Refer to my third post :) clearer and the picture in my second post
This is extremely vague. "current .... flows BACK to an ammeter"?
Spend some time producing a circuit diagram and post it here!
Furthermore, if this is part of a HW/Coursework question, please do it in the HW/Coursework forum.
Its a simple setup as shown( this is what i thought of so it's not a homework qn)
I put arrows to show a case where current will flow through the voltmeter ( unlike normal case where an ideal voltmeter has an infinite resistance and assumed to have no current through it)
You might want to look up Kirchoff's Rule, also. I think you are trying to ask what the ammeter readings will be in that circuit diagram, it can be answered by the said computational method. Intuitively, at the split, the current reading should be lower, this is a direct consequence of the law of conservation of charges.
The voltmeter in parallel to the load will indeed decrease the total circuit resistance thus increasing circuit current. Also the Ammeter will have a small resistance in series with the load decreasing the circuit current. To compute the total circuit current you will need to know the resistance of your meters then just compute the total circuit resistance.
Comparing a case when no current flow through the voltmeter and when current flows through it -- why would there be an decrease in the reading of the voltmeter from the former to the latter case?
Using V=IR, when I increases due to current flowing through voltmeter but initially there us none, the voltmeter reading should increase.
Using the same equation V=IR, when resistance of the voltmeter has decreased(no longer infinite resistance voltmeter to allow current to flow through it) then considering that factor alone the voltmeter reading should decrease.
Why is the latter case the correct one?- where voltmeter reading decrease (to be less than the emf) from a case of no current through voltmeter to a case where there is current through the voltmeter (the explanation or formula used may be wrong)
And i may have to rephrase my first qn- would the ammeter reading change from a case where no voltmeter is attached and the circuit is in series to a case where a voltmeter is attached as shown above and current PASSES through it?
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