Question

Ouestion 1 (10%) The circuit shown in Fig. 1 is at steady state before t-0s. Determine v(t) and i(t) for>0. 3Ω 60 |(t) 12 u() 22 ) 12Ω Fig. 1

Answer 1

The unit step function is defined as follbws: for t20 u(t) = for to The circuit is at steady state before t=0s. At steady state, the inductor acts as short circuited. Since the value of unit step is zero for t<0, the value of input voltage source will also be zero. Thus, the circuit is source free-circuit and current flowing through inductor is zero. At 120, transform the circuit diagram in the Laplace domain. The Laplace transform of the input voltage source is, v(t) = 12u(t) V(s) = 12 The inductive impedance Zc is, Zc = sl = 3s

Redraw the modified circuit diagram in the s-domain as shown in Figure 1. 32 622 3s 22 0 (3) 352 00 ) + av 1203 22&v(s) Figure 1

From Figure 1, apply the nodal analysis at node 1. V-12 S V-0 V-VS) V-VS - +- + + 24-0. K-VC), K-VO)-, s-12, 13, 1-10.K-10- 6 12 * 3 " 3s SV, -12 - - 65 V, V-VS) V-VS +- +- - +- 12 3 35 + + - +- 0 -+ - 6s 12 - 3 - 3 - 3 3s (2** *)K-45 ")" (o)==0 (7s+4)V, - 4(5+1)V (s)– 24 = 0 ...... (1) From Figure 1, apply the nodal analysis at node 2. V, -V(s). V, -V(s)_V(s)-0 3 3s

(11 1 - t - V = - + - +- (3 353 35 2 ) (s+1). (2s +2+3s V = (35) OS (4 ) K = (25 +2+35)+() 2(s+1)V, = (55+2)V(s) ni + 20) (55 +2 Substitute PT V (s) for V, in equation (1). ute 2(s+1) (7674) (55 +2) (7s+4) 611_V(s)– 4(s +1)V (s) – 24 = 0 (35s2 +14s +20s +8)V (s)-8(52 +1+2s)V (s) = 48(s+1) (27s+185)V (s )= 48(5+1)

V (s) = 1 48(s+1) +18s 48(s+1) 16(5+1) А — + В S -

+ |2| 3| 2-3 + 1/2-3 - + +- 043 Ano=22 Ble 2 = S+1

16 A - +- S 21 + Take the inverse Laplace transform.

Observe Figure 1, the value of current 1(s) is, 1(s)=V-V(s) 3s Substitute PST4V(s) for V,. 2(5+1) (55+2) y(s)-V(s) 1(3)=2(s+1)" (S)-V(s) 35 (55 +2), 2(s +1)) y(0) 6s(s+1) PV() 6s (5+1) 3s 2(5+1)(s)

Substitute 1(s)=1 48(s +1) for V(s). me 27s+ +186" I(S) = 2(5+1)| 2782 +185) = 278 +185 48(5+1) - 24 24

Take the inverse Laplace transform. 10)((26 ) -1-7)(OA