
The relation between an A.C. voltage source and time in SI units is: $ V=120\sin (100\pi t)\cos (100\pi t) $ volt value of peak voltage and frequency will be respectively:
(A) 120 volt and 100 Hz
(B) $ \dfrac{120}{\sqrt{2}} $ volt and 100 Hz
(C) 60 volt and 200 Hz
(D) 60 volt and 100 Hz
Answer
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Hint
It is known that alternating current describes the flow of charge that changes direction periodically. As a result, the voltage level also reverses along with the current. AC is used to deliver power to houses, office buildings, and many other places. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. Based on this concept we have to solve this question.
Complete step by step answer
From the equation, $ \mathrm{V}=120 \sin (100 \pi \mathrm{t}) \cos (100 \pi \mathrm{t}) $
We can calculate,
$ \mathrm{V}=60 \sin (200 \pi \mathrm{t}) \mathrm{V} \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta) $
Comparing it with standard equation, we get,
$ \mathrm{V}=\mathrm{V}_{0} sin\omega t $
We get, $ \mathrm{V}_{0}=60 \mathrm{V} $
and $ \omega=200 \pi $ or $ 2 \pi v=200 \pi $ or $ v=100 \mathrm{Hz} $
Therefore, the correct answer is Option (D).
Note
We should know that Peak voltage is the highest point or highest value of voltage for any voltage waveform. It is a power quality issue that occurs when devices that use Pulse Width Modulation, such as a variable frequency drive, are added to a power system. Peak-to-peak voltage, VPP, is a voltage waveform which is measured from the top of the waveform, called the crest, all the way down to the bottom of the waveform, called the trough. So peak-to-peak voltage is just the full vertical length of a voltage waveform from the very top to the very bottom. The peak value is the highest voltage that the waveform will ever reach, like the peak is the highest point on a mountain. The RMS (Root-Mean-Square) value is the effective value of the total waveform. It is equal to the level of the DC signal that would provide the same average power as the periodic signal.
It is known that alternating current describes the flow of charge that changes direction periodically. As a result, the voltage level also reverses along with the current. AC is used to deliver power to houses, office buildings, and many other places. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. Based on this concept we have to solve this question.
Complete step by step answer
From the equation, $ \mathrm{V}=120 \sin (100 \pi \mathrm{t}) \cos (100 \pi \mathrm{t}) $
We can calculate,
$ \mathrm{V}=60 \sin (200 \pi \mathrm{t}) \mathrm{V} \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta) $
Comparing it with standard equation, we get,
$ \mathrm{V}=\mathrm{V}_{0} sin\omega t $
We get, $ \mathrm{V}_{0}=60 \mathrm{V} $
and $ \omega=200 \pi $ or $ 2 \pi v=200 \pi $ or $ v=100 \mathrm{Hz} $
Therefore, the correct answer is Option (D).
Note
We should know that Peak voltage is the highest point or highest value of voltage for any voltage waveform. It is a power quality issue that occurs when devices that use Pulse Width Modulation, such as a variable frequency drive, are added to a power system. Peak-to-peak voltage, VPP, is a voltage waveform which is measured from the top of the waveform, called the crest, all the way down to the bottom of the waveform, called the trough. So peak-to-peak voltage is just the full vertical length of a voltage waveform from the very top to the very bottom. The peak value is the highest voltage that the waveform will ever reach, like the peak is the highest point on a mountain. The RMS (Root-Mean-Square) value is the effective value of the total waveform. It is equal to the level of the DC signal that would provide the same average power as the periodic signal.
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