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GUJCET · Physics · Alternating Current
At time \(t=0\) second, voltage of an AC Generator starts from 0 V and becomes 2 V at time \(t=\frac{1}{100 \pi}\) second. The voltage keeps on increasing up to 100 V , after which it starts to decrease. Find the frequency of the generator.
- A 100 Hz
- B 1 Hz
- C 2 Hz
- D 5 Hz
Answer & Solution
Correct Answer
(B) 1 Hz
Step-by-step Solution
Detailed explanation
(B) 1Hz
\(\quad \mathrm{V}=\mathrm{V}_{m} \sin \omega t\)
\(
\begin{aligned}
2 & =100 \sin (2 \pi v t) \\
2 & =100 \sin \left(2 \pi v \times \frac{1}{100 \pi}\right) \\
\frac{2}{100} & =\sin \left(\frac{v}{50}\right) \\
\frac{1}{50} & =\frac{v}{50}\left(\because \sin \left(\frac{v}{50}\right) \approx \frac{v}{50}\right) \\
\nu & =1 \mathrm{~Hz}
\end{aligned}
\)
\(\quad \mathrm{V}=\mathrm{V}_{m} \sin \omega t\)
\(
\begin{aligned}
2 & =100 \sin (2 \pi v t) \\
2 & =100 \sin \left(2 \pi v \times \frac{1}{100 \pi}\right) \\
\frac{2}{100} & =\sin \left(\frac{v}{50}\right) \\
\frac{1}{50} & =\frac{v}{50}\left(\because \sin \left(\frac{v}{50}\right) \approx \frac{v}{50}\right) \\
\nu & =1 \mathrm{~Hz}
\end{aligned}
\)
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