A) \[5.5\text{ }H\]
B) \[1.5\text{ }H\]
C) \[2.5\text{ }H\]
D) \[9.5\text{ }H\]
Correct Answer: A
Solution :
\[{{I}_{0}}=\frac{E}{R}=\frac{5}{50}=0.1A\] \[I=60\,mA=60\times {{10}^{-3}}A,\,t=0.1\] Now, \[I={{I}_{0}}\left( 1-{{e}^{\frac{R}{L}t}} \right)\] \[\therefore \,\,\,60\times {{10}^{-3}}=0.1\left( 1-{{e}^{-\frac{50}{L}\times 0.1}} \right)=0.1\left( 1-{{e}^{-\frac{5}{L}}} \right)\] or \[1-{{e}^{-5/L}}=0.6\] \[\therefore \,\,\,{{e}^{-5/L}}=1-0.6=0.4=\frac{4}{10}\] or \[{{e}^{5/L}}=10/4\] Taking log of both sides \[\frac{5}{L}=2.303[{{\log }_{10}}10-{{\log }_{10}}4]\] \[=2.303[1.0000-0.6021]=2.303\times 0.3979=0.9164\]\[\therefore \,\,L=\frac{5}{0.9164}=5.5H\]
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