A galvanometer gives full deflection when a current of 0-3 mA is passed through it- The resistance of galvanometer is 100 - Now to convert it into anammeter- a shunt of 1 m - is connected- The new range of this ammeter is-A- 3 mAB- 30 mAC- 3 AD- 30 A

The correct option is D 30A (n 1) i_g.S = i_g.G S =G/n 1 n 1 nbsp; =100/10^ 3 =10^5 n = 10^5 new range = n old range =10^5 × 0.3 × 10^ 3 =30 A ...

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Byju's Answer

Standard XII

Physics

Ammeter

A galvanomete...

Question

A galvanometer gives full deflection when a current of 0.3 mA is passed through it. The resistance of galvanometer is $100 Ω$ . Now to convert it into an ammeter, a shunt of $1 m Ω$ is connected. The new range of this ammeter is:

3mA

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30mA

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30A

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Solution

The correct option is D 30A

$(n - 1) i_{g} . S = i_{g} . G$
$S = \frac{G}{n - 1}$
$n - 1 = \frac{100}{10^{- 3}} = 10^{5}$
$n = 10^{5}$
new range = n {old range}
$= 10^{5} \times 0.3 \times 10^{- 3} = 30 A$

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A galvanometer gives full deflection when a current of 0.3 mA is passed through it. The resistance of galvanometer is 100Ω. Now to convert it into an ammeter, a shunt of 1mΩ is connected. The new range of this ammeter is:

The correct option is D 30A (n−1)ig.S=ig.G S=Gn−1 n−1=10010−3=105 n=105 new range = n {old range} =105×0.3×10−3=30A

A galvanometer gives full deflection when a current of 0.3 mA is passed through it. The resistance of galvanometer is $100 Ω$ . Now to convert it into an ammeter, a shunt of $1 m Ω$ is connected. The new range of this ammeter is:

The correct option is D 30A

$(n - 1) i_{g} . S = i_{g} . G$
$S = \frac{G}{n - 1}$
$n - 1 = \frac{100}{10^{- 3}} = 10^{5}$
$n = 10^{5}$
new range = n {old range}
$= 10^{5} \times 0.3 \times 10^{- 3} = 30 A$