A) 69 V
B) 20 V
C) 23 V
D) 27 V
Correct Answer: C
Solution :
Given, changing flux is \[\phi =2{{t}^{2}}+3t+4\] On differentiating with respect to time \[|e|=\left| \frac{d\phi }{dt} \right|=\frac{d(2{{t}^{2}}+3t+4)}{dt}\] \[=4t+3+0\] At \[t=5\] \[e={{\left| \frac{d\phi }{dt} \right|}_{t\,=\,5s}}=4\times 5+3=23\,V\]You need to login to perform this action.
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