Rectifier Meter

  • Context: Engineering 
  • Thread starter Thread starter PhysicsTest
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the operation and calculations related to a rectifier meter, specifically focusing on the relationship between voltage and current measurements, the use of a moving coil meter, and the implications of rectified sinewave inputs. Participants explore various aspects of the meter's functionality, including full-scale deflection, average current, and RMS values.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express confusion about the need for the peak sine input to calculate the full-scale reading of the meter, suggesting that 1 mA is the resolution of the meter.
  • One participant describes the meter as a moving coil meter with an internal resistance of 10 Ω and discusses the relationship between current and magnetic force, questioning whether Vrms applies to a rectified sinewave.
  • Another participant emphasizes the importance of understanding conversions between average, RMS, and peak values in AC measurements, noting that the 120V household supply represents multiple measurements.
  • There is a mention of integral calculus in relation to finding the average value of a curve, indicating a potential method for resolving some of the confusion about the calculations involved.
  • One participant calculates parameters for a full-wave bridge rectifier and expresses difficulty in understanding the relationship between voltage and current, particularly regarding the 1 mA full-scale reading.
  • Another participant states that without knowing the applied voltage, it is challenging to decipher the problem, suggesting that the problem statement may be incomplete.
  • One participant clarifies that the 1 mA full-scale deflection corresponds to a time average voltage of 5.01 Vdc, and calculates the corresponding Vrms for the rectified sinewave.
  • Another participant confirms the calculation of Vrms in relation to Vdc, providing a specific ratio for the relationship.

Areas of Agreement / Disagreement

Participants express various levels of understanding and confusion regarding the calculations and relationships involved, indicating that multiple competing views remain. There is no clear consensus on the interpretation of the problem or the calculations required.

Contextual Notes

Some participants note the potential incompleteness of the problem statement, which may affect the ability to resolve the questions raised. The discussion involves assumptions about voltage and current relationships that are not fully clarified.

PhysicsTest
Messages
260
Reaction score
27
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,
1766241962631.webp

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?
 
Physics news on Phys.org
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?
 
  • Like
Likes   Reactions: berkeman
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
 
Are you familiar with integral calculus? Like finding the average value of an arbitrary curve.
 
1766295568441.webp
Fig: Full bridge rectifier output

1766295060880.webp

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.
 
  • Like
Likes   Reactions: DaveE
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.
 
Last edited:
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
 
  • #10
PhysicsTest said:
Vrms = 1.11 * Vdc = 1.11*5.01 V = 5.56V
Correct.

Vrms / Vdc = ½ ⋅ π / √2 = 1.11072
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
7K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K