
#1
Jun313, 08:55 PM

P: 296

let R_{g} be the resistance of the galavanometer. A formula in my book states that V=I_{g}(R_{g}+R) my doubt is: here I_{g}(R_{g}+R) is the potential drop across the two resistors of the voltmeter. But how can this include the potential drop by the circuit element across which the voltmeter is connected. 



#2
Jun413, 12:59 AM

Sci Advisor
P: 4,003

While you have the voltmeter in circuit, it is in parallel with whatever it is connected to and may affect that voltage slightly.
There will be a voltage across the voltmeter which is the same as the voltage in the circuit. This voltage causes a current to flow in the galvanometer which is calibrated to give a correct reading for whatever voltage it is connected to. This calibration is done by adjusting the series resistor so that the combined voltmeter resistance causes exactly the right current to flow through the galvanometer for the applied voltage to match the voltage indicated on the voltmeter dial. 



#3
Jun413, 02:09 AM

P: 296

Is the formula correct?




#4
Jun413, 02:19 AM

Sci Advisor
P: 4,003

concept of voltmeter
Yes, the formula is correct.
Why do you have a problem with it? Assuming the resistance of the voltmeter is large compared with the resistance it is being measured across, it will draw sufficient current to make the meter needle deflect and give a voltage reading. Did you read the explanation I gave above? 



#5
Jun413, 02:28 AM

P: 296

I am weak at concepts of current electricity.
Here i am not understanding is i_{g}(R_{g}+R) will give me voltage drop by the resistors of the voltmeter itself. so the voltage measured by this formula will give me the voltage drop by its own resistors.How can i get the voltage drop done by a circuit component by using this formula. 



#6
Jun413, 03:00 AM

Sci Advisor
P: 4,003

The formula is correct, but maybe not appropriate in that form.
If you rearrange it to read Ig = V / (Rg + R) it becomes more useful. The resistors are constant in value, so the current depends only on the voltage. This is a voltmeter so the starting point must be the voltage. As you change the voltage, the current changes and so does the deflection on the galvanometer needle. 



#7
Jun413, 04:09 AM

Sci Advisor
PF Gold
P: 11,341

The only way to measure the V and I reliably in a circuit is to leave the meters in place all the time. If you remove the meters then the Vs and Is all round the circuit will change minutely. 


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