Title: Default for 14.2 The Mean Value and the Root-Mean-Square Value
Course: HELM Workbooks 2023-2024

Average value of a function

Suppose a time-varying function \(f(t)\) is defined on the interval \(a\le t\le b\). The area, \(A\), under the graph of \(f(t)\) is given by the integral \(\displaystyle A=\int_a^b f(t)\,\mathrm{d}t\). This is illustrated in the Figure below.

[Image: Left: the area under the curve f(t) from x=a to x=b. Right: a rectangle from x=a to x=b with height m and with equal are]

On the Figure we have also drawn a rectangle with base spanning the interval \(a\le t\le b\) and which has the same area as that under the curve. Suppose the height of the rectangle is \(m\). Then \[\text{area of rectangle} = \text{area under curve} \Rightarrow\quad m(b-a) = \int_a^b f(t)\,\mathrm{d}t\quad \Rightarrow\quad m = \frac{1}{b-a}\int_a^b f(t)\,\mathrm{d}t\] The value of \(m\) is the mean value of the function across the interval \(a\le t\le b\).

Key Point

The mean value of a function \(f(t)\) in the intervall \(a\le t\le b\) is \(\displaystyle \frac{1}{b-a}\int_a^b f(t)\,\mathrm{d}t\).

The mean value depends upon the interval chosen. If the values of \(a\) or \(b\) are changed, then the mean value of the function across the interval from \(a\) to \(b\) will in general change as well.

Example

Find the mean value of \(f(t)=t^2\) over the interval \(1\le t\le 3\).

Solution

Using the Key Point above with \(a=1\) and \(b=3\) and \(f(t)=t^2\) \[\text{mean value}~=~\frac{1}{b-a}\int_a^b f(t)\,\mathrm{d}t ~=~ \frac{1}{3-1}\int_1^3 t^2 \,\mathrm{d}t ~=~ \frac{1}{2} \left[\frac{t^3}{3}\right]_1^3 ~=~ \frac{13}{3}\]

Exercise

Find the mean value of \(f(t)=[value]\) over the interval \([value]\le t\le [value]\).

Use the Key Point with \(a=[value]\) and \(b=[value]\) to write down the required integral:

mean value\(~=~\) \(\displaystyle \int\) |  | \(~\) \(\ \mathrm{d}t\)

Now evaluate the integral:

mean value \(~=~\)
Feedback: Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\]

Engineering Example

Sonic boom

Introduction

Impulsive signals are described by their peak amplitudes and their duration. Another quantity of interest is the total energy of the impulse. The effect of a blast wave from an explosion on structures, for example, is related to its total energy. This Example looks at the calculation of the energy on a sonic boom. Sonic booms are caused when an aircraft travels faster than the speed of sound in air. An idealized sonic-boom pressure waveform is shown in the figure below where the instantaneous sound pressure \(p(t)\) is plotted versus time \(t\). This wave type is often called an N-wave because it resembles the shape of the letter N. The energy in a sound wave is proportional to the square of the sound pressure.

[Image: An idealized sonic-boom pressure waveform (an N-wave).]

Problem in words

Calculate the energy in an ideal N-wave sonic boom in terms of its peak pressure, its duration and the density and sound speed in air.

Mathematical statement of problem

Represent the positive peak pressure by \(P_0\) and the duration by \(T\). The total acoustic energy \(E\) carried across unit area normal to the sonic-boom wave front during time \(T\) is defined by \[E ~=~ <\! p(t)^2\!> T /\rho c\qquad \text{(1)}\] where \(\rho\) is the air density, \(c\) the speed of sound and the time average of \([p(t)]^2\) is \[ <\!p(t)^2\!> ~=~ \frac{1}{T}\int_0^T p(t)^2\,\mathrm{d}t\qquad \text{(2)}\]

1. Find an appropriate expression for \(p(t)\).
2. Hence show that \(E\) can be expressed in terms of \(P_0,\ T,\ \rho\) and \(c\) as \(\displaystyle E ~=~\frac{TP_0^2}{3\rho c}\).

