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Root mean square (rms) values

The root mean square (rms) of a quantity is a type of average.

It gives meaningful summary information about data that can take positive and negative values. It is frequently used for describing voltage and current generated by an a.c.

The rms current $$I_{\text{rms}}$$ is equal to the root of the mean of the squares of the values for the current over a full period (or a whole-number multiple of periods). An equivalent value can be calculated for voltage.

The rms current/voltage is, in general, not equivalent to the mean current/voltage. It is only equal to the mean when the current/voltage is constant and positive (i.e. in a direct current, making the rms value completely unnecessary).

The mean is not very meaningful for a function that turns negative, and so the root mean square is commonly used.
The mean is not very meaningful for a function that turns negative, and so the root mean square is commonly used.

The general equation for the rms of a function $$y$$ is: $$$y_{\text{rms}}=\sqrt{\langle y^{2} \rangle} = \left (\frac{1}{T}\int_0^T y^2(t)\,\mathrm{d}t\right )^{\frac{1}{2}}$$$

The symbols $$\langle y\rangle$$ denote the mean of a function.

The $$\int$$ sign denotes the integral of the function $$y$$. The integral gives the area between the curve and an axis. For the case of the rms, the integral in the formula gives the area between the curve of the $$y^{2}$$ function and the t-axis.

The rms values are used more often than the peak values to describe different alternating currents. Many different waveforms can have the same peak value but different rms values. However, the rms determines the "average" power delivered over time.

The rms current and voltage depends on the waveform. The rms current and voltage are equal to the direct current and voltage that dissipate the same average energy in a resistor as the alternating current.

This is why rms values are used in the power and voltage ratings of appliances.

For a sinusoidal a.c., the rms current $$I_{\text{rms}}$$ and the rms voltage $$V_{\text{rms}}$$ are given by: $$$\begin{align*}& I_{\text{rms}}=\frac{I_{0}}{\sqrt 2}& V_{\text{rms}}=\frac{V_{0}}{\sqrt 2}\end{align*}$$$ This is because the mean value of the sine squared function occurs at $$30^{\circ}$$.

The root mean square voltage of a sinusoidal a.c. is lower than the peak voltage.
The root mean square voltage of a sinusoidal a.c. is lower than the peak voltage.

For a square-wave a.c., $$$\begin{align*}& I_{\text{rms}}=I_{0} & V_{\text{rms}}=V_{0}\end{align*}$$$ This is because the squared value of a square-wave function is constant.

In a square-wave a.c, the rms voltage is equal to the peak voltage.
In a square-wave a.c, the rms voltage is equal to the peak voltage.

The mean power loss $$\langle P \rangle$$ is the average power dissipated in a resistor over a period. The mean power loss in a resistor of resistance $$R$$ is related to the rms current/voltage by:$$$\begin{align*}\langle P \rangle&=\langle I^{2}R\rangle=\langle\frac{V^{2}}{R}\rangle\\&=\langle I^{2}\rangle R=\frac{\langle V^{2}\rangle}{R}\\\langle P \rangle &=I_{\text{rms}}^{2}R=\frac{V_{\text{rms}}^{2}}{R}\end{align*}$$$ In other words, the mean power loss in a resistor for an a.c. is the same as that for a steady direct current which has current and voltage values equivalent to the rms current and voltage (i.e. when $$I_{\text{d.c.}}=I_{\text{rms}}$$, $$V_{\text{d.c.}}=V_{\text{rms}}$$, $$P_{\text{d.c.}}=\langle P_{\text{a.c.}}\rangle$$).

For a sinusoidal a.c.: $$$\begin{align*}&&I_{\text{rms}}=\frac{I_{0}}{\sqrt{2}}\quad&\text{and}\quad\langle P \rangle=I_{\text{rms}}^{2}R\\&\Rightarrow & \langle P \rangle &=\frac{I_{0}^{2}}{2}R\\&& \langle P \rangle &=\frac{P_{0}}{2}\end{align*}$$$ For a square-wave a.c.:$$$\begin{align*}&& I_{\text{rms}}=I_{0} \quad &\text{and} \quad \langle P \rangle =I_{\text{rms}}^{2}R\\&\Rightarrow &\langle P \rangle &=I_{0}^{2}R\\&& \langle P \rangle &=P_{0}\end{align*}$$$ $$P_{0}$$ is the peak power loss.