Rectifier Meter

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Homework Statement
A 1-mA dc meter whose resistance is 10 Ω is calibrated to read rms volts when used in a bridge circuit with semiconductor diodes. The effective resistance of each element may be considered to be zero in the forward direction and infinite in the inverse direction. The sinusoidal input voltage is applied in series with a 5-kΩ resistance. What is the full-scale reading of this meter?
Relevant Equations
Irms = Im/sqrt(2); Idc = 2*Im / Pi for full wave rectifier.
I am confused with the question it falls under the topic of Rectifier meter,
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I feel the peak of the sine input need to be provided to calculate the full scale reading of the meter, 1mA is the resolution of the meter. Any hint to solve this problem?
 
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The meter is a moving coil meter, with an internal resistance of 10 Ω. That is the resistance of the coil and the hairsprings.
The magnetic force on the coil, countered by the hairsprings, is proportional to current.
The current meter is being used with a 5-kΩ series resistance to measure voltage.
For a sinewave, Vrms = Vpeak / √2 ;
Is that also true of a rectified sinewave?
What is the time average force of the coil against the hairsprings?
 
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PhysicsTest said:
I feel the peak of the sine input need to be provided to calculate the full scale reading of the meter
Yes. But you could give it a variable name, like V, and write an equation.

A galvanometer like this one will measure the average value of the current through it. RMS is used for other things. However, we always use the RMS value to describe AC sinewave source amplitude. So the 120V in your house is actually all of these measurements: 120V, 120Vrms, 170Vpk, 340Vp-p, 0Vdc.
You need to learn these conversions sooner or later, maybe now is a good time.
https://www.tek.com/en/documents/application-note/fundamentals-ac-power-measurements
 
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Fig: Full bridge rectifier output

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I calculated above parameters for full wave bridge rectifier, so if Vm of the sine wave increases/decreases the Vdc changes accordingly and the Vrms.
The corresponding current is
1766295235406.webp

Beyond this i am unable to understand the 1mA and full-scale reading. Converting from voltage to current, current to voltage all messing up.
 
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DaveE said:
Are you familiar with integral calculus? Like finding the average value of an arbitrary curve.
Yes i am familiar.
 
PhysicsTest said:
Beyond this i am unable to understand the 1mA and full-scale reading.
Without knowing the applied voltage, I don't think anyone can decipher this. I suspect that the problem statement isn't complete somehow. As to the voltage and current, that will just be ohms law, except for a scaling factor (the resistance) they will be the same.

PS: Oops, I misread the question there is a relatively simple answer as @Baluncore described below.
 
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The 1 mA FSD meter will have Full-Scale-Deflection, 100%, when an average current of 1 mA flows.
The total resistance is 5kΩ + 10Ω = 5.010kΩ ;
1mA * 5.010kΩ = 5.01 Vdc FSD.
What rectified Vrms sinewave, has a time average voltage of 5.01V ?
 
Baluncore said:
The 1 mA FSD meter will have Full-Scale-Deflection, 100%, when an average current of 1 mA flows.
The total resistance is 5kΩ + 10Ω = 5.010kΩ ;
1mA * 5.010kΩ = 5.01 Vdc FSD.
What rectified Vrms sinewave, has a time average voltage of 5.01V ?
Vrms = 1.11 * Vdc = 1.11*5.01 V = 5.56V