Mathematical analysis

1. The interval of integration needed to compute (2) is \([0, T].\) Therefore it is necessary to find an expression for \(p(t)\) only in this interval. The figure above shows that, in this interval, the dependence of the sound pressure \(p\) on the variable \(t\) is linear, i.e. \(p(t) = a t + b.\)
From the figure above also \(p(0) = P_0\) and \(p(T) = - P_0\). The constants \(a\) and \(b\) are determined from these conditions.
At \(t = 0,\ a \times 0 + b = P_0\) implies that \(b = P_0\).
At \(t = T,\ a \times T + b = - P_0\) implies that \(a = -2P_0/T.\)
Consequently, the sound pressure in the interval \([0, T]\) may be written \(\displaystyle p(t) = \frac{-2P_0}{T}\ t + P_0.\)
2. This expression for \(p(t)\) may be used to compute the integral (2) \[\begin{aligned} \frac{1}{T}\int_0^T p(t)^2 \,\mathrm{d}t&~=\frac{1}{T}\int_0^T\ \left(\frac{-2P_0}{T}\ t+P_0\right)^2 \,\mathrm{d}t =\frac{1}{T}\ \int_0^T\ \left(\frac{4P_0^2}{T^2}\ t^2-\frac{4P_0^2}{T}\ t+P_0^2\right) \,\mathrm{d}t\\ \\ &~=\frac{1}{T}\left[\frac{4P_0^2}{3T^2}\ t^3-\frac{2P_0^2}{T}\ t^2+P_0^2 t\right]_0^T\\ \\ &~=\frac{P_0^2}{T}\left(\frac{4}{3T^2}\ T^3-\frac{2}{T}\ T^2+T\right)-0=P_0^2/3.\end{aligned}\] Hence, from Equation (1), the total acoustic energy \(E\) carried across unit area normal to the sonic-boom wave front during time \(T\) is \(E = \dfrac{TP_0^2}{3\rho c}.\)

Interpretation

The energy in an N-wave is given by a third of the sound intensity corresponding to the peak pressure multiplied by the duration.

Exercise

Calculate the mean value of the given functions across the specified interval.

1. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
2. \(f(x)=[value]\) across \([[value],[value]]\)
mean value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
4. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
5. \(f(z)=[value]\) across \([[value],[value]]\)
mean value \(=\)
Feedback: Remember the Key Point, mean value \(\displaystyle = \frac{1}{b-a} \int_a^b f(t)\ \mathrm{d}t\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-([value])} \int_[value]^[value] [value] \ \mathrm{d}x\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 5. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(z)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] ~\ \mathrm{d}z\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\]

Exercise

Calculate the mean value of the given functions over the specified interval.

1. \(f(x)=[value]\) across \([[value],[value]]\)
mean value \(=\)
2. \(f(x)=\displaystyle[value]\) across \([[value],[value]]\)
mean value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
4. \(f(z)=[value]\) across \([[value],[value]]\)
mean value \(=\)
Feedback: Remember the Key Point, mean value \(\displaystyle = \frac{1}{b-a} \int_a^b f(t) \ \mathrm{d}t\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}x\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=\displaystyle [value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-([value])} \int_[value]^[value] [value] \ \mathrm{d}x\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(z)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value]\ \mathrm{d}z\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &= [value] \end{aligned}\]

Exercise

Calculate the mean value of the following:

1. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
2. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
4. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
5. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
6. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
7. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
Feedback: Remember the Key Point, mean value \(\displaystyle = \frac{1}{b-a} \int_a^b f(t)\ \mathrm{d}t\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-([value])\right]\\ &= [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-([value])} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-([value])\right]\\ &= [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-\left([value]\right)\right]\\ &= [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 5. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 6. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-[value]\right]\\ &= [value] \end{aligned}\] 7. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= [value]\left[[value]\right]_[value]^[value]\\ &= [value]\left[[value]-\left([value]\right)\right]\\ &= [value] \end{aligned}\]

