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Question: The peak value of an alternating emf E given by \(E = {E_0}\cos \omega t\) is \(10V\) and frequency ...

The peak value of an alternating emf E given by E=E0cosωtE = {E_0}\cos \omega t is 10V10V and frequency is 50Hz50Hz. At time t=(1600)st = \left( {\dfrac{1}{{600}}} \right)s, the instantaneous value of e.m.f is
(A)10V\left( A \right)10V
(B)53V\left( B \right)5\sqrt 3 V
(C)5V\left( C \right)5V
(D)1V\left( D \right)1V

Explanation

Solution

The instantaneous equation of an alternating emf is known to us so first convert the angular frequency into regular frequency as ω=2πf\omega = 2\pi f and from that regular frequency value is known to us. Now using that frequency we can now calculate angular frequency. Finally putting all the given values and calculated value in the instantaneous equation of emf to find its value.

Complete step by step answer:
As per the problem we have The peak value of an alternating emf E given by E=E0cosωtE = {E_0}\cos \omega t is 10V10V and frequency is 50Hz50Hz. At time t=(1600)st = \left( {\dfrac{1}{{600}}} \right)s
Now the instantaneous equation of the alternating emf is given to us as,
E=E0cosωt(1)E = {E_0}\cos \omega t \ldots \ldots \left( 1 \right)
Where,
The peak value of the alternating emf is E0=10V{E_0} = 10V.
Angular frequency of the alternating emf is ω\omega .
Where,
ω=2πf\omega = 2\pi f
Where f is the regular frequency that is equal to 50Hz50Hz.
Now putting frequency value in the above angular frequency will be,
ω=2π(50Hz)=100πHz\omega = 2\pi \left( {50Hz} \right) = 100\pi Hz
The time period of the alternating emf is t=(1600)st = \left( {\dfrac{1}{{600}}} \right)s.
Now putting all these value in equation (1)\left( 1 \right) we will get,
E=10Vcos(100π)(1600)=10Vcos(π6)E = 10V\cos \left( {100\pi } \right)\left( {\dfrac{1}{{600}}} \right) = 10V\cos \left( {\dfrac{\pi }{6}} \right)
We know, cos(π6)=32\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}
Hence the instantaneous emf will be,
E=1032V=53VE = \dfrac{{10\sqrt 3 }}{2}V = 5\sqrt 3 V
Therefore the correct option is (B).

Note: Remember that a peak voltage is the highest point or we can say highest value of voltage for any voltage waveform. Another important point to know is that when a closed coil is radiated in a uniform magnetic field with its axis perpendicular to the magnetic field then the magnetic flux linked with the coil changes and an induced emf and current start flowing. And this will give an equation that is the equation for the instantaneous value of an induced emf.