in an ammeter 0.2% of main current passes through tha galvanometer. if resistance of galvanometer is G the resistance of ammeter will be.......✌️
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in an ammeter 0.2% of main current passes through tha galvanometer. if resistance of galvanometer is G the resistance of ammeter will be.......✌️
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.
please choose as brainliest
Verified answer
Here, resistance of the galvanometer = G
Current through the galvanometer,
[tex]I_{G} = 0.2\% \: of \: I\: = \frac{0.2}{100} I = \frac{1}{500} I[/tex]
•: Current through the shunt
[tex] = > I_{S} = I - I_{G} \\ = > I_{S} = I- \frac{1}{500} I = \frac{499}{500} I[/tex]
As shunt ans galvanometer are in parallel,
[tex] I_{G}G= I_{S}S[/tex]
[tex] (\frac{1}{500} I)G = ( \frac{499}{500} )sS= > S= \frac{g}{499} \\ [/tex]
Resistance of ammeter [tex] R_{A}[/tex]is
[tex] = > \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[/tex]
Hope its help uh❤