A Bs C Vo s s 1 s2 6s 10 s 1 s2 6s 10 10 A s2 6s 10 B s2 s C s 1 Equating coefficients s2 0 A B ------ B -A s1 0 6A B C 5A C ------ C -5A s0 10 10A C 5A ------ A 2 B -2 C -10 2 2s 10 2 2 s 3 4 o s s 1 s2 6s 10 s 1 s 3 2 12 s 3 2 12 vo t 2e-t - 2e-3t cos t - 4e-3t sin t V Chapter 16 Solution .
