# Conversion of a galvanometer into a voltmeter

When we are given a question regarding conversion of a galvanometer into a voltmeter,we are given full scale deflection current and initial resistance of the galvanometer...then we are given the converted voltmeter full scale reading and then asked the voltmeter's resistance.. For this we use V=IR and simply put the given value of V and the full scale deflection current from earlier

This is what I am unable to understand why would the full scale deflection current remain same?Isn't it a completely new circuit with new values of V,I and R?

A mathematical explanation would be much appreciated

During your experiment, you don't modify the galvanometer at all. So the initial specification must continue to hold, that it gives a full-scale deflection at some current $$I$$.

Your modification (presumably) involves placing resistors in series with the galvanometer. Since there are no parallel branches, all current in the circuit must go through the galvanometer. This changes the resistance of the circuit and therefore the voltage necessary to create this current. But you haven't changed the required current for the instrument to work.

Therefore to make the galvanometer give a full-scale deflection, you must (still) present it with current $$I$$, and that current must be in the rest of the circuit as well.

[Why does the] same value of I deflects the galavanometer before and after it is converted into a voltmeter ?

This is how galvanometers work. Their deflection depends on the specifics of their construction and the current through them. As this experiment is not modifying the galvanometer itself, we know that at the end of the experiment, we will need to send a current $$I$$ through the instrument to get a full scale deflection.

The question then is how does current $$I$$ in the instrument correspond to other values in the circuit? That's what the experiment is trying to show. What has to happen in the circuit to tie together a given $$V$$ (from the problem spec) and a given $$I$$ (from the design of the instrument)? And the answer is that adding some resistance into the circuit will do this job.

The circuit is different now, but it is designed so that a specific voltage will generate a specific current. $$I$$ is not random.

• But can we prove this mathematically? – Schwarz Kugelblitz Aug 22 '19 at 20:13
• What do you want to prove, that $I$ deflects the galvanometer (which you're given), or that the current is constant through the entire circuit? – BowlOfRed Aug 22 '19 at 20:14
• Same value of I deflects the galavanometer before and after it is converted into a voltmeter – Schwarz Kugelblitz Aug 22 '19 at 20:15
• This technique does not convert the galvanometer into a voltmeter (it's still a galvanometer at the end). It makes the circuit into a voltmeter. The specifications on the galvanometer remain true. – BowlOfRed Aug 22 '19 at 20:17
• So it's a physical fact of the system without a mathematical description? – Schwarz Kugelblitz Aug 22 '19 at 20:19