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