Question 3.30: In the circuit of Fig. 1, the switch remains in position-1 f......

Case i : Switch in position-1
In position-1, the circuit should have attained steady state because the switch is closed for a long time. The steady state condition of the given circuit with switch in position-1 is shown in Fig. 2. Here a steady current of $I_0$ flows through the inductance. With reference to Fig. 2, we can write,

$I _0=\frac{10}{2}=5 A$

This current $I_0$ will be the initial current when the switch is changed from position-1 to position-2.

Case ii : Switch in position-2
Let, i (t) be the current through the circuit when the switch is closed to position-2.

Let, $I ( s )= \mathcal{L}  \{i( t )\}$

The time domain and s-domain circuits with switch in position-2 are shown in Figs 3 and 4, respectively.

With reference to Fig. 4, we can write,

The roots of the quadratic, s² + 5s + 25 = 0 are,

Since the roots are complex conjugate, the response will be damped sinusoid. Let us rearrange the terms of denominator polynomial of I (s).

Let us take the inverse Laplace transform of I (s).

$\mathcal{L} ^{-1}\{ I ( s )\}=5 \times \mathcal{L} ^{-1}\left\{\frac{( s +2.5)}{( s +2.5)^2+4.33^2}\right\}-2.8868 \times L ^{-1}\left\{\frac{4.33}{( s +2.5)^2+4.33^2}\right\} \\ \therefore \quad i( t )=5 e ^{-2.5 t } \cos 4.33 t -2.8868 e ^{-2.5 t } \sin 4.33 t \\\quad= e ^{-2.5 t }[\cos 4.33 t \times 5-\sin 4.33 t \times 2.8868]$          ……..(1)

Let us construct a right-angled triangle with 5 and 2.8868 as two sides as shown in Fig. 5. With reference to Fig. 5, we can write,

$\tan \phi=\frac{2.8868}{5}=0.5774 \\ \therefore \phi=\tan ^{-1} 0.5774=30^{\circ}$
Also,$\cos \phi=\frac{5}{5.7735} \quad \Rightarrow \quad 5=5.7735 \cos \phi \\ \therefore \quad 5=5.7735 \cos 30^{\circ}      …..(2) \\ \sin \phi=\frac{2.8868}{5.7735} \Rightarrow 2.8868=5.7735 \sin \phi \\ \therefore 2.8868=5.7735 \sin 30^{\circ}$            …..(3)
Using equations (2) and (3), equation (1) can be written as,