A) OA
B) 0.6 A
C) 0.4 A
D) 0.2 A
Correct Answer: A
Solution :
[a]: Here,\[\phi =4{{t}^{2}}-4t+1\] Induced current, \[|\varepsilon |=\frac{d\phi }{dt}\] \[=\frac{d}{dt}(4{{t}^{2}}-4t+1)=8t-4\] Induced current, \[I=\frac{|\varepsilon |}{R}=\frac{8t-4}{10}\] At \[t=\frac{1}{2}s,I=\frac{8\times \frac{1}{2}-4}{10}=0A\]You need to login to perform this action.
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