Question

Prove that for A.C. current: ${P_{av}} = {V_{r.m.s.}} \times {I_{r.m.s}}\cos \phi $.

Answer 1

Hint:To prove the formula for average power, we will use the concept of instantaneous power. The instantaneous power absorbed by an element is the product of the instantaneous voltage across the element and the instantaneous current through it. The average power is the average of the instantaneous power over one period.

Complete step by step answer:
We know that instantaneous voltage and instantaneous current in ac circuit are given by the following formulas:
\[v = {V_m}\sin \omega t\]
where, \[v\]is the instantaneous voltage, \[{V_m}\]is the peak voltage, \[\omega \]is the frequency and \[t\] is the time.
\[i = {I_m}\sin \left( {\omega t + \phi } \right)\], where, \[i\]is the instantaneous current, \[{I_m}\] is the peak current, \[\omega \]is the frequency ,\[t\] is the time and \[\phi \] is the phase difference between voltage and current.
Now, as per the definition of instantaneous power, it is given by the formula
\[
p = vi \\
\Rightarrow p = \left( {{V_m}\sin \omega t} \right)\left( {{\operatorname{I} _m}\left( {\sin \omega t + \phi } \right)} \right) \\
\Rightarrow p = {V_m}{\operatorname{I} _m}\sin \omega t \cdot \sin \left( {\omega t + \phi } \right) \\ \]
Now, we will use the trigonometric property:
\[\sin A\sin B = \dfrac{1}{2}\cos (A - B) - \dfrac{1}{2}\cos (A + B)\]
\[
\Rightarrow p = \dfrac{{{V_m}{\operatorname{I} _m}}}{2}\left( {\cos (\omega t - \omega t - \phi ) - \cos (\omega t + \omega t + \phi )} \right) \\
\Rightarrow p = \dfrac{{{V_m}{\operatorname{I} _m}}}{2}\left( {\cos ( - \phi ) - \cos (2\omega t + \phi )} \right) \\
 \]
We know that $\cos \left( { - \phi } \right) = \cos \phi $
\[
p = \dfrac{{{V_m}{\operatorname{I} _m}}}{2}\left( {\cos \phi - \cos (2\omega t + \phi )} \right) \\
\Rightarrow p = \dfrac{{{V_m}{\operatorname{I} _m}}}{2}\cos \phi
\dfrac{{{V_m}{\operatorname{I} _m}}}{2}\cos (2\omega t + \phi ) \\ \]
Here, the first term has no element of time, so it gives a fixed value for a specific value of phase angle. This is called average power dissipated in the load. This average power is mostly called real or active power. So the equation for the average power becomes
\[
{P_{av}} = \dfrac{{{V_m}{I_m}}}{2}cos\phi \\
\Rightarrow {P_{av}} = \dfrac{{{V_m}}}{{\sqrt 2 }}\dfrac{{{I_m}}}{{\sqrt 2 }}cos\phi \\ \]
We know the relation between the rms and peak values of voltage and current is:
\[{V_{rms = }}\dfrac{{{V_m}}}{{\sqrt 2 }}\] and \[{I_{rms = }}\dfrac{{{I_m}}}{{\sqrt 2 }}\]
\[ \therefore{P_{av}} = {V_{rms}} \times {I_{rms}}cos\phi \]

Hence, it is proved that for A.C. current: ${P_{av}} = {V_{r.m.s.}} \times {I_{r.m.s}}\cos \phi $.

Note:We have seen that the instantaneous power is made up of two cosine terms. The second term frequency is twice as that of current or voltage. This term has a component of time in it, which causes the cosine wave to oscillate and its average value over a cycle is zero. That is why no energy is transferred due to the second term to the generic load attached to the ac source. Energy only floats back and forth between load and source. Therefore, we have considered only the first component of the equation for calculating average power.