A) 0.12 A
B) 0.38 A
C) 0.34 A
D) 0.12 A
Correct Answer: C
Solution :
Induced potential is given by \[\varepsilon =-\frac{d\phi }{dt}=-\frac{d}{dt}(4{{t}^{2}}-5t+1)\] \[=(2\times 4t-5)=5-8t\] At \[t=0.20s\] \[=3.4V\] So, induced current will be \[I=\frac{\varepsilon }{R}=\frac{3.4}{10}=0.34A\]You need to login to perform this action.
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