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.
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.
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?
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.
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.
The circuit element doesn't count in this. the measured V is the Potential difference across the terminals. If the (total) resistance of the voltmeter is significantly low then the current it draws may affect the voltage situation at that part of the circuit (it is another resistance in parallel with the circuit element). This is why you need a voltmeter to have a high enough internal resistance for any current it takes to be 'negligible'. Remember, measurement of anything will disturb the system. The same problem exists with an Ammeter. All ammeters have some finite series resistance (non-zero) so the current passing through them will introduce a voltage drop, which affects the current.. 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.