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\[\tau =\frac{L}{{{R}_{ab}}}=\frac{10}{2.5}=4\]sYou need to login to perform this action.
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