A) \[200\pi \,volt\]
B) \[\frac{10\pi }{3}volt\]
C) \[\frac{4\pi }{3}\,volt\]
D) \[\frac{2}{3}\]volt
Correct Answer: C
Solution :
\[e=NBA\omega ;\ \omega =2\pi f=2\pi \times \frac{2000}{60}\] \[\therefore \ e=50\times 0.05\times 80\times {{10}^{-4}}\times 2\pi \times \frac{2000}{60}=\frac{4\pi }{3}\]
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