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Question: The relationship between A. C. voltage and time in SI unit is \(V = 120\sin (100\pi t)\cos (100\pi t...

The relationship between A. C. voltage and time in SI unit is V=120sin(100πt)cos(100πt)V = 120\sin (100\pi t)\cos (100\pi t) Volt value of peak voltage and frequency will be:
(A) 120 volt and 100Hz120{\text{ }}volt{\text{ }} and {\text{ }}100Hz
(B) 1202Volt and 100Hz\dfrac{{120}}{{\sqrt 2 }}Volt{\text{ }} and {\text{ }}100Hz
(C) 60 volt and 200Hz60{\text{ }}volt{\text{ }} and {\text{ }}200Hz
(D) 60 volt and100Hz60{\text{ }}volt{\text{ }} and 100Hz

Explanation

Solution

To solve this type of question we use the following formula and concept.
Voltage generally represented as V=V0sinωtV = {V_0}\sin \omega t
ω=2πν\omega = 2\pi \nu Where  ω\;\omega is angular frequency, VV 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=120sin(100πt)cos(100πt)V = 120\sin (100\pi t)\cos (100\pi t)
As we know the general equation of voltage is V=V0sinωtV = {V_0}\sin \omega t so let us convert the given voltage into its general representation so that we can compare.
V=120sin(100πt)cos(100πt)V = 120\sin (100\pi t)\cos (100\pi t)
Let us use the formula 2sinAcosB=sin(A+B)+sin(AB)2\sin A\cos B = \sin (A + B) + \sin (A - B).
V=60(sin((100πt+100πt))+sin(100π100πt)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(sin(200πt)+sin(0))V = 60\left( {\sin \left( {200\pi t} \right) + \sin (0)} \right)
Hence, we get the following form of voltage.
V=60sin(200πt)V = 60\sin (200\pi t) ………..(1)
Now, let us compare this with the general form of voltage which is following.
V=V0sinωtV = {V_0}\sin \omega t ………….(2)
Let us compare equation (1) and (2) and we get the following.
V0=60,ω=200π{V_0} = 60,\omega = 200\pi …………..(3)
Now let us find the linear frequency using the following formula.
ω=2πν\omega = 2\pi \nu
Let us now substitute the values.
200π=2πν200\pi = 2\pi \nu
Let us simplify it.
200=2ν200 = 2\nu
ν=100s1=100Hz\Rightarrow \nu = 100{s^{ - 1}} = 100Hz
Therefore, required frequency and peak value of voltage is 100Hz100Hz and 6060 respectively.
Hence, option (D) 60volt60volt and 100Hz100Hz 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.
V0=2Vr.m.s.{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.