
The relationship between A. C. voltage and time in SI unit is $V = 120\sin (100\pi t)\cos (100\pi t)$ Volt value of peak voltage and frequency will be:
(A) \[120{\text{ }}volt{\text{ }} and {\text{ }}100Hz\]
(B) $\dfrac{{120}}{{\sqrt 2 }}Volt{\text{ }} and {\text{ }}100Hz$
(C) \[60{\text{ }}volt{\text{ }} and {\text{ }}200Hz\]
(D) \[60{\text{ }}volt{\text{ }} and 100Hz\]
Answer
590.7k+ views
Hint: To solve this type of question we use the following formula and concept.
Voltage generally represented as $V = {V_0}\sin \omega t$
$\omega = 2\pi \nu $ Where\[\;\omega \] is angular frequency, $V$ is general frequency.
Complete step-by-step answer:
Let us first write the information given in the question.
Relation between AC voltage and time is given below.
$V = 120\sin (100\pi t)\cos (100\pi t)$
As we know the general equation of voltage is $V = {V_0}\sin \omega t$ so let us convert the given voltage into its general representation so that we can compare.
$V = 120\sin (100\pi t)\cos (100\pi t)$
Let us use the formula $2\sin A\cos B = \sin (A + B) + \sin (A - B)$.
\[V = 60\left( {\sin \left( {\left( {100\pi t + 100\pi t} \right)} \right) + \sin (100\pi - 100\pi t} \right)\]
Let us further simplify it.
$V = 60\left( {\sin \left( {200\pi t} \right) + \sin (0)} \right)$
Hence, we get the following form of voltage.
$V = 60\sin (200\pi t)$ ………..(1)
Now, let us compare this with the general form of voltage which is following.
$V = {V_0}\sin \omega t$ ………….(2)
Let us compare equation (1) and (2) and we get the following.
${V_0} = 60,\omega = 200\pi $ …………..(3)
Now let us find the linear frequency using the following formula.
$\omega = 2\pi \nu $
Let us now substitute the values.
$200\pi = 2\pi \nu $
Let us simplify it.
$200 = 2\nu $
$ \Rightarrow \nu = 100{s^{ - 1}} = 100Hz$
Therefore, required frequency and peak value of voltage is \[100Hz\] and $60$ respectively.
Hence, option (D) \[60volt\] and \[100Hz\] is the correct option.
Note: Peak voltage is the maximum possible voltage for any voltage waveform.
Peak value can be calculated by the r.m.s value using the following formula.
${V_0} = \sqrt 2 {V_{r.m.s.}}$. Root means square (r.m.s.) values are calculated over one cycle.
All the AC voltages observed in daily life are r.m.s. values.
Voltage generally represented as $V = {V_0}\sin \omega t$
$\omega = 2\pi \nu $ Where\[\;\omega \] is angular frequency, $V$ is general frequency.
Complete step-by-step answer:
Let us first write the information given in the question.
Relation between AC voltage and time is given below.
$V = 120\sin (100\pi t)\cos (100\pi t)$
As we know the general equation of voltage is $V = {V_0}\sin \omega t$ so let us convert the given voltage into its general representation so that we can compare.
$V = 120\sin (100\pi t)\cos (100\pi t)$
Let us use the formula $2\sin A\cos B = \sin (A + B) + \sin (A - B)$.
\[V = 60\left( {\sin \left( {\left( {100\pi t + 100\pi t} \right)} \right) + \sin (100\pi - 100\pi t} \right)\]
Let us further simplify it.
$V = 60\left( {\sin \left( {200\pi t} \right) + \sin (0)} \right)$
Hence, we get the following form of voltage.
$V = 60\sin (200\pi t)$ ………..(1)
Now, let us compare this with the general form of voltage which is following.
$V = {V_0}\sin \omega t$ ………….(2)
Let us compare equation (1) and (2) and we get the following.
${V_0} = 60,\omega = 200\pi $ …………..(3)
Now let us find the linear frequency using the following formula.
$\omega = 2\pi \nu $
Let us now substitute the values.
$200\pi = 2\pi \nu $
Let us simplify it.
$200 = 2\nu $
$ \Rightarrow \nu = 100{s^{ - 1}} = 100Hz$
Therefore, required frequency and peak value of voltage is \[100Hz\] and $60$ respectively.
Hence, option (D) \[60volt\] and \[100Hz\] is the correct option.
Note: Peak voltage is the maximum possible voltage for any voltage waveform.
Peak value can be calculated by the r.m.s value using the following formula.
${V_0} = \sqrt 2 {V_{r.m.s.}}$. Root means square (r.m.s.) values are calculated over one cycle.
All the AC voltages observed in daily life are r.m.s. values.
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