# Average Power of Sinusoidal Voltage   Whatsapp  For any load in a sinusoidal ac network, the voltage across the load and the current through the load will vary in a sinusoidal nature. The questions then arise, How does the power to the load determined by the product v-i vary, and what fixed value can be assigned to the power since it will vary with time? Fig. 1: Determining the power delivered in a sinusoidal ac network.
If we take the general case depicted in [Fig. 1] and use the following for v and i:
$$v = V_m \sin (wt + \theta_v)$$
$$i = I_m \sin (wt + \theta_i)$$
then the power is defined by
$$p = vi = V_m \sin (wt + \theta_v) I_m \sin (wt + \theta_i)$$
$$p = vi = V_m I_m\sin (wt + \theta_v) \sin (wt + \theta_i)$$
Using the trigonometric identity
$$\sin A \sin B = {\cos (A-B) - \cos(A+B) \over 2}$$
the function $\sin(wt + \theta_v)$ $\sin(wt + \theta_i)$ becomes
$$\sin(wt + \theta_v) \sin(wt + \theta_i) = {\cos (wt + \theta_v-wt - \theta_i) - \cos(wt + \theta_v + wt + \theta_i) \over 2}$$
$$= {\cos (\theta_v - \theta_i) - \cos (2wt + \theta_v + \theta_i) \over 2}$$
so that
$$p = {V_m I_m \over 2}\cos (\theta_v - \theta_i) - {V_m I_m \over 2}\cos (2wt + \theta_v + \theta_i)$$
A plot of v, i, and p on the same set of axes is shown in [Fig. 2]. Fig. 2: Defining the average power for a sinusoidal ac network.
the voltage or current. The average value of this term is zero over one cycle, producing no net transfer of energy in any one direction.
The first term in the preceding equation, however, has a constant magnitude (no time dependence) and therefore provides some net transfer of energy. This term is referred to as the average power, the reason for which is obvious from[ Fig. 2]. The average power, or real power as it is sometimes called, is the power delivered to and dissipated by the load. It corresponds to the power calculations performed for dc networks. The angle ($\theta_v - \theta_i$) is the phase angle between v and i. Since $\cos(-\alpha) = \cos \alpha$,
the magnitude of average power delivered is independent of whether v leads i or i leads v.
Defining v as equal to |$\theta_v - \theta_i$|, where | | indicates that only the magnitude is important and the sign is immaterial, we have
$$\bbox[10px,border:1px solid grey]{P = {V_m I_m \over 2} \cos \theta} \, (watts, W) \tag{1}$$
where P is the average power in watts. This equation can also be written
$$P = {V_m \over \sqrt{2}}{I_m \over \sqrt{2}} \cos \theta$$
or, since
$$V_{eff} = {V_m \over \sqrt{2}} \, \text{and} \,I_{eff}= {I_m \over \sqrt{2}}$$
$$\bbox[10px,border:1px solid grey]{P = V_{eff} I_{eff} \cos \theta }$$
Let us now apply Eqs. (1) and (2) to the basic R, L, and C elements.

### Resistor

In a purely resistive circuit, since v and i are in phase, |$\theta_v - \theta_i$| $= \theta = 0$, and $\cos 0 = 1$, so that
$$\bbox[10px,border:1px solid grey]{P = {V_m I_m \over 2} = V_{eff} I_{eff} } \tag{3}$$
Or, since
$$I_{eff} = {V_{eff}\over R}$$
then
$$\bbox[10px,border:1px solid grey]{P = {V_{eff}^2\over R} = {I_{eff}^2 R}}\, (W) \tag{4}$$

### Inductor

In a purely inductive circuit, since v leads i by $90^\circ$, |$\theta_v - \theta_i$| $= \theta = 90^\circ$. Therefore,
$$P = {V_m I_m \over 2} \cos 90 = {V_m I_m \over 2} (0) = 0 W$$
The average power or power dissipated by the ideal inductor (no associated resistance) is zero watts.

### Capacitor

In a purely capacitive circuit, since i leads v by $90^\circ$, |$\theta_v - \theta_i$| $= \theta = |-90^\circ|$. Therefore,
$$P = {V_m I_m \over 2} \cos 90 = {V_m I_m \over 2} (0) = 0 W$$
The average power or power dissipated by the ideal capacitor (no associated resistance) is zero watts.

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