Ammeter and voltmeter resistance

Click For Summary
SUMMARY

The discussion focuses on calculating the internal resistances of an ammeter (Ra) and voltmeter (Rv) based on their readings in two circuit configurations. In the first configuration, the ammeter reads 9.901A and the voltmeter reads 100V, leading to a calculated Ra of approximately 0.099 ohms. In the second configuration, the ammeter reads 9.911A and the voltmeter reads 99.01V, resulting in a calculated Rv of approximately 9901 ohms. Participants emphasize the importance of understanding the ideal characteristics of these instruments, specifically that the voltmeter has infinite resistance in parallel with its internal resistance.

PREREQUISITES
  • Understanding of Ohm's Law
  • Knowledge of series and parallel circuits
  • Familiarity with the characteristics of ammeters and voltmeters
  • Basic circuit analysis techniques
NEXT STEPS
  • Study the principles of parallel resistor calculations
  • Learn about the ideal characteristics of measuring instruments in circuits
  • Explore advanced circuit analysis techniques using Kirchhoff's laws
  • Investigate the impact of instrument loading on circuit measurements
USEFUL FOR

Students in electrical engineering, electronics technicians, and anyone involved in circuit design and analysis who seeks to understand the effects of measurement instruments on circuit performance.

tommoturbo
Messages
33
Reaction score
0

Homework Statement



A little help with this one please



The power in a 10ohm resister when connected to a 100v supply is measured by an ammeter and voltmeter. There are two possible ways of connecting the instruments (as seen in the attachment)

The values of Ra and Rv represent the resistance of the ammeter and volmeter respectfully.



When connected as in Fig a The voltmeter reads 100v and ammeter reads 9.901A

When connected as in Fig b The voltmeter reads 99.01v and ammeter reads 9.911A


Calculate the values of Ra and Rb


Not sure where to start with this one but what i have so far is





Homework Equations



i know that the ammeter resistanse will be tiny and the voltmeter should be really large


The Attempt at a Solution


i calculated the voltage drop across the 10 ohm resistor



Vd=IxR

99.01V=9.901x10



i get Ra to be .99ohms



and Rb to be 1000.1ohms



can really put my working out down its long winded and a bit of guess work, working back to give figures,



there is another part to the question about calculating percentage error but i can do that fine as i didnt need to know Ra or Rv



if anyone can shed some light on a preferred method of calculating them please let me know



many thanks ian
 

Attachments

  • ammeter.JPG
    ammeter.JPG
    19.8 KB · Views: 810
Physics news on Phys.org
I stared at your circuit for a good few seconds, trying to figure out how the heck you were going to analyze it, and how you could possibly get the numbers that you were getting. Then I realized what was wrong with it: you should remember that a real voltmeter has an ideal voltmeter (with infinite resistance) in parallel with its internal resistance, not in series as your pictures indicate.

A real ammeter has an ideal ammeter (with zero resistance) in series with its internal resistance, as you show in your diagrams.

I'll get you started. In figure A, there should be four parallel branches:
1) the battery
2) the branch including the ammeter
3) the voltmeter's internal resistance
4) the ideal voltmeter with infinite resistance (i.e. no current flows through it--you can show this by connecting it to the rest of the circuit using dashed lines, such that it's connected, but not really connected)

The remainder of the question is left as an exercise to the reader!
 
Thanks that makes a bit more sense

what i have now is Ra to = 0.099ohms

as I = Vx R
9.901=100 x R

R= 100/9.901 10.0999 OHMS We know 1 resistor is 10 ohms therefore Ra = .099ohms

Now fig b I am assuming you need to use both figures to determine Ra and Rv

im still not sure about how to calc the value of rv

ive had a go and got a current of 0.01A flowing towards V of 99.01V giving a resistance of Rv=10k??

does that sound right?
 
tommoturbo said:
Thanks that makes a bit more sense

what i have now is Ra to = 0.099ohms

as I = Vx R
9.901=100 x R

R= 100/9.901 10.0999 OHMS We know 1 resistor is 10 ohms therefore Ra = .099ohms

Now fig b I am assuming you need to use both figures to determine Ra and Rv

im still not sure about how to calc the value of rv

ive had a go and got a current of 0.01A flowing towards V of 99.01V giving a resistance of Rv=10k??

does that sound right?

You've got Ra, mostly (I rounded to 0.100 ohm, but that might just be me, after getting 0.099901 ohms). However, start by redraw that circuit in B properly (with Rv in parallel with an ideal voltmeter). You'll find out that you know what the voltage across Rv is. And remember where the current flowing through the ammeter flows through (you do not need to use the value of Ra in circuit B)
 
I have calculated Rv to be 9901 Ohms

the voltage drop across Rv and Rl add up to the measured voltage of 99.01v

I3 which i worked out as the current flowing to the voltmeter (in practice it is about zero) is 0.01 amps which appliing ohms law

Rv =V/i 99.01/0.01 = 9901 ohms

i think!
 
tommoturbo said:
Thanks that makes a bit more sense

what i have now is Ra to = 0.099ohms

as I = Vx R
9.901=100 x R

R= 100/9.901 10.0999 OHMS We know 1 resistor is 10 ohms therefore Ra = .099ohms

Now fig b I am assuming you need to use both figures to determine Ra and Rv

im still not sure about how to calc the value of rv

ive had a go and got a current of 0.01A flowing towards V of 99.01V giving a resistance of Rv=10k??

does that sound right?

You've got Ra, mostly (I rounded to 0.100 ohm, but that might just be me, after getting 0.099901 ohms). However, start by redraw that circuit in B properly (with Rv in parallel with an ideal voltmeter). You'll find out that you know what the voltage across Rv is. And remember where the current flowing through the ammeter flows through (you do not need to use the value of Ra in circuit B). And remember what the formula for a parallel resistor is!
 
tommoturbo said:
I have calculated Rv to be 9901 Ohms

the voltage drop across Rv and Rl add up to the measured voltage of 99.01v

I3 which i worked out as the current flowing to the voltmeter (in practice it is about zero) is 0.01 amps which appliing ohms law

Rv =V/i 99.01/0.01 = 9901 ohms

i think!

That's the value I get. However, I used a different method (also more complicated, since I took another three lines to unwrap parallel resistance). Good on you for recognizing the shortcut!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 1 ·
Replies
1
Views
753
  • · Replies 2 ·
Replies
2
Views
662
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K