A) 0.12 V
B) 2.4 V
C) 0.02 V
D) 1.2 V
Correct Answer: C
Solution :
(c) 0.02 V Give, \[\phi =(2{{t}^{2}}+4t+6)mWb\] As \[\varepsilon =\frac{d\phi }{dt}=\frac{d}{dt}(2{{t}^{2}}+4t+6)\times {{10}^{-3}}\] \[=(4t+4)\times {{10}^{-3}}V\] At \[t=4\,\,s,\,\,\varepsilon =(4\times 4+4)\times {{10}^{-3}}V\] \[=20\times {{10}^{-3}}V=0.02\,V\]You need to login to perform this action.
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