Infinite resistance of a Voltmeter

AI Thread Summary
A voltmeter must have a high resistance compared to the circuit it measures to avoid altering the circuit's current and potential difference (P.D.). If the voltmeter's resistance is low, significant current can flow through it, affecting the measurements. This principle emphasizes that an observer can disturb the system being measured, so minimal disturbance is crucial. To illustrate this, comparing circuits with high and low resistance voltmeters can clarify the impact on voltage readings. Understanding this concept is essential for accurate measurements in electrical circuits.
shehri
Messages
30
Reaction score
0
While studying about 'Potentiometer' it's said :
"The resistance of the voltmeter be large compared to the circuit resistance across which the voltmeter is connected.Otherwise an appreciable current will flow through voltmeter which will alter the circuit current & the P.D to be measured". Can anybody explain this concept?.Thanks.
 
Last edited:
Physics news on Phys.org
The concept is actually straight forward. You need to show some thought about your question before we can help you. (see https://www.physicsforums.com/showthread.php?t=94379)

If you are truly lost, you might begin by defining some of the terms or if it is a measuring device, how do you normally use this in the lab. Have you done any research about this, on the web? (hint: use search engines)..
 
Last edited:
Think of two circuits each with a battery and a two equal resistors of say 1kohm, one where you are measuring the voltage across a resistor with a high resistance voltmeter say 1Mohm and one where you measure with a low resistance voltmeter say 1ohm. Draw out the circuits and see if you can work out what the voltage measured in each case would be and what the actual voltage should be.
 
shehri said:
While studying about 'Potentiometer' it's said :
"The resistance of the voltmeter be large compared to the circuit resistance across which the voltmeter is connected.Otherwise an appreciable current will flow through voltmeter which will alter the circuit current & the P.D to be measured". Can anybody explain this concept?.Thanks.

Because "an observer disturbs the system he/she is trying to observe", it is important to include a measuring device in a such way that only minimal change to the original system results. In the voltmeter case, you are providing the current an alternative pathway to flow once you have introduced the voltmeter in the circuit. Hence, in order to minimise the amount of currrent using this path (created by the inclusion of the voltmeter) you want the resistance of the voltmeter to be very large compared to the circuit component it is measuring. Higher resistance means that most of the current will follow the original path and hence leads to minimal change to the circuit conditions.
 
I know both of you, are trying to be helpful to shehri, and you have not earned "homework helper" badges yet. It would be more helpful to the person asking the question, to show they have given some thought to them and not just seeking answers from the PF community.

You are encouraged to read this https://www.physicsforums.com/showthread.php?t=94379 which outlines what is expected of posters to the homework section.
 
Thanks for clearing me about 'infinite resistance' in a simple manner.Especially 'mjsd'.Thanks once again.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top