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# Can someone explain - Current Electricity - JEE Main-4

A galvanometer of resistance $100\Omega$ has 50 divisions on its scale and has sensitivity of $20 \mu A/$ division. It is to be converted to a voltmeter with three ranges, of $0-2V$$0-10V$ and $0-20V$. The appropriate circuit to do so is :

• Option 1)

$R_{1}=2000\Omega$

$R_{2}=8000\Omega$

$R_{3}=10000\Omega$

• Option 2)

$R_{1}=19900\Omega$

$R_{2}=9900\Omega$

$R_{3}=1900\Omega$

• Option 3)

$R_{1}=1900$

$R_{2}=8000$

$R_{3}=10000$

• Option 4)

$R_{1}=1900$

$R_{2}=9900$

$R_{3}=19900$

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Conversion of galvanometer into voltameter -

Connected a large Resistance $R$  in series

- wherein

$Ig =50\times20=1mA$

option (2) can not ever be appropriate circuit because we can not measure high voltage i.e 20V before smaller voltage i.e 2V, because if the higher voltage is at the end then the resistance is large and hence more range.

Now checking option (1), (2), & (3)

$Ig(100+R_{1})=2V$

$100+R_{1}=2\times10^{3}$

$\Rightarrow R_{1}=1900\Omega ...................$option (1) is false

$\Rightarrow Ig (100+R_{1}+R_{2})=10V$

$\Rightarrow R_{2}=10000-100-1900$

$R_{2}=8000$

$\Rightarrow Ig(100+R_{1}+R_{2}+R_{3})=20V$

$R_{3}=20000-8000-1900-100$

$R_{3}=10000\Omega$

Option 1)

$R_{1}=2000\Omega$

$R_{2}=8000\Omega$

$R_{3}=10000\Omega$

Option 2)

$R_{1}=19900\Omega$

$R_{2}=9900\Omega$

$R_{3}=1900\Omega$

Option 3)

$R_{1}=1900$

$R_{2}=8000$

$R_{3}=10000$

Option 4)

$R_{1}=1900$

$R_{2}=9900$

$R_{3}=19900$

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