The current sensitivity of voltmeter of moving coil type like PMMC voltmeter is defined as the deflection per unit current in the coil. If Ɵ be the angular deflecting of needle of voltmeter for current I through the moving coil, then sensitivity of voltmeter is given as
S = Ɵ / I
You might think that, why we are expressing the sensitivity of voltmeter in terms of current through the moving coil. For getting answer to this question, I would suggest you to read the basics of PMMC type instrument.
As the deflecting torque in PMMC type instrument is given as
Td = (NBLd)I
Where N = Number of turns
L = Length of moving coil
d = width of moving coil
For a given instrument, number of turns in the coil, length and width of moving coil and magnetic flux density are constant, therefore we can assume that
G = NBLd = Constant
Therefore, deflecting torque Td = GI
But this deflecting torque is controlled by spring torque, hence
K Ɵ = Td
KƟ = GI where K is spring constant.
Ɵ / I = G/K
Sensitivity of voltmeter S = Ɵ / I = G/K
Thus the sensitivity of voltmeter of moving coil type can be increased either by increasing the value of G or by reducing the value of K. Now, for a given coil of given cross-sectional area i.e. A = Ld and given magnetic flux density, the value of G can be increased by using many number of turns of thin wire coil. Again, the value of spring constant can be reduced by using light flat spring and the coil assembly lightly pivoted. Thus it can be stated that, the more the sensitivity of voltmeter the more will be the resistance of its coil as many number of turns are wound with fine wire.
The sensitivity of voltmeter specified on the meter dial specifies the resistance of the meter for one volt range. The sensitivity of voltmeter is also defined as
S = 1/Ifs
= 1/Im Ω/V
Where Ifs = Current required for full scale deflection
If the sensitivity of voltmeter is given and you need to find the total resistance of the meter then you multiply the sensitivity with the voltage range to get the resistance. The sensitivity of voltmeter also depend on the circuit resistance which is known as loading effect. To better understand the sensitivity of voltmeter, let us consider a simple example.
Meter A have a range of 0-10 V and multiplier resistance of 18 kΩ. Meter B has a range of 0-300 V and multiplier resistance of 298 kΩ. Which meter is more sensitive assuming both meters movement have the same resistance of 2 kΩ?
Total resistance of meter = 18 + 2 = 20 kΩ
Range of Meter = 10 V
Thus current required for full scale deflection = 10 / 20 = 0.5 A
Therefore sensitivity = 1/0.5 = 2 Ω/V
Total resistance of meter = 298 + 2 = 300 kΩ
Range of Meter = 300 V
Thus current required for full scale deflection = 300 / 300 = 1 A
Therefore sensitivity = 1/1 = 1 Ω/V
Thus we observe that meter A is more sensitive as it requires less current for a given deflection. Here we have assumed full scale deflection.