# Power in AC Circuit

Here we investigate on the cases of power delivered to the components of an ac circuit. You could have guessed that the energy stored in the capacitor and inductor are later retrieved, there is no power delivered on these components.

## Power in Resistor

The power in resistor is as we all know $p_\text{av} = I_{\text{rms}}^2R$. In case of a sinusoidally changing current, we use the rms value equivalent to dc current for power calculations. The power input in the resistance is spent in heating the material or by radiation, and therefore, we usually, think it as the energy is wasted. So, some sources call it *energy is dissipated*, but we don't do so. We say that the energy is transferred or delivered to some other energy, as the word "dissipated" might suggest the disappearance of energy violating the law of conservation of energy.

## Power in Inductor

We all know that the energy stored in an inductor is retrieved, that is no energy is lost in an inductor. So, the energy stored in the inductor due to the increase in current in one half cycle is retrieved in another half cycle and delivered back to the source, therefore there is no energy loss and the average power over a full cycle is zero. If $v$ and $i$ are the instantaneous values of voltage and current respectively, the product $v\,i$ is the instantaneous value of power for the inductor. Note that $v$ and $i$ are $\pi/2 \,\text{rad}$ out of phase, and therefore the power through the inductor is positive during one half cycle and negative during another half cycle and the average power is zero.

## Power in Capacitor

The charge builds up in a capacitor in a half cycle and goes back to the source in another half cycle. Therefore, there is no power input to the capacitor. Similar to the inductor's case, the energy stored in the electric field is transferred back to the source. Similar to the inductor's case, the average power over a full cycle is zero.

## Power in General AC Circuit

There is not much interest in how much power is delivered to individual components. What's more interesting is how much power is delivered to an entire circuit with inductor, capacitor and resistor. So, here, we find the average power in an ac circuit. Let the sinusoidal current is $i = I\sin(\omega\,t)$, and the voltage across the circuit is $v = V\sin(\omega\,t + \phi)$ which is equal to the terminal voltage of the source. The voltage is at some phase $\phi$ with the current. You could also represent in terms of cosine function instead of sine function. The power delivered to the circuit is

\[p = v\,i = IV\sin\omega\,t\sin(\omega\,t + \phi) \]

We know the trigonometric identity $\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)$. So, you can rewrite the above equation as

\[p = IV\sin^2\omega \,t \cos \phi + IV\sin \omega\,t\cos\omega\,t\sin\phi\]

The average of $\sin^2\omega\,t$ is $1/2$ and the average of $\sin\omega t$ or $\cos\omega t$ is zero, therefore, the average power is

\[p_\text{av} = \frac{1}{2}IV\cos\phi = I_\text{rms}V_\text{rms}\cos\phi\]

The value $\cos \phi$ is called power factor. Note that there is no power delivered to the inductor and capacitor, and the value $V_\text{rms}\cos\phi$ is the component of rms voltage in phase with the current (you can see this RLC series circuit). Therefore all power goes into the resistance. If there is a pure resistance, $\phi = 0$ and $\cos\phi = 1$ and $p_\text{av} = I_\text{rms}V_\text{rms}$. For the pure inductor and capacitor $\phi = \pm \pi/2 \,\text{rad}$, therefore $p_\text{av} = 0$.

We don't want a low value of power factor. The low value of power factor means large $\phi$. The above expression shows that if the power factor is low, large current is required to maintain the same power, that is we don't want large phase difference between current and voltage (large value of $\phi$). When $\phi$ is zero, the power factor is $1$, and it is the same as there is no inductor and capacitor.

If we have the lower value of power factor (large $\phi$), the charge flows in the transmission lines simply to store and empty the energy in inductor or capacitor, and that leads to power loss, and further large current is required to maintain the same power, which no one wants to happen. If the circuit has motors with large inductances we want the power factor close to $1$ as much as possible. The easy way to solve this is by connecting an appropriate capacitor in parallel with the inductive load.