KCC's Quizzes AQQ285 about analog voltmeter expertise

The average DC value of a full-wave rectified sinewave is Vpk-pk/pi = 2Vpk/pi = 2x1.414Vrms/pi = 0.9Vrms

So, the 1/0.9 = 1.11 scale factor is needed to convert from the AC to the DC scale

The square wave signal has a 20ms period but the duty cycle isn't specified.  However, it looks to be about 2/3rds, so the average DC value of the signal shown looks to be 25V, so you would read 27.78V on the AC scale.

1 . The form factor is the ratio of the RMS value divided by the half-period mean value of the measured signal.

1.11=ratio of RMS/average, and the meter will read around 27.75V for the square wave

1. This comes from Form Factor, which is used to interpolate the rectfied AC signal to its AC RMS equivalent. To get the form factor, divide the AC RMS equivalent of a particular waveform to its DC average. In this case, the waveform is expected to be a sine, FF = ( 1/sqrt(2)) / (1/pi) = 1.11

2. Let's assume that the signal is 50 V 50% of the time, and -25V for the rest ( their periods were not given ) . The average then is
               Vave = ((50 - 25) / 2 ) + 25 V = 37.5 V.
    Multiplying this average by 1.11, this gives:
              Vrms_meas = 41.625 V

The meter hopefully will show 46.28 V on the AC-scale

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1) For a full wave rectified sinewave, the AC scale should ideally read an RMS value of Vmax/sqrt(2). This same signal will read an average value of 2*Vmax/Pi on the DC scale. The position of the needle will line up with each reading so:

[2*Vmax/Pi]*SF = Vmax/sqrt(2)
SF = Pi/2*sqrt(2) = 1.1107...

2) The rectified average of the 50Hz pulse train wil be 25*(DC+1), where DC is the fraction of time that 50 volts is present. A graphical estimation of DC from the waveform image is 0.7 so the reading on the AC scale will be 25*(1.7)*1.11 = 47.175 volts.

(1)  The provided image of an analog meter does not appear to show a factor of 1.11 between the DC and AC scales. If there was such a difference, then at the right end of the scale one of the scales would be 10 V and the other would be close to 11 V. On the other hand, I have owned and use the best analog meter ever: Sanwa N-501. And it shows the scale is compressed at the zero end (left end of the scale).  I am guessing but I expect this is related to three things: The voltage drop across the rectifiers, the parasitic capacitance of the rectifiers which will cause a frequency dependent effect, and likewise the inductive effect of the moving coil which will also be frequency dependent.

(2) For this question the ratio of -25V and +50V is not given, so an actual value can't be given, but can be calculated.

As the AC input is rectified, the effective waveform is switching between 25V and 50V.

Here is the formula: ((50 * time in ms at 50V) + (25 * time in ms at -25V))/20

For example: time at -25V is 5 ms, time at 50V is 15ms -> ((50 * 15) + (25 * 5))/20 = 43.75 V which is the time weighted average.

As we don't know how the multimeter rectifies the AC input (germanium or silicon diodes for example, or half/full bridge) we can't
take that into consideration. But the multimeter does take into consideration for creating the scale, so I'll stick with 43.75V

1. The factor 1.11 comes from the ratio between the Vrms and the Average (Vavg) values (after rectification) of a pure sinewave ("Form Factor=FF"). Actually you measure Vavg, so you need to multiply it by 1.11 to get the RMS.

2. I believe you need to know the duty cycle to answer question nr.2. You can calculate the FF with t1 (portion of period at +50V) and t2 (portion of the period at -25V) as parameters. By using the definition of RMS and AVG after rectification and after some algebra, you get:

Vrms= sqrt (1/T (2500*t1 + 625*t2))

Vavg= 1/T (50*t1 + 25*t2)

where T=t1+t2

For example, if t1=14ms and t2=6ms, which seems similar to the provided periodic signal, you get FF= Vrms/Vavg= 1.03. In this case, the AC reading, which incorporates a factor of 1.11, would provide a slightly higher value for Vrms.   

Appologize: the rectangular signal has a duty cycle 1/3 low and 2/3 high

Corrected waveform: