Question

in an ammeter 0.2% of main current passes through tha galvanometer. if resistance of galvanometer is G the resistance of ammeter will be.......✌️​

Answer 1

The resistance of ammeter will be 1/499 G.

Explanation:

For ammeter:

0.0002 l x G = 0.9981 x rg

rg = 0.002 / 0.998 G

That is

rg = 0.002004

   = 1/499 G

Hence the resistance of ammeter will be 1/499 G.

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Answer 2

Here, resistance of the galvanometer = G

Current through the galvanometer,

$I_{G} = 0.2\% \: of \: I\: = \frac{0.2}{100} I = \frac{1}{500} I$

•: Current through the shunt

$= > I_{S} = I - I_{G} \\ = > I_{S} = I- \frac{1}{500} I = \frac{499}{500} I$

As shunt ans galvanometer are in parallel,

$I_{G}G= I_{S}S$

$(\frac{1}{500} I)G = ( \frac{499}{500} )sS= > S= \frac{g}{499}$

Resistance of ammeter $R_{A}$is

$= > \frac{1}{ R_{A}} = \frac{1}{G} + \frac{1}{S} \\ = > \frac{1}{ R_{A} } = \frac{1}{G} + \frac{1}{ \frac{G}{499} } = \frac{500}{G} \\ = > R_{A} = \frac{1}{500} G$

Hope its help uh❤