In full scale deflection current in galvanometer of `100 ohm` resistance is 1 mA. Resistance required in series to convert it into voltmeter of range 10 V.

Options

A. lodoform
B. `4000 Omega`
C. `4600 Omega`
D. `4900 Omega`

Answer

D

Explanation

A galvanometer is converted into a voltmeter by connecting a resistance in series with the galvanometer as show n in the circuit diagram,

where. R is resistance of the resistor connected in series.
Given, galvanometer resistance, `R_(G) = 100 Omega`
Voltmeter range, `V_("max") = 50V` and full deflection current `I_(fl) = 10 mA`
So, by applying the KVL in above circuit diagram,
`V_(AB)=100I_(fl)+RI_(fl)`
`!V_(AB)=(100+R)I_(fl)`
`therefore` For a 50 V voltmeter range there must be,
`V_(AB)=50 V` and `I_(fl)=10 mA`
Now, substituting the values of `V_(AB)` and `I_(fl)` in Eq. (i)
we get,
`50=(100+R)10xx10^(-3)`
`thereforeR=4900 Omega`
Hence, the resistance to be connected in series, to get a voltmeter of range 50 V is 4900 `Omega`.