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  • A galvanometer needs 6.0V at full scale deflection and is graded according to its resistance per volt at full scale deflection as <img alt="7000 \Omega \mathrm{V}^{-1}" src="https://entranceco

A galvanometer needs 6.0V at full scale deflection and is graded according to its resistance per volt at full scale deflection as 7000 \Omega \mathrm{V}^{-1}. How will you convert it into a voltmeter that needs 30V at full scale deflection?

Option: 1

171428


Option: 2

174682


Option: 3

17328


Option: 4

48792


Answers (1)

best_answer

Hence, resistance per volt = 7000 \Omega V^{-1}

It means for 1 volt potential difference, the resistance of galvanometer is 7000 \Omega.

So current for full scale deflection I_g=\frac{1}{7000} \mathrm{~A}=0.14 \times 10^{-3} \mathrm{~A}

In order to convert it into a voltmeter of range 0 to 30V, let a resistance R be connected in series with it. Then on applying an extra potential difference = 30V-6V = 24V

The current through it for full scale deflection is again 0.14 \times 10^{-3} \mathrm{~A} 

i.e. I_g=0.14 \times 10^{-3} \mathrm{~A}

R I_g=24 \mathrm{~V} \text { or } R=\frac{24}{I_g}=\frac{24}{0.14 \times 10^{-3}}

=171.428 \times 10^3=171428\Omega

Thus, to convert a given galvanometer (of range 0 to 6V) into a voltmeter (of range 0 to 30V) a resistance of 171428\Omega should be connected in series.

Posted by

Sanket Gandhi

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