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# A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions the resistance in series should be:

A
4450Ω
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B
5050Ω
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C
5550Ω
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D
6050Ω
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Solution

## The correct option is A 4450ΩA galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.Let V be the potential difference to be measured by galvanometer.To do so, a resistance R of such a value is connected in series with the galvanometer so that if a potential difference V is applied across the terminals A and B, a current Ig flows through the galvanometer.Now, total resistance of voltmeter =G+RFrom Ohm's law, Ig=VG+Ror G+R=VIgor R=VIg−GHere,The current through the galvanometer=32950+50=10−3ATo reduce the deflection from 30 divisions to 20 divisions, the required current=2030×10−3=23×10−3AThe required resistance R is given by3R+50=23×10−3or R=4450Ω   Suggest Corrections  0      Similar questions  Related Videos   Series and Parallel Combination of Resistors
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