Question

# The necessary current for full scale deflection is $$4mA$$ for any galvanometer of resistance $$99\Omega$$. What would you do to convert it to an ammeter of range $$0$$ to $$6A$$?

Solution

## Given $$G=99\Omega$$Current at full deflection $$=4mA$$Range of Ammeter $$I=6A$$To connect a galvanometer into an ammeter, a low resistance is to added in parallel$$S=\cfrac{{I}_{g}G}{I-{I}_{g}}=\cfrac{4\times {10}^{-3}\times 99}{6-(4\times {10}^{-3})}$$$$S=\cfrac{4\times 99\times {10}^{-3}}{(6000-4)\times {10}^{-3}}=\cfrac{396}{5996}=6.6\times {10}^{-2}\Omega$$PhysicsNCERTStandard XII

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