Exercise

Calculate the mean value of the following to 4 decimal places:

1. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
2. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
mean value \(=\)
Feedback: Remember the Key Point, mean value \(\displaystyle = \frac{1}{b-a} \int_a^b f(t) \ \mathrm{d}t\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-[value]} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-([value])} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-[value]\right]\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{mean value }&= \frac{1}{[value]-([value])} \int_[value]^[value] [value] \ \mathrm{d}t\\ &= \frac{1}{[value]}\left[[value]\right]_[value]^[value]\\ &= \frac{1}{[value]}\left[[value]-([value])\right]\\ &\approx [value] \end{aligned}\]

Root-mean-square value of a function

If \(f(t)\) is defined on the interval \(a\le t\le b\), the mean-square value is given by the expression: \[ {\frac{1}{b-a}\int_a^b [f(t)]^2\,\mathrm{d}t}\] This is simply the mean value of \([f(t)]^2\) over the given interval. The related quantity: the root-mean square (r.m.s.) value is given by the following formula.

Key Point

\[\text{r.m.s. value } =~ \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2\ \mathrm{d}t}\]

The r.m.s. value depends upon the interval chosen. If the values of \(a\) or \(b\) are changed, then the r.m.s. value of the function across the interval from \(a\) to \(b\) will in general change as well. Note that when finding an r.m.s. value the function must be squared before it is integrated.

Example

Find the r.m.s. value of \(f(t)=t^2\) across the interval from \(t=1\) to \(t=3\).

Solution

\(\displaystyle \text{r.m.s. }=~\sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2\ \mathrm{d}t} ~=~ \sqrt{\frac{1}{3-1} \int_1^3 [t^2]^2\ \mathrm{d}t} ~=~ \sqrt{\frac{1}{2} \int_1^3 t^4\ \mathrm{d}t} ~=~ \sqrt{\frac{1}{2} \left[\frac{t^5}{5}\right]_1^3} \approx 4.92\)

Example

Calculate the r.m.s value of \(f(t)=\sin t\) across the interval \(0\le t\le 2\pi\).

Solution

Here \(a=0\) and \(b=2\pi\) so r.m.s. \(\displaystyle =~\sqrt{\frac{1}{2\pi} \int_0^{2\pi} \sin^2 t ~\ \mathrm{d}t}\).

The integral of \(\sin^2 t\) is performed by using trigonometric identities to rewrite it in the alternative form \(\dfrac{1}{2}(1-\cos 2t)\). This technique was described before, as we hope you will remember.

\[\text{r.m.s. value }=~ \sqrt{\frac{1}{2\pi} \int_0^{2\pi} \frac{(1-\cos 2t)}{2}\ \mathrm{d}t}~=~ \sqrt{\frac{1}{4\pi} \left[t-\frac{(\sin 2t)}{2}\right]_0^{2\pi}} ~=~ \sqrt{\frac{1}{4\pi}(2\pi)} ~=~ \sqrt{\frac{1}{2}} \approx 0.707\]

In the previous Example the amplitude of the sine wave was 1, and the r.m.s. value was 0.707. In general, if the amplitude of a sine wave is \(A\), its r.m.s value is 0.707\(A\).

Key Point

The r.m.s value of any sinusoidal waveform taken across an interval of width equal to one period is 0.707 \(\times\) amplitude of the waveform.

Engineering Example

Electrodynamic meters

Introduction

A dynamometer or electrodynamic meter is an analogue instrument that can measure d.c. current or a.c. current up to a frequency of 2 kHz. A typical dynamometer is shown in the figure below.

