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If a galvanometer has full scale deflect...

If a galvanometer has full scale deflection current `I_(g)` and resistance G. A shunt resistance `R_(A)` is used to convert it into an ammeter of range `I_(o)` and a resistance `R_(V)` is connected in series to convert it into a voltmeter of range `V_(0)` such that `V_(0)=I_(0)G` then `R_(A)R_(V)` and `(R_(A))/(R_(V))` respectively are:

A

`R_(A)R_(V)=G^(2)((I_(0)-I_(g))/(I_(g)))` and `(R_(A))/(R_(V))=((I_(g))/(I_(0)-I_(g)))^(2)`

B

`R_(A)R_(V)=G^(2)` and `(R_(A))/(R_(v))=(I_(g))/((I_(0)-I_(g)))`

C

`R_(A)R_(v)=G^(2)` and `(R_(A))/(R_(v))=((I_(g))/(I_(0)-I_(g)))^(2)`

D

`R_(A)R_(V)=G^(2)((I_(g))/(I_(0)-I_(g)))` and `(R_(A))/(R_(V))=((I_(0)-I_(g))/(I_(g)))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • A galvanometer of resistance 15 ohm gives full scale deflection for a current of 2 mA. Calculate the shunt resistance needed to convert it into an ammeter of range 0 to 5 A

    A
    `0.712ohm`
    B
    `1.203ohm`
    C
    `0.006ohm`
    D
    `0.015 ohm`

  • 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.

    A
    `0.99 K ohm`
    B
    `9.9 K ohm`
    C
    `9.8 K ohm`
    D
    `10 K ohm`

  • A galvanometer of resistance 100Omega gives full scale deflection for the current of 0.1 A . The series resistance required to convert it into voltmeter of range 100 V is

    A
    `800Omega`
    B
    `900Omega`
    C
    `700Omega`
    D
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