Correct Answer - Option 3 : 100 Hz
Concept:
Let the voltage and current equation be
\({\rm{v}} = {{\rm{V}}_m}\sin \omega t\) and \({\rm{i}} = {{\rm{I}}_m}\sin \left( {\omega t - \varphi } \right)\)
Instantaneous power at any instant is
\({\rm{p}} = {\rm{v}} \times {\rm{i}}\)
\({\rm{p}} = {\rm{\;}}{{\rm{V}}_m}\sin \omega t \times \;{{\rm{I}}_m}\sin \left( {\omega t - \varphi } \right)\)
\({\rm{p}} = {\rm{\;}}{{\rm{V}}_m}{\rm{\;}}{{\rm{I}}_m}[\sin \omega t \times \;\sin \left( {\omega t - \varphi } \right)]\)
Using trigonometric identities, to solve
\(\sin a\; \times \sin b = \;\frac{1}{2}\;\left[ {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right]\)
\({\rm{p}} = \frac{{{{\rm{V}}_m}{\rm{\;}}{{\rm{I}}_m}}}{2}\left[ {\cos \varphi - \cos \left( {2\omega t - \varphi } \right)} \right]\)
\({\rm{p}} = {\rm{VIcos\varphi }} - {\rm{VI}}\cos \left( {2\omega t - \varphi } \right)\)
Where V and I are the root-mean-squared (RMS) values of the sinusoidal waveforms of voltage and current respectively.
- The above equation shows that the instantaneous AC power is the sum of two different terms.
- The first term is the average value of the instantaneous power.
- The second term is a time-varying sinusoid whose frequency is equal to twice the angular frequency of the supply due to the 2ω part of the term.
Therefore, Instantaneous power is double the frequency of applied voltage.
Calculation:
Since the frequency of the AC main supply power is 50 Hz
Therefore, the frequency of instantaneous power consumed by the boiler is = 2 × 50 = 100 Hz
