16.1.2 Root-Mean-Square Value

If a fixed resistor R is connected across the AC power supply, the current I flowing in the resistor is also sinusoidal.

\displaystyle \begin{aligned}I&={{I}_{{pk}}}\sin \omega t\\&=\frac{{{{V}_{{pk}}}}}{R}\sin \omega t\end{aligned}

Since the current through the resistor is not constant, the instantaneous power dissipated in the resistor also varies with time. However, in many applications, we are only interested in the average power \left\langle P \right\rangle .

Using \left\langle {} \right\rangle to denote “the (time) average of”, we can write

\displaystyle \begin{aligned}\left\langle P \right\rangle &=\left\langle {{{I}^{2}}R} \right\rangle \\&=\left\langle {{{I}^{2}}} \right\rangle R\\&={{I}_{{rms}}}^{2}R\end{aligned}

Note that

  • \displaystyle \left\langle {{{I}^{2}}} \right\rangle =\frac{1}{T}\int_{0}^{T}{{{{{({{I}_{0}}\sin \omega t)}}^{2}}\text{ }dt}} is called the mean-square current. It is the average value of I2 over time.
  • {{I}_{{rms}}}=\sqrt{{\left\langle {{{I}^{2}}} \right\rangle }} is called the root-mean-square (rms) current. It is the square root of the average value of I2 over time.
  • The root-mean-square value can be thought as the DC equivalent value, when it comes to average power dissipation. For example, a 3.0 A rms current passing through a 2.0 W resistor results in an average power dissipation of \left\langle P \right\rangle ={{I}_{{rms}}}^{2}R={{(3.0)}^{2}}(2.0)=18\text{ W}. This is the same as passing a 3.0 A constant current.

For sinusoidal functions, there is a simple relationship between the peak value and the rms value.

\displaystyle {{I}_{{rms}}}=\frac{{{{I}_{{pk}}}}}{{\sqrt{2}}}

This relationship can be derived simply by inspecting the I2 graph! As the I2 value rises and falls between 0 and Ipk2, the symmetry of the sine-square graph makes it obvious that the average value is exactly \displaystyle \frac{{{{I}_{{pk}}}^{2}}}{2}. If the mean-square value is \displaystyle \frac{{{{I}_{{pk}}}^{2}}}{2}, the root-mean-square value is of course \displaystyle \sqrt{{\frac{{{{I}_{{pk}}}^{2}}}{2}}}=\frac{{{{I}_{{pk}}}}}{{\sqrt{2}}}.

We can repeat the exact same argument for a sinusoidal voltage. So,

\displaystyle {{V}_{{rms}}}=\frac{{{{V}_{{pk}}}}}{{\sqrt{2}}}

Concept Test

3217

3204

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s