Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentEasy2 minQB

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In an ammeter, $5%$ of the main current passes through the galvanometer. If resistance of the galvanometer is $G,$ the resistance of ammeter will be:

Answer:

Solution:

As potential drop across both branches would be the same, we get $I_{S} S = I_{g} G$

$(\frac{95}{100} I) S = (\frac{5 I}{100}) G$

$\Rightarrow S = \frac{G}{19}$

Now, resistance of ammeter will be

$R_{A} = \frac{SG}{S + G} = \frac{\frac{G ^{2}}{19}}{\frac{20 G}{19}}$

$\Rightarrow R_{A} = \frac{G}{20}$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument

⏱ 2mℹ️ Source: QB

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