A) 500 sec
B) 20 sec
C) 35 milli sec
D) 1 milli sec
Correct Answer: D
Solution :
Peak current in the circuits \[{{i}_{0}}=\frac{12}{6}=2A\] Current decreases from 2A to 1A i.e., becomes half in time \[t=0.693\frac{L}{R}=0.693\times \frac{8.4\times {{10}^{-3}}}{6}=1milli\,sec.\]
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