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

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

For 0 ≤ t ≤ 4, S_{1} is closed so that the 4-Ω and 6-Ω resistors are in series. Hence, assuming for now that S_{1} is closed forever,

i(\infty)=\frac{40}{4+6}=4 A , \quad R_{ Th }=4+6=10 \Omega
\tau=\frac{L}{R_{ Th }}=\frac{5}{10}=\frac{1}{2} s
Thus,
i(t)=i(\infty)+[i(0)-i(\infty)] e^{-t / \tau}
=4+(0-4) e^{-2 t}=4\left(1-e^{-2 t}\right) 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

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

\frac{40-v}{4}+\frac{10-v}{2}=\frac{v}{6} \quad \Longrightarrow \quad v=\frac{180}{11} V
i(\infty)=\frac{v}{6}=\frac{30}{11}=2.727 A

The Thevenin resistance at the inductor terminals is

and

Hence,

i(t)=i(\infty)+[i(4)-i(\infty)] 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}, \quad \tau=\frac{15}{22}
=2.727+1.273 e^{-1.4667(t-4)},        t ≥ 4
Putting all this together,

At t = 2,

At t = 5,