Troubleshooting Voltmeter Reading w/ Resistance 5*10^4

Click For Summary
SUMMARY

The discussion centers on troubleshooting a voltmeter reading with a resistance of 5*10^4 Ohms when connected between point b and ground. The user, Josh, attempts to calculate the voltage using the formula V = I_fs (R_c + R_s) but encounters discrepancies due to an additive constant. The solution involves recognizing that the voltmeter's resistance is in parallel with a 200k-ohm resistor, necessitating the calculation of equivalent resistance to determine the correct voltage ratio between points a and b. Ultimately, Josh resolves his confusion with the assistance of forum members.

PREREQUISITES
  • Understanding of Ohm's Law and voltage calculations
  • Familiarity with parallel resistor combinations
  • Knowledge of voltmeter specifications and usage
  • Basic circuit analysis techniques
NEXT STEPS
  • Study the calculation of equivalent resistance in parallel circuits
  • Learn about the impact of voltmeter resistance on circuit measurements
  • Explore advanced applications of Ohm's Law in circuit analysis
  • Investigate common errors in voltage measurement techniques
USEFUL FOR

Electrical engineers, physics students, and anyone involved in circuit design or troubleshooting voltmeter readings will benefit from this discussion.

joshanders_84
Messages
23
Reaction score
0
This is another problem I am having troubles with. Here's the prompt:

What is the reading of a voltmeter with the proper range and with resistance 5*10^4 Ohms when connected between point b and ground?

I know that for voltmeters, V = I_fs (R_c + R_s), and I tried solving for V using the I through the system as I_fs, which I solved as (400V/300000) = .001333, and then using 5*10^4 and 200000 as the resistors, but it says my answer is off by an additive constant. I don't understand what I am dong wrong...all help is appreciated. Thanks
Josh
 

Attachments

  • grounded.jpg
    grounded.jpg
    4 KB · Views: 473
Physics news on Phys.org
If the voltmeter wasn't there, or was perfect, then the voltage at b would be 2/3 of that at a.

Now when the voltmeter is present, its resistance is in parallel with the 200k-ohm resistor, so you can calculate the equivalent resistance of the parallel combination, and get a new ratio for the voltages at a and b.
 
  • Like
Likes   Reactions: berkeman
I don't understand how that helps...sorry. I just don't get what I am doing wrong still, or how the above would help me go further...
 
o i ci I got it now, thank you
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
26
Views
5K
  • · Replies 25 ·
Replies
25
Views
11K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
5
Views
4K