A) \[2\times {{10}^{4}}V\]
B) \[1.2\times {{10}^{4}}V\]
C) \[2\times {{10}^{-4}}V\]
D) None of these
Correct Answer: A
Solution :
Amount of magnetic flux linked with inductor is \[\phi =Li\] Now, the emf induced in the inductor is given by \[e=-\frac{d\phi }{dt}=\frac{d}{dt}(Li)\] Or \[e=-L\frac{di}{dt}\] \[|e|=-L\frac{di}{dt}\] Here, induced current\[=\frac{V}{R}=\frac{10}{5}\] \[=2A\] Circuit switches off in millisecond or \[dt=1\times {{10}^{3}}s\] and \[L=10H\] \[\therefore \] Induced emf in inductor is \[|e|=10\times \frac{2}{1\times {{10}^{-3}}}\] \[|e|=2\times {{10}^{4}}V\]
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