Question

In the circuit shown in the following figure, the switch is kept closed for a long time and then opened at t = 0.
The values of the current just before opening the switch $(t=0^{-})$ are, respectively.

• A. $\frac{3}{4}Aand1A$
• B. $\frac{7}{6}Aand\frac{5}{2}A$
• C. $1Aand\frac{7}{6}A$
• D. 1 A and 1 A

Answer 1

The correct option is C $1Aand\frac{7}{6}A$

$att=0^{-}$
The inductor will be short,


$\therefore i=\frac{30}{20+10}A=1A$

and current through inductor at $t=0^{+}$

$I_L=1A$

From the figure,

Applying KVL in outer loop.

$30=(20+20)I+20i$ ....(i)

As the current through the inductor does not change instantaneously. The current the inductor will be 1 A at t = $0^{+}$

$i=(I+1)$

$30=40I+20I+20$

$I=\frac{10}{60}=\frac{1}{6}$

Hence current,

$i=I+1=\frac{1}{6}+1=\frac{7}{6}A$