Question

What is the frequency of power in the a.c circuit when connected to a capacitor? Given \[f = 50{\text{ Hz}}\] and \[V = 220{\text{ V}}\].
A. \[{\text{ 50 Hz}}\]
B. \[{\text{ 100 Hz}}\]
C. \[{\text{ 60 Hz}}\]
D. Zero \[{\text{Hz}}\]

Answer 1

Hint: Total Power consumed in the circuit is the product of voltage of circuit and the current flowing in the circuit. In A.c circuit we can find the instantaneous power of circuit by multiplying the instantaneous voltage and current. Also the alternating current or alternating voltage has some frequency.

Formula used:
\[P = VI\]
Where, \[P = \] Total power consumed, \[V = \] Voltage and \[I = \] Current.

Complete step by step answer:
In an A.C circuit the instantaneous power of the circuit is calculated by multiplying instantaneous voltage and instantaneous current. This can be represented as:
\[P = VI\]

Now we can calculated instantaneous voltage as:
\[V = {V_ \circ }\sin (wt)\] _______________\[(1)\]
Similarly instantaneous value of current can be calculated as:
\[I = {I_ \circ }\sin (wt)\]
But in a capacitor the current leads to its original value by a phase angle of \[\dfrac{\pi }{2}\] or \[{90^ \circ }\] . Thus it can be represented as:
\[I = {I_ \circ }\sin \left( {wt{\text{ + }}\dfrac{\pi }{2}} \right)\] _____________\[(2)\]

By using equation \[(1)\] and \[(2)\] , we can write the power as,
\[P = VI\]
\[\Rightarrow P = {V_ \circ }\sin (wt){\text{ }} \times {\text{ }}{I_ \circ }\sin \left( {wt{\text{ + }}\dfrac{\pi }{2}} \right)\]
We can write, \[\sin \left( {wt{\text{ + }}\dfrac{\pi }{2}} \right){\text{ = cos(}}wt{\text{)}}\] . Therefore equation can be deduced as:
\[P = {V_ \circ }\sin (wt){\text{ }} \times {\text{ }}{I_ \circ }\cos (wt)\]
On rearranging the terms we get the result as:
\[P = {V_ \circ }{I_ \circ }\sin (wt){\text{ }} \times \cos (wt)\]

On multiplying and dividing by two we get the result as:
\[P = \dfrac{{{V_ \circ }{I_ \circ }}}{2}{\text{ }} \times {\text{ 2}}\sin (wt) \times \cos (wt)\]
We know that, \[{\text{2}}\sin (wt) \times \cos (wt){\text{ = }}\sin \left( {2wt} \right)\]
\[P = \dfrac{{{V_ \circ }{I_ \circ }}}{2}{\text{ }} \times {\text{ sin(2}}wt{\text{)}}\]
Therefore we can observe that magnitude of power will be equal to \[\dfrac{{{V_ \circ }{I_ \circ }}}{2}\] and its frequency will be equal to \[2w\] , which is double of original frequency. According to the question the frequency is \[f = 50{\text{ Hz}}\]. Therefore the frequency of power will be \[2 \times 50 = 100{\text{ Hz}}\].

Hence the correct option is B.

Note: It must be noted that \[w = 2\pi f\]. Power consumption gets reduced to \[\dfrac{{{V_ \circ }{I_ \circ }}}{2}\] when the capacitor is used in a.c circuit and frequency becomes double of original frequency. In an inductive circuit the current lags by \[\dfrac{\pi }{2}\]. Since the current leads in the capacitive circuit this is why it is used to store charge.