# Voltmeter with low resistance measuring emf

Discussion in 'Homework Help' started by Manager1234, Jun 28, 2013.

1. ### Manager1234 Thread Starter New Member

Jun 7, 2013
4
0

here is the circuit.

I said that r (internal resistnce of battery) decreases because of the resistors in parallel rule. I also said that the voltmeter with low resistance draws more current.
are these correct?

2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Draw the circuit with real models for the meters and then figure out what the E and r values would be if the meter readings were taken at face value. How do those compare to the actual values? Which meter's non-ideal resistance dominates the error?

3. ### Manager1234 Thread Starter New Member

Jun 7, 2013
4
0
am i right in thinking i can use the resistors in parallel equation rule to say that the value for the internal resistance has decreased????

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
No, because the resistors aren't in parallel. In order to be in parallel, they have to have the same voltage across each. Is that the case here?

5. ### Manager1234 Thread Starter New Member

Jun 7, 2013
4
0
I meant the voltmeter and little r(internal resistance)? does the 1/v=1/v(1)+1/v(2) apply. as in the exam i said the resistance of r falls if the voltmeter resistance is low.

6. ### screen1988 Member

Mar 7, 2013
310
3
Hi,
I think they aren't in parallel. Because in addition r there is also ε in series with it.
ε and r is in parallel with voltmeter.
What do you mean by this? Do you mean 1/R = 1/R1 + 1/R2?
Let's call the resistances of ammeter and voltmeter are Ra and Rv respectively.
Then how can you determine ε and r basing on readings from ammeter and voltmeter?

7. ### screen1988 Member

Mar 7, 2013
310
3
This is how I determine electromotive force ε by using voltmeter.
First when you connect your battery open circuit with voltmeter is in parallel with the battery:

Model the voltmeter with resistance Rv.
Then the voltage across the voltmeter is V= ε*Rv/(r + Rv)
If Rv = ∞ then V = ε
But if Rv is finite, then V < ε

PS: I am not sure if the exercise is supposed to use the circuit given to determine ε and r. If so, I think that you need to model ammeter and voltmeter with resistances Ra and Rv respectively.
Then you determine the ε and r from the readings I and V from ammeter and voltmeter. ε and r may be a function of R, r, Rv, Ra, V, I.
Then you consider the case when Rv is infinite or finite and Ra is zero or finite.
That is what I thought, maybe it is wrong!!!

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Last edited: Jun 29, 2013
8. ### circuitfella11 Member

May 10, 2013
56
5
your answer was correct.. parallel resistance (with the voltmeter) lowers the resistance, but often times its negligible, depends on their resistance gap..it is slightly lower than the original resistance or slightly lower than the resistance of the voltmeter, if the resistance of both is in large gap..if they're almost the same, it lowers almost half.. whichever case, they do follow the parallel equation rule.. r always goes lower..

--cheers..

9. ### vk6zgo Active Member

Jul 21, 2012
677
85
"r" is the internal resistance of the battery.
If we call the internal resistance of the meter Rv ,then;

With"S" in screen1988's circuit open, no current is drawn by the rest of the circuit,so it can be ignored.
We now have a voltage divider in series with the ideal source "ε".

ε[Rv/(Rv+r)].

With"S" closed,it is no longer a simple voltage divider,as the voltage drop across r is dependent on the circuit current through R as well as that through Rv.

Last edited: Jul 2, 2013
10. ### deco75943531 New Member

Jul 2, 2013
5
0
I do not know this can not generate electricity, the test results to know

11. ### vk6zgo Active Member

Jul 21, 2012
677
85
As I hinted at in the last part of my previous posting,Rv is in parallel with R,but it is always in series with r.

With a high impedance voltmeter,say,a DMM with an input impedance of around 10 MΩ,Rv is so large in comparison with r & R,that it may be neglected.
Now,with S open,the open circuit value of ε may be determined.

If we now adjust R so that the voltage across it is ε /2 volts,measuring the value of R will give the value of r.