A galvanometer coil has a resistance of `12Omega` and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V ?

Options

A. 6012
B. 5980
C. 5985
D. 5988

Answer

D

Explanation

Here, current capacity of galvanometer is,
`I_(g)=3xx10^(-3)A`
Series resistance required to convert above galvanometer into voltmeter of voltage capacity
V = 18 volt is,
`R_(S)=V/I_(g)-G`
`thereforeR_(S)=18/(3xx10^(-3))-12=5988Omega`
Thus, we should connect `5988Omega` resistance in series with above galvanometer and then we should calibrate its scale from 0 to 18 volt in order to prepare required voltmeter.