Question

If the frequency of alternating $emf$ is $50Hz$, then the time taken to attain maximum positive value from maximum negative value of $emf$ is equal to
$\begin{aligned} & A)\dfrac{1}{100}s \\\ & B)\dfrac{1}{200}s \\\ & C)100s \\\ & D)200s \\\ \end{aligned}$

Answer 1

For an alternating $emf$ of particular frequency, time taken to attain maximum positive value of $emf$ from maximum negative value of $emf$ is equal to half its time period. Time period of an alternating wave is equal to the reciprocal of frequency. It is nothing but the time required for the wave to complete one cycle.

Formula used: $1)T=\dfrac{1}{f}$
$2)t=\dfrac{T}{2}$

Complete step by step answer:
We know that the time period of an alternating wave of a particular frequency is the time required for the wave to complete one cycle. Time period of a wave is mathematically equal to the reciprocal of frequency of the wave. It is given by
$T=\dfrac{1}{f}$
where
$T$ is the time period of an alternating wave
$f$ is the frequency of the alternating wave
Let this be equation 1.
Now, time taken to attain maximum positive amplitude of an alternating wave from its maximum negative amplitude is equal to half the time period of an alternating wave. If $t$ denotes the same, then, $t$ is given by
$t=\dfrac{T}{2}$
where
$t$ is the time taken to attain maximum positive amplitude from maximum negative amplitude of an alternating wave
$T$ is the time period of the alternating wave
Let this be equation 2.
Coming to our question, we are provided with an alternating $emf$ of frequency $50Hz$. We are required to determine the time taken for the alternating $emf$ to attain maximum positive value of $emf$ from maximum negative value of $emf$.
Using equation 1, time period of alternating $emf$ is given by
$T=\dfrac{1}{f}=\dfrac{1}{50Hz}=0.02s$
Clearly, time taken by alternating $emf$ to complete one cycle is equal to $0.02s$.
Let this be equation 3.
Substituting equation 3 in equation 2, we have
$t=\dfrac{T}{2}=\dfrac{0.02s}{2}=0.01s=\dfrac{1}{100}s$
Here, $t$ is the time taken by alternating $emf$ to attain maximum positive value of $emf$ from maximum negative value of $emf$.
Clearly, time taken by the given alternating $emf$ of frequency $50Hz$ to attain maximum positive $emf$ from maximum negative $emf$ is equal to $\dfrac{1}{100}s$.

So, the correct answer is “Option A”.

Note: Students can visualise the sinusoidal waveform of alternating $emf$ to arrive at the answer. When we observe the waveform carefully, we can understand that time taken for the wave to attain its maximum positive value from its maximum negative value is half the time taken by the wave to complete one cycle $\left( {{t}_{\max (-)\to \max (+)}}=\dfrac{T}{2} \right)$. Similarly, one can deduce that the time taken for the wave to attain its maximum positive value or maximum negative value from zero is quarter the time taken by the wave to complete one cycle $\left( {{t}_{0\to \max (+/-)}}=\dfrac{T}{4} \right)$. Hence, one complete cycle of wave can be thought of as four sections, each of time $\dfrac{T}{4}$.