Design a Multirange Voltmeter with Variable Series Resistances

  • Thread starter Thread starter hsiao
  • Start date Start date
  • Tags Tags
    Design Voltmeter
Click For Summary

Homework Help Overview

The discussion revolves around designing a multirange voltmeter capable of measuring voltages of 20.0V, 50.0V, and 100V using a galvanometer with a resistance of 60.0 ohms and a full-scale deflection current of 1.00 mA. Participants explore the necessary series resistances to achieve full-scale deflection for each voltage range.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need for a series resistor to limit current through the galvanometer and explore the implications of using multiple voltage ranges. There is a suggestion to tackle each voltage requirement individually, and questions arise regarding the correct configuration of resistors.

Discussion Status

Some participants have provided calculations for the required series resistances for each voltage setting. Guidance has been offered regarding the use of a switching mechanism to select the appropriate series resistance for the desired voltage range. There is an acknowledgment of potential overcomplication in the original approach.

Contextual Notes

Participants note the challenge of managing multiple variables in the problem setup and express uncertainty about the interpretation of the question, particularly regarding the circuit design and component values.

hsiao
Messages
4
Reaction score
0

Homework Statement


"Design a multirange voltmeter capable of full-scale deflection for 20.0V, 50.0V, and 100 V. Assume that the meter movement is a galvanometer that has a resistance of 60.0 ohms and gives a full-scale deflection for a current of 1.00 mA"

Homework Equations


V = IR
Kirchoff's Law's

The Attempt at a Solution


I know that the current through the galvanometer must be 1.00mA, meaning i'll need a large resistor in series with the galvanometer. I would be able to solve this no problem if only one voltage was needed, it is the variable voltages that throw me. So far, I believe i have a voltmeter (galv + large resistor) in parallel with a small resistor (so that most of the current flows through the small). I've attempted kirchhoffs and equating v/r for each voltage, but there are too many variables for me to solve. Help? Thanks in advance

Example of a similar, simpler problem:
"A galvanometer requires a current of 1.5mA for full-scale deflection and has a resistance of 75 ohm, may be used to measure currents of much larger values. Calculate the value of the shunt resistor that enables the meter to be used to measure a current of 1 A and full-scale deflection"
Solution:
V = IR = (75)(1.5x10^-3) = (1-1.5x10^-3)R
R = .113 ohms
 
Last edited:
Physics news on Phys.org
Forget about the parallel resistor, that is for current measurements and is not needed for voltage measurements.

Try doing things 1 voltage at a time. For 20.0V, what does the series resistor need to be?
 
Ok, so in order to have full-scale deflection w/ 20.0 V:

V = IR
20/(1x10^-3) = 20000 ohms (<--total resistance)
20000 - 60 = 19940 ohms for the resistor in series with the 60 ohm galvanometer.

Likewise, for 50 V the resistor would need to be 49940 ohm and for 100 V, 99940 ohm

I'm not sure how to use that information to help me solve the problem, perhaps I am misinterpreting what the question is asking, I was trying to work the problem so that, as a solution, I have a drawing of a circuit with defined values for each component. Can anyone give further guidance?
 
Last edited:
Well, you have the 60-ohm galvanometer. And you have 3 different series-resistances, but the circuit should use only one of those at any given time, depending on the desired scale.

All that's missing, it would seem, is some means of switching among the different series resistances.
 
Redbelly98 said:
All that's missing, it would seem, is some means of switching among the different series resistances.

OH! Switches! Thank you Redbelly98, that was just the tip I needed. I definitely overcomplicated this problem.

I will updated this post with the solution presented in class once this is reviewed to compare with. Thanks!
 

Similar threads

Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K