Chapter 7, Problem 1P

Use the differential equation approach to find i(t) for $t>0$ in the network in Fig. P7.l.

To determine

The current $i\left(t\right)$ for $t>0$ in the given circuit using the differential equation.

Answer to Problem 1P

$i_L\left(t\right)=2e^{-3t}A$

Explanation of Solution

Given information:

Given circuit is shown below

Fig: Given circuit diagram

Calculation:

Although for ^-$t=0^{-}$ inductor acts as short circuit, now draw the circuit for $t=0^{-}$ .

Fig: Circuit diagram for $t=0^{-}$

$i_L\left(0^{-}\right)=\frac{12}{6}=2A$

Now, draw the circuit when switch is opened at t =0.

Fig: circuit diagram for $t>0$

Apply KVL for loop 1

$\begin{array}{l}2\frac{d{i}_L\left(t\right)}{dt}+6i_L\left(t\right)=0\\ \frac{d{i}_L\left(t\right)}{dt}=-3i_L\left(t\right)\end{array}$

Standard from of solution of above equation is

$i_L\left(t\right)=K_{2}e{}^{-\frac{t}{\tau }}$

Plugging $i_L\left(t\right)$ into above equation.

$\begin{array}{l}-\frac{K_2{e}^{-\frac{t}{\tau }}}{\tau }=-3K_2{e}^{-\frac{t}{\tau }}\\ \tau =\frac{1}{3}\end{array}$

Therefore,

$\begin{array}{l}{i}_L\left(t\right)=K_2{e}^{-\frac{t}{\frac{1}{3}}}\\ i_L\left(t\right)=K_2{e}^{-3t}\left(1\right)\\ \because \text{at}\text{t}=0^{-}\text{,}{i}_L=2A\\ \therefore i_L\left(0\right)=K_2{e}^{-3\times 0}\\ K_2=2\end{array}$

Now, from equation (1),

$i_L\left(t\right)=2e^{-3t}A$