Instantaneous and Average Power Formula

ElectronElectronics & ElectricalDigital Electronics

Single Phase System

Instantaneous Power

The instantaneous power in an AC circuit is defined as the product of instantaneous voltage (v) across the element and instantaneous current (i) through the element and is denoted by lower case letter p.

Instantaneous Power, $\mathrm{p=v\times\:i}$

Since, the values of instantaneous voltage and instantaneous current changes from instant to instant, thus the instantaneous power changes with time. The instantaneous power (p) is measured in watts. The instantaneous power may be positive or negative. A positive instantaneous power means power flows from source to load whereas negative instantaneous power means power flows from load to source.

Instantaneous Power Formula

Case 1 – Pure Resistive Circuit

In case of pure resistive circuit, the instantaneous value current and voltage being,

$$\mathrm{v=V_{m}\sin\omega\:t}$$

$$\mathrm{i=I_{m}\sin\omega\:t}$$

Therefore, the instantaneous power will be,

$$\mathrm{p=v\times\:i=(v_{m}\sin\omega\:t)\times\:(I_{m}\sin\omega\:t)}$$

$$\mathrm{\Rightarrow\:p=\frac{V_{m}I_{m}}{2}-\frac{V_{m}I_{m}}{2}\cos2\omega\:t\:\:\:...(1)}$$

Case 2 – Pure Inductive Circuit

For pure inductive circuit, the equation of instantaneous voltage and current is given by,

$$\mathrm{v=V_{m}\sin\omega\:t}$$

$$\mathrm{i=I_{m}\sin(\omega\:t-90)}$$

Hence, the instantaneous power will be,

$$\mathrm{p=v\times\:i=(V_{m}\sin\omega\:t)\times\:(I_{m}\sin(\omega\:t-90))}$$

$$\mathrm{\Rightarrow\:p=-\frac{V_{m}I_{m}}{2}\sin2\omega\:t\:\:\:...(2)}$$

Case 3 – Pure Capacitive Circuit

The equation of instantaneous voltage and current for a pure capacitive circuit is given by,

$$\mathrm{v=V_{m}\sin\omega\:t}$$

$$\mathrm{i=I_{m}\sin(\omega\:t+90)}$$

Therefore, the instantaneous power will be,

$$\mathrm{p=v\times\:i=(V_{m}\sin\omega\:t)\times\:(I_{m}\sin(\omega\:t+90))}$$

$$\mathrm{\Rightarrow\:p=\frac{V_{m}I_{m}}{2}\sin2\omega\:t\:\:\:...(3)}$$

From the equations (1), (2) and (3) it can be noted that the instantaneous power in a single phase system varies from zero to maximum values at twice the supply frequency and also it may be positive or negative.

Average Power

The average power is defined as the average of instantaneous power over one cycle and is denoted by upper case letter P. It is also measured in watts.

Avergae Power, p = Avg. of p over one cycle

$$\mathrm{p=\frac{1}{2\pi}\int_{0}^{2\pi}p\:d\omega\:t\:\:\:\:...(4)}$$

Average Power Formula

Case 1 – Pure Resistive Circuit

$$\mathrm{\mathrm{p=\frac{1}{2\pi}\int_{0}^{2\pi}p\:d\omega\:t=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{V_{m}I_{m}}{2}-\frac{V_{m}I_{m}}{2}\cos2\omega\:t\:d\omega\:t}}$$

$$\mathrm{\Rightarrow\:p=\frac{V_{m}I_{m}}{2}=\frac{V_{m}}{\sqrt{2}}\times\:\frac{I_{m}}{\sqrt{2}}=VI\:\:\:...(5)}$$

Where, V and I are the RMS values of voltage and current respectively.

Case 2 – Pure Inductive Circuit

$$\mathrm{p=\frac{1}{2\pi}\int_{0}^{2\pi}p\:d\omega\:t=\frac{1}{2\pi}\int_{0}^{2\pi}-\frac{V_{m}I_{m}}{2}\cos2\omega\:t\:d\omega\:t=0\:\:\:...(6)}$$

Hence, average power absorbed by a pure inductor is zero.

Case 3 – Pure Capacitive Circuit

$$\mathrm{p=\frac{1}{2\pi}\int_{0}^{2\pi}p\:d\omega\:t=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{V_{m}I_{m}}{2}\cos2\omega\:t\:d\omega\:t=0\:\:\:...(7)}$$

Hence, average power absorbed by a pure capacitor is also zero.

Three Phase System

Instantaneous Power

As we know, the single phase instantaneous power (for lagging power factor load) is given by,

$$\mathrm{p=\frac{V_{m}I_{m}}{2}\cos\varphi-\frac{V_{m}I_{m}}{2}\cos(2\omega\:t-\varphi)}$$

In terms of RMS values, it becomes,

$$\mathrm{p=VI\cos\varphi-VI\cos(2\omega\:t-\varphi)}$$

Now, the instantaneous power in three phases (RYB –phase sequence) can be written as,

$$\mathrm{p_{R}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi)}$$

$$\mathrm{p_{Y}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi-120^{\circ})}$$

$$\mathrm{p_{H}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi+120^{\circ})}$$

Therefore, total instantaneous power in three phase system is given by,

$$\mathrm{p=P_{R}+P_{Y}+P_{H}}$$

$$\mathrm{p=3V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\begin{bmatrix}\cos(2\omega\:t-\varphi)\\+\cos(2\omega\:t-\varphi-120^{\circ})\\ +\cos(2\omega\:t-\varphi+120^{\circ})\end{bmatrix}}$$

$$\mathrm{\because\begin{bmatrix}\cos(2\omega\:t-\varphi)\\+\cos(2\omega\:t-\varphi-120^{\circ})\\ +\cos(2\omega\:t-\varphi+120^{\circ})\end{bmatrix}=0}$$

Thus,

$$\mathrm{p=3V_{ph}I_{ph}\cos\varphi\:\:\:...(8)}$$

$$\mathrm{\Rightarrow\:p=\sqrt{3}V_{L}I_{L}\cos\varphi\:\:\:...(9)}$$

The eqns. (8) & (9) shows that the 3 phase instantaneous power is constant and is not the function of supply frequency.

Average Power

By the definition of average power, we obtain,

$$\mathrm{P=\frac{1}{2\pi}\int_{0}^{2\pi}3V_{ph}I_{ph}\cos\varphi\:d\omega=3V_{ph}I_{ph}\cos\varphi}$$

$$\mathrm{\Rightarrow\:P=3V_{ph}I_{ph}\cos\varphi=\sqrt{3}V_{L}I_{L}\cos\varphi\:\:\:...(10)}$$

Therefore, in case of a three system, the average power and instantaneous power are same.

raja
Published on 05-Jul-2021 07:33:51
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