Answer
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Hint: Unlike the DC current which moves in the same direction an AC electric current reverses its direction in a cyclic manner. The value of current starts from zero and reaches a peak positive value, then it reverses back to the zero value and reaches a negative value and the cycle continues. Generally, an AC waveform is represented by a sine wave.
Complete step by step solution:
The standard equation of voltage for an AC electric current represented by a sinusoidal waveform is given by,
$V = {V_o}\sin \omega t$ …………. (1)
Where V is the value of the voltage at any instant ‘t’, Vo is the value of the peak voltage and is the frequency of the alternating current.
Now, the given relation between an AC voltage source and time is given by,
$V = 120\sin (100\pi t)\cos (100\pi t)V$ ............ (2)
We know that $\operatorname{Sin} 2\theta = 2\operatorname{Sin} \theta \operatorname{Cos} \theta $. Using this trigonometric identity we can rewrite equation (2) as,
$V = 60\sin (200\pi t)$ …………. (3)
Comparing, equation (1) & (3) we get,
The value of peak voltage, Vo = 60 v
And the value of frequency of the alternating current, $\omega = 200\pi $ …….. (4)
Also, we know that frequency in hertz is given by, $\omega = 2\pi f$ ……….. (5)
Hence, by equating equation (4) & equation (5) we have, $2\pi f = 200\pi $
I.e. f = 100 Hz
As the value of peak voltage is Vo= 60 volt; and the value of alternating current frequency is f = 100 Hz. we can say that option D is the correct answer.
Note: Most of the electromechanical alternators and physical phenomena produce sine waves, but there are also other waveforms that are of alternating nature. In electronic circuits, other forms of alternating current waveforms are also formed. The other form of alternating non-sinusoidal waveforms are square wave, triangle wave and smooth wave.
Complete step by step solution:
The standard equation of voltage for an AC electric current represented by a sinusoidal waveform is given by,
$V = {V_o}\sin \omega t$ …………. (1)
Where V is the value of the voltage at any instant ‘t’, Vo is the value of the peak voltage and is the frequency of the alternating current.
Now, the given relation between an AC voltage source and time is given by,
$V = 120\sin (100\pi t)\cos (100\pi t)V$ ............ (2)
We know that $\operatorname{Sin} 2\theta = 2\operatorname{Sin} \theta \operatorname{Cos} \theta $. Using this trigonometric identity we can rewrite equation (2) as,
$V = 60\sin (200\pi t)$ …………. (3)
Comparing, equation (1) & (3) we get,
The value of peak voltage, Vo = 60 v
And the value of frequency of the alternating current, $\omega = 200\pi $ …….. (4)
Also, we know that frequency in hertz is given by, $\omega = 2\pi f$ ……….. (5)
Hence, by equating equation (4) & equation (5) we have, $2\pi f = 200\pi $
I.e. f = 100 Hz
As the value of peak voltage is Vo= 60 volt; and the value of alternating current frequency is f = 100 Hz. we can say that option D is the correct answer.
Note: Most of the electromechanical alternators and physical phenomena produce sine waves, but there are also other waveforms that are of alternating nature. In electronic circuits, other forms of alternating current waveforms are also formed. The other form of alternating non-sinusoidal waveforms are square wave, triangle wave and smooth wave.
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