
The average half-cycle value of a sine wave with a 40V peak is
A. 25.48V
B.6.37V
C.14.14V
D.50.96V
Answer
553.2k+ views
Hint: As a first step you could recall the definition of the quantity that is, the average half cycle value of a sine wave that we are supposed to find here. Then you could remember a relation for the average value in terms of the peak voltage for which we have the value given in the question. Now by doing simple substitution, we will get the answer.
Formula used:
Expression for average value,
${{V}_{avg}}=\dfrac{2{{V}_{P}}}{\pi }$
Complete answer:
In the question, we are asked to find the average half cycle value of a sine wave that has a peak of 40V.
In order to find the answer for this, we have to first understand what exactly the average voltage of a periodic wave is. For any wave be it sine wave, square wave or triangular wave, when we represent the waveform with respect to time, the quotient of the area under it will give the average voltage of any periodic waveform. From the definition you may realize that for a full cycle, the average or the mean value of alternating voltage will be zero as this will be the sum of the area under the positive and negative half cycles that cancels each other. You may now recall that the expression relating the average voltage with the peak voltage is given by the expression,
${{V}_{avg}}=\dfrac{2{{V}_{P}}}{\pi }$
Where, ‘${{V}_{P}}$’ is the peak voltage.
Now, we could substitute the given value of peak voltage in the above expression to get the required average half-cycle value of a sine wave.
${{V}_{avg}}=\dfrac{2\times 40}{\pi }$
$\Rightarrow {{V}_{avg}}=\dfrac{80}{\pi }$
$\therefore {{V}_{avg}}=25.48V$
Therefore, we found the average half cycle value of the given sine wave to be 25.48V.
Hence, option A is found to be the correct answer.
Note:
The average value of an alternating quantity which is often referred as the mean value is not a function of either the frequency or the phase angle. This value only depends on the magnitude of the waveform. Also, the above expression where we found the average value from the peak value just by multiplying with a constant value applies only for sinusoidal waves.
Formula used:
Expression for average value,
${{V}_{avg}}=\dfrac{2{{V}_{P}}}{\pi }$
Complete answer:
In the question, we are asked to find the average half cycle value of a sine wave that has a peak of 40V.
In order to find the answer for this, we have to first understand what exactly the average voltage of a periodic wave is. For any wave be it sine wave, square wave or triangular wave, when we represent the waveform with respect to time, the quotient of the area under it will give the average voltage of any periodic waveform. From the definition you may realize that for a full cycle, the average or the mean value of alternating voltage will be zero as this will be the sum of the area under the positive and negative half cycles that cancels each other. You may now recall that the expression relating the average voltage with the peak voltage is given by the expression,
${{V}_{avg}}=\dfrac{2{{V}_{P}}}{\pi }$
Where, ‘${{V}_{P}}$’ is the peak voltage.
Now, we could substitute the given value of peak voltage in the above expression to get the required average half-cycle value of a sine wave.
${{V}_{avg}}=\dfrac{2\times 40}{\pi }$
$\Rightarrow {{V}_{avg}}=\dfrac{80}{\pi }$
$\therefore {{V}_{avg}}=25.48V$
Therefore, we found the average half cycle value of the given sine wave to be 25.48V.
Hence, option A is found to be the correct answer.
Note:
The average value of an alternating quantity which is often referred as the mean value is not a function of either the frequency or the phase angle. This value only depends on the magnitude of the waveform. Also, the above expression where we found the average value from the peak value just by multiplying with a constant value applies only for sinusoidal waves.
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