question_answer

The value of time constant for given circuit is

A)  1s                                          

B)  2 s

C)  3 s                                         

D)  4 s

Correct Answer: D

Solution :

 After short circuiting, the battery \[{{R}_{ab}}=\frac{(2+3)\times 5}{(2+3+5)}=2.5\Omega \] \[2\Omega \]and\[3\Omega \]are in series and this combination is in parallel with\[5\Omega \] Now, time constant\[\tau =\frac{L}{{{R}_{ab}}}=\frac{10}{2.5}=4\]s