4 Engineering Example 3

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 Figure 7.

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:

$T=I_1{I}_2\frac{dM}{d\theta }$

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.

Figure 7 :

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 Key Point 4 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 $\frac{1}{\sqrt{2}}$ . 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

$V_{\text{peak}}=240\times \sqrt{2}=339.4V$

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 $V=V_{\text{peak}}sin\left(\theta \right)$ where $\theta =\omega t$ where $\omega$ is the angular frequency and $t$ is time we find that during the positive part of the cycle

$I_{\text{rms}}^2=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi }{\left(\frac{339.4sin\left(\theta \right)}{600}\right)}^{2}d\theta$

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

$I_{\text{rms}}^2=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_{\pi }^{2\pi }{\left(\frac{339.4sin\left(\theta \right)}{1400}\right)}^{2}d\theta$

 Therefore over an entire cycle

$I_{\text{rms}}^2=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi }{\left(\frac{339.4sin\left(\theta \right)}{600}\right)}^{2}d\theta +\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_{\pi }^{2\pi }{\left(\frac{339.4sin\left(\theta \right)}{1400}\right)}^{2}d\theta$

 We can calculate this value to find $I_{\text{rms}}^2$ and therefore $I_{\text{rms}}$ .

Mathematical analysis

$I_{\text{rms}}^2=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_0^{\pi }{\left(\frac{339.4sin\left(\theta \right)}{600}\right)}^{2}d\theta +\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_{\pi }^{2\pi }{\left(\frac{339.4sin\left(\theta \right)}{1400}\right)}^{2}d\theta$

$I_{\text{rms}}^2=\frac{339.4^2}{2\pi \times 10000}\left({\int \; <!--nolimits\; -->}_0^{\pi }\frac{{sin}^2\left(\theta \right)}{36}d\theta +{\int \; <!--nolimits\; -->}_{\pi }^{2\pi }\frac{{sin}^2\left(\theta \right)}{196}d\theta \right)$

Substituting the trigonometric identity ${sin}^2\left(\theta \right)\equiv \frac{1-cos\left(2\theta \right)}{2}$ we get

Interpretation

The reading on the meter would be 0.31 A.

Exercises

1. Calculate the r.m.s values of the given functions across the specified interval.
   1. $f\left(t\right)=1+t$ across $\left[0,2\right]$
   2. $f\left(x\right)=2x-1$ across $\left[-1,1\right]$
   3. $f\left(t\right)=t^2$ across $\left[0,1\right]$
   4. $f\left(t\right)=t^2$ across $\left[0,2\right]$
   5. $f\left(z\right)=z^2+z$ across $\left[1,3\right]$
2. Calculate the r.m.s values of the given functions over the specified interval.
   1. $f\left(x\right)=x^3$ across $\left[1,3\right]$
   2. $f\left(x\right)=\frac{1}{x}$ across $\left[1,2\right]$
   3. $f\left(t\right)=\sqrt{t}$ across $\left[0,2\right]$
   4. $f\left(z\right)=z^3-1$ across $\left[-1,1\right]$
3. Calculate the r.m.s values of the following:
   1. $f\left(t\right)=sint$ across $\left[0,\frac{\pi }{2}\right]$
   2. $f\left(t\right)=sint$ across $\left[0,\pi \right]$
   3. $f\left(t\right)=sin\omega t$ across $\left[0,\pi \right]$
   4. $f\left(t\right)=cost$ across $\left[0,\frac{\pi }{2}\right]$
   5. $f\left(t\right)=cost$ across $\left[0,\pi \right]$
   6. $f\left(t\right)=cos\omega t$ across $\left[0,\pi \right]$
   7. $f\left(t\right)=sin\omega t+cos\omega t$ across $\left[0,1\right]$
4. Calculate the r.m.s values of the following functions:
   1. $f\left(t\right)=\sqrt{t+1}$ across $\left[0,3\right]$
   2. $f\left(t\right)=e^t$ across $\left[-1,1\right]$
   3. $f\left(t\right)=1+e^t$ across $\left[-1,1\right]$

Answer