A) 500s
B) 25s
C) 35ms
D) 1ms
Correct Answer: D
Solution :
[d] Using \[I={{I}_{0}}(1-{{e}^{-t/\tau }})\] But \[{{I}_{0}}=\frac{V}{R}\] and \[\tau =\frac{L}{R}\] \[\therefore \,\,\,\,I=\frac{V}{R}(1-{{e}^{-Rt/L}})=\frac{12}{6}\left[ 1-{{e}^{-6t/8.4\times {{10}^{-3}}}} \right]\] =1 (given) \[\therefore \,\,\,t=0.97\times {{10}^{-3}}\,s\approx 1ms\]
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