A) 6.7 mA
B) 0.67 mA
C) 100 mA
D) 67 mA
Correct Answer: B
Solution :
\[i=\frac{15}{0.15\times {{10}^{+3}}}\] \[15\times {{10}^{-2}}\times {{10}^{+3}}\] \[1={{10}^{-1}}{{e}^{-1\frac{Rt}{L}}}\] \[=\frac{1}{150}\] \[\frac{0.15\times {{10}^{3}}}{0.03}\times {{10}^{-3}}\] \[\frac{1}{1500}\] \[=1\] \[{{10}^{-2}}\] \[=6.67\times {{10}^{-4}}\] \[=0.667\times {{10}^{-3}}\] \[=0.667mA\]You need to login to perform this action.
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