A) 272V
B) 320V
C) 415V
D) 471V
Correct Answer: D
Solution :
\[i={{i}_{0}}\sin \omega t={{i}_{0}}\sin 2\,\pi nt\] \[{{E}_{\max }}=M\frac{di}{dt}=N.\frac{d}{dt}({{i}_{0}}.\sin 2\pi nt)\] \[=M{{i}_{0}}.2\pi n\,\,\cos 2\pi \,nt\] \[\therefore \] \[{{E}_{\max }}=M{{i}_{0}}.2\pi n\] \[=1.5\times 1\times 2\pi \times 50\] \[=150\pi V=471\,V\]You need to login to perform this action.
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