It consists of a circular dynamic coil positioned in a magnetic field produced by two wound circular stator coils connected in series with each other. The torque \(T\) on the moving coil depends upon the mutual inductance between the coils given by: \[ T=I_1I_2\frac{\mathrm{d}M}{\mathrm{d}\theta}\] where \(I_1\) is the current in the fixed coil, \(I_2\) the current in the moving coil and \(\theta\) is the angle between the coils. The torque is therefore proportional to the square of the current. If the current is alternating the moving coil is unable to follow the current and the pointer position is related to the mean value of the square of the current. The scale can be suitably graduated so that the pointer position shows the square root of this value, i.e. the r.m.s. current.

[Image: Representation of an electrodynamic meter.]

Problem in words

A dynamometer is in a circuit in series with a 400 \(\Omega\) resistor, a rectifying device and a 240 V r.m.s alternating sinusoidal power supply. The rectifier resists current with a resistance of 200 \(\Omega\) in one direction and a resistance of 1 k\(\Omega\) in the opposite direction. Calculate the reading indicated on the meter.

Mathematical Statement of the problem

We know from the Key Point in the text that the r.m.s. value of any sinusoidal waveform taken across an interval equal to one period is 0.707 \(\times\) amplitude of the waveform. Where 0.707 is an approximation of \(\dfrac{1}{\sqrt{2}}\). This allows us to state that the amplitude of the sinusoidal power supply will be: \[ V_{\text{peak}}=\frac{V_{\text{rms}}}{\frac{1}{\sqrt{2}}}=\sqrt{2}V_{\text{rms}}\] In this case the r.m.s power supply is 240 V so we have \[ V_{\text{peak}}=240\times \sqrt{2}=339.4\ V\] During the part of the cycle where the voltage of the power supply is positive the rectifier behaves as a resistor with resistance of 200 \(\Omega\) and this is combined with the 400 \(\Omega\) resistance to give a resistance of 600 \(\Omega\) in total. Using Ohm’s law \[ V=IR\Rightarrow I=\frac{V}{R}\] As \(V=V_{\text{peak}}\sin(\theta)\) where \(\theta=\omega t\) where \(\omega\) is the angular frequency and \(t\) is time we find that during the positive part of the cycle \[ I_{\text{rms}}^2=\frac{1}{2\pi}\int_0^\pi\left(\frac{339.4\sin(\theta)}{600}\right)^2\ \mathrm{d}\theta\] During the part of the cycle where the voltage of the power supply is negative the rectifier behaves as a resistor with resistance of 1 k\(\Omega\) and this is combined with the 400 \(\Omega\) resistance to give 1400 \(\Omega\) in total. So we find that during the negative part of the cycle \[ I_{\text{rms}}^2=\frac{1}{2\pi}\int_\pi^{2\pi}\left(\frac{339.4\sin(\theta)}{1400}\right)^2\ \mathrm{d}\theta\] Therefore over an entire cycle \[ I_{\text{rms}}^2=\frac{1}{2\pi}\int_0^\pi \left(\frac{339.4\sin(\theta)}{600}\right)^2\ \mathrm{d} \theta+\frac{1}{2\pi}\int_\pi^{2\pi}\left(\frac{339.4\sin(\theta)}{1400}\right)^2\ \mathrm{d}\theta\] We can calculate this value to find \(I_{\text{rms}}^2\) and therefore \(I_{\text{rms}}\).

