MHT CET · Physics · Alternating Current
An alternating e.m.f. is \(\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}\). In what time the e.m.f. will have half its maximum value, if ' \(\mathrm{e}\) ' starts from zero? ( \(\mathrm{T}=\) time period)
- A \(\frac{\mathrm{T}}{12}\)
- B \(\frac{T}{16}\)
- C \(\frac{T}{4}\)
- D \(\frac{T}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{T}}{12}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}\)
If \(\mathrm{e}=\frac{\mathrm{e}_0}{2}\) then \(\frac{\mathrm{e}_0}{2}=\mathrm{e}_0 \sin \omega \mathrm{t}\)
\(\begin{aligned} & \sin \omega \mathrm{t}=\frac{1}{2} \\ & \omega \mathrm{t}=30^{\circ}=\frac{\pi}{6} \mathrm{rad} \\ & \frac{2 \pi}{\mathrm{T}} \cdot \mathrm{t}=\frac{\pi}{6} \\ & \mathrm{t}=\frac{\mathrm{T}}{12}\end{aligned}\)
If \(\mathrm{e}=\frac{\mathrm{e}_0}{2}\) then \(\frac{\mathrm{e}_0}{2}=\mathrm{e}_0 \sin \omega \mathrm{t}\)
\(\begin{aligned} & \sin \omega \mathrm{t}=\frac{1}{2} \\ & \omega \mathrm{t}=30^{\circ}=\frac{\pi}{6} \mathrm{rad} \\ & \frac{2 \pi}{\mathrm{T}} \cdot \mathrm{t}=\frac{\pi}{6} \\ & \mathrm{t}=\frac{\mathrm{T}}{12}\end{aligned}\)
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