The voltage meter has some resistance and as such will always allow some current
ID: 1444125 • Letter: T
Question
The voltage meter has some resistance and as such will always allow some current to flow. How would it affect your data if the resistance used in the circuit were near to the resistance in the voltmeter (i.e. voltage sensor)? If you had a current sensor (ammeter) instead of a voltmeter, how could we get the same data without using a voltmeter? Be specific! Do you think it would be easier or more complicated? Would it be more or less accurate?
ALSO Can anyone PLEASE click on the click and check whether the highlighted portion looks right, NOT looking for answers but please say whetehr there is a wrong answer.
https://docs.google.com/document/d/13j3ipqcWS4JX6ldRLVLq5-6PwalEPaCwYcJhwXZ0mpg/edit?usp=sharing
Explanation / Answer
I'll answer now to the first question(s). I'll come back later on to the link you give, read it carefully, and tell what's about. I hope you can wait few hours (I have an emergency now). So, your first issue:
1. voltmeter influence in DC circuits:
The voltmeter is always connected in parallel with the component under test. Thus, any current through the voltmeter will contribute to the overall current in the tested circuit, affecting the voltage being measured.
A perfect voltmeter has infinite resistance, in such a way it draws no current from the circuit under test (no current passes through it). The voltmeter’s own resistance alters the resistance ratio of the voltage divider circuit, consequently affecting the voltage being measured.
In the present case you'll have half of the main current through the circuit passing through your voltmeter (its resistance: RV/R/2); the voltage measured by the voltmeter will be: U(V)=IR/2, with I=initial current through the resistor of interest.
There are good practical solutions to this problem.
2. ammeter influence in DC circuits:
The ammeter is always connected in series with the component under test. In comparison with the voltmeter, the ideal ammeter has zero internal resistance (it has to drop as little voltage as possible when the current passes through it). If it has a resistance, the measured intensity will be lower than the 'real ' one.
If RA=R, the intensity will he half than the intensity before introducing the ammeter in the circuit.
There are nice practical solutions also in this case. Tell me if you need any.
About the link you provided, and your request: it's a lab issue.You're asking me to check your work and to say if it's correct or not. This is impossible without knowing the detailed procedure (what and how did you measure, in which conditions and prerequisites, the instruments (you have there a certain software, a certain waveform, etc.).
Anyway, considering the values you obtained, in comparison with the theoretical values, you have large errors. Too large.
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