Mathematical analysis

\[ I_{\text{rms}}^2=\frac{1}{2\pi}\int_0^\pi\left(\frac{339.4\sin(\theta)}{600}\right)^2\ \mathrm{d}\theta+\frac{1}{2\pi}\int_\pi^{2\pi}\left(\frac{339.4\sin(\theta)}{1400}\right)^2\ \mathrm{d}\theta \]\[ I_{\text{rms}}^2=\frac{339.4^2}{2\pi\times 10000} \left(\int_0^\pi\frac{\sin^2(\theta)}{36}\ \mathrm{d}\theta+ \int_\pi^{2\pi}\frac{\sin^2(\theta)}{196}\ \mathrm{d}\theta\right)\] Substituting the trigonometric identity \(\sin^2(\theta)\equiv \dfrac{1-\cos(2\theta)}{2}\) we get \[\begin{aligned} I_{\text{rms}}^2&=\frac{339.4^2}{4\pi\times 10000}\left(\int_0^\pi \frac{1-\cos(2\theta)}{36}\ \mathrm{d}\theta+\int_\pi^{2\pi}\frac{1-\cos(2\theta)}{196}\ \mathrm{d}\theta\right) \\ &=\frac{339.4^2}{4\pi\times 10000}\left(\left[\frac{\theta}{36}-\frac{\sin(2\theta)}{72} \right]_0^\pi+\left[\frac{\theta}{196}-\frac{\sin(2\theta)}{392}\right]_\pi^{2\pi}\right) \\ &=\frac{339.4^2}{4\pi\times 10000}\left(\frac{\pi}{36}+\frac{\pi}{196}\right)=0.0946875\ A^2 \\ I_{\text{rms}}&=0.31\ A\ \text{to}\ 2 \ \text{d.p.}\end{aligned}\]

Interpretation

The reading on the meter would be 0.31 A.

Exercise

Calculate the r.m.s. value of the given functions across the specified interval, to 4 decimal places.

1. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
2. \(f(x)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
4. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
5. \(f(z)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
Feedback: Remember the Key Point, r.m.s. value \(\displaystyle = \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2 \ \mathrm{d}t}\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] [[value]]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-([value])} \int_[value]^[value] [[value]]^2 \ \mathrm{d}x}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-\left([value]\right)\right]}\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] [[value]]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] [[value]]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 5. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(z)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] [[value]]^2 \ \mathrm{d}z}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\]

Exercise

Calculate the r.m.s. value of the given functions over the specified interval, to 4 decimal places.

1. \(f(x)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
2. \(f(x)=\displaystyle[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
4. \(f(z)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
Feedback: Remember the Key Point, r.m.s. value \(\displaystyle = \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2\ \mathrm{d}t}\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}x}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(x)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-([value])} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}x}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-\left([value]\right)\right]}\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(z)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}z}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\]

Exercise

Calculate the r.m.s. value of the following, to 4 decimal places.

1. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
2. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
Feedback: Remember the Key Point, r.m.s. value \(\displaystyle = \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2 \ \mathrm{d}t}\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-\left([value]\right)\right]}\\ &\approx [value] \end{aligned}\]

Exercise

Calculate the r.m.s. value of the following to 4 decimal places:

1. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
2. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
Feedback: Remember the Key Point, r.m.s. value \(\displaystyle = \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2 \ \mathrm{d}t}\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-([value])} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-([value])} \int_[value]^[value] \left[[value]\right]^2 \ \mathrm{d}t}\\ &= \sqrt{\frac{1}{[value]}\left[[value]\right]_[value]^[value]}\\ &= \sqrt{\frac{1}{[value]}\left[[value]-\left([value]\right)\right]}\\ &\approx [value] \end{aligned}\]

Exercise

Calculate the r.m.s. value of the following, to 4 decimal places where appropriate:

1. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
2. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
3. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
4. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
5. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
6. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
7. \(f(t)=[value]\) across \([[value],[value]]\)
r.m.s. value \(=\)
Feedback: WORKED SOLUTION Remember the Key Point, r.m.s. value \(\displaystyle = \sqrt{\frac{1}{b-a} \int_a^b [f(t)]^2 dt}\). 1. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 2. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 3. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 4. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 5. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 ~dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 6. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 ~dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-[value]\right]}\\ &\approx [value] \end{aligned}\] 7. Using the Key Point with \(a=[value]\), \(b=[value]\) and \(f(t)=[value]\) we have: \[\begin{aligned}\text{r.m.s. value }&= \sqrt{\frac{1}{[value]-[value]} \int_[value]^[value] \left[[value]\right]^2 ~dt}\\ &= \sqrt{[value]\left[[value]\right]_[value]^[value]}\\ &= \sqrt{[value]\left[[value]-\left([value]\right)\right]}\\ &\approx [value] \end{aligned}\]