Question 7.13: At t = 0 , switch 1 in Fig. 7.53 is closed, and switch 2 is......

We need to consider the three time intervals t ≤ 0 , 0 ≤ t ≤ 4 , and t ≥ 4 separately. For t < 0 ,  switches S_{1}   and S_{2}   are open so that i = 0 .  Since the inductor current cannot change instantly,

i(0^{-} ) = i(0) = i(0^{+} ) = 0

For 0 ≤ t ≤ 4 , S_{1}   is closed so that the 4-Ω  and 6-Ω  resistors are in series. (Remember, at this time S_{2} , is still open.) Hence, assuming for now that  S_{1} is closed forever,

i(∞) = \frac{40}{4 + 6} = 4 A ,      R_{Th} = 4 + 6 =   10 Ω

\tau = \frac{L}{R_{Th}} = \frac{5}{10} = \frac{1}{2} s

Thus,

i(t) = i(∞) + [i(0) – i(∞)] e^{-t/\tau}

= 4  + (0 – 4) e^{-2t} = 4( 1 – e^{-2t} ) A ,      0 ≤ t ≤ 4 

For t ≤ 4 , S_{2}   is closed; the 10-V voltage source is connected, and the circuit changes. This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is

i(4) = i(4^{-}) = 4( 1 – e^{-8} ) \simeq 4 A

To find i(∞)  let be the voltage at node P in Fig. 7.53. Using KCL,

\frac{40 – v}{4} + \frac{10 – v}{2} = \frac{v}{6}    ⇒    v= \frac{180}{11} V

i(∞) = \frac{v}{6} = \frac{30}{11} = 2.727 A

The Thevenin resistance at the inductor terminals is

R_{Th} = 4 \parallel 2 + 6 = \frac{4 × 2}{6} + 6 = \frac{22}{3}Ω

and

\tau = \frac{L}{R_{Th}} = \frac{5}{\frac{22}{3}}= \frac{15}{22} s

Hence,

i(t) = i(∞) + [ i(4) – i(∞)] e^{-(t-4)/\tau} ,   t ≥ 4

We need ( t- 4)  in the exponential because of the time delay. Thus,

i(t) = 2.727  + (4 – 2.727) e^{-(t – 4)/ \tau} ,        \tau = \frac{15}{22}

= 2.727 + 1.273 e^{-1.4667(t – 4)} ,           t ≥ 4

Putting all this together,

At t = 2,

i(2) = 4 ( 1 – e^{-4} ) = 3.93  A

At t = 5 ,

i(5) = 2.727 + 1.273 e^{-1.4667} = 3.02  A