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An alternating electromotive force is given by $e = 100\sin 314t$, the value of the time within which the electromotive force will be half of its maximum value is
(A) $0.16\;\sec $
(B) $0.16\;\sec .$
(C) $0.0016\;\sec .$
(D) $1.6\;\sec .$

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Last updated date: 13th Jun 2024
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Answer
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Hint Here, compare the given expression for e.m.f. with the standard formula for e.m.f $e = {e_0}\sin \omega t$ to calculate the peak value of electromotive force. Once peak value is obtained, simplify as per the given condition in the question.
Formula Used: Here we will be using the formula $e = {e_0}\sin \omega t$, where ${e_0}$ will be peak value of electromotive force, $t$ is the time and $\omega $ is the frequency.

Complete step by step solution
Given expression for an alternating electromotive force is $e = 100\sin 314t$.
As we know that $e = {e_0}\sin \omega t$ is the standard formula for e.m.f.
Now, equate the given formula to get the peak value of electromotive force.
${e_0}\sin \omega t = 100\sin 314t$
From the above, the maximum value of electromotive force is $100$.
As the electromotive force reduced to half of its maximum value, then
$ \Rightarrow \dfrac{1}{2} \times 100 = 100\sin 314t$
Now, write the left hand side of expression in terms of sine trigonometric function and cancel out the same terms at both sides to reduce the equation.
As we know that $\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$, then
Now, compare the two values and write the expression at the right hand side in terms of $\pi $.
$ \Rightarrow \dfrac{\pi }{6} = 100 \times \pi \times t$
Take the same variables at the same side of the equation.
Divide both sides of the equation by the coefficient of $t$.
$ \Rightarrow t = \dfrac{1}{{600}}\;{\text{sec}}{\text{.}} = 0.0016\;{\text{sec}}$

So, option (c) is correct answer

Additional informationThe electromotive force equals the potential difference between the terminals of the cell when the cell is in open circuit. It is abbreviated as emf or $e$. The maximum value of the electromotive force can be determined by its peak value. In an alternating current source, its peak value is defined as the maximum value of current in either direction of the cycle.

Note
The maximum value of any function is its peak value. Knowledge of Trigonometric values must be necessary for comparison to determine the value of time.