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  • (a) Briefly explain how a galvanometer is converted into an ammeter. (b) A galvanometer coil has a resistance of and it shows full-scale deflection for a cu

(a) Briefly explain how a galvanometer is converted into an ammeter.

(b) A galvanometer coil has a resistance of 15\Omega and it shows full-scale deflection for a current of 4 mA. Convert it into an ammeter of range 0 to 6 A.

 

 
 
 
 
 

Answers (1)

(a) By connecting a small resistance called Shunt(s) in parallel to the coil of a galvanometer. a galvanometer can be converted into an ammeter.

       

According to formula,

        S=\frac{I_{g}G}{I-I_{g}}

Where, G= resistance

            I_{g}= Current in the galvanometer.

(b) The formula for the shunt is given as :

        S=\frac{I_{g}G}{I-I_{g}}

We have given ;

    G=15\Omega ,\; \; \; Ig=4\times 10^{-3}A

    I=6A ,\; \; \; IgG=(I-I_{g})S

    S=\frac{4\times 10^{-3}\times 15}{6.4\times 10^{-3}}

    S=0.01\Omega

Posted by

Safeer PP

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