A galvanometer having coil resistance shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be:

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Q.
A galvanometer having coil resistance $10 Ω$ shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be: [2024]

1 $3 \times 10^{- 3} Ω$

2 $4.85 \times 10^{- 3} Ω$

3 $3.75 \times 10^{- 3} Ω$

4 $2.75 \times 10^{- 3} Ω$

Ans.
(3)

Given $G = 10 Ω, I_{g} = 3 mA and I = 8 A$

In case of conversion of galvanometer into ammeter.

We have $I_{g} G = (I - I_{g}) S$

$S = \frac{I_{g} G}{I - I_{g}} = \frac{(3 \times 10^{- 3}) \times 10}{8 - 0.003} = 3.75 \times 10^{- 3} Ω$

Solution

Q.
A galvanometer having coil resistance $10 Ω$ shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be: [2024]

1 $3 \times 10^{- 3} Ω$

2 $4.85 \times 10^{- 3} Ω$

3 $3.75 \times 10^{- 3} Ω$

4 $2.75 \times 10^{- 3} Ω$

Ans.
(3)

Given $G = 10 Ω, I_{g} = 3 mA and I = 8 A$

In case of conversion of galvanometer into ammeter.

We have $I_{g} G = (I - I_{g}) S$

$S = \frac{I_{g} G}{I - I_{g}} = \frac{(3 \times 10^{- 3}) \times 10}{8 - 0.003} = 3.75 \times 10^{- 3} Ω$

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Solution

Q. A galvanometer having coil resistance 10Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be: [2024]

1 3×10-3Ω

2 4.85×10-3Ω

3 3.75×10-3Ω

4 2.75×10-3Ω

Ans. (3) Given G=10Ω, Ig=3 mA and I=8A In case of conversion of galvanometer into ammeter. We have IgG=(I-Ig)S S=IgGI-Ig=(3×10-3)×108-0.003=3.75×10-3Ω

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Q.
A galvanometer having coil resistance $10 Ω$ shows a full scale deflection for a current of 3 mA. For it to measure a current of 8A, the value of the shunt should be: [2024]

1 $3 \times 10^{- 3} Ω$

2 $4.85 \times 10^{- 3} Ω$

3 $3.75 \times 10^{- 3} Ω$

4 $2.75 \times 10^{- 3} Ω$

Ans.
(3)

Given $G = 10 Ω, I_{g} = 3 mA and I = 8 A$

In case of conversion of galvanometer into ammeter.

We have $I_{g} G = (I - I_{g}) S$

$S = \frac{I_{g} G}{I - I_{g}} = \frac{(3 \times 10^{- 3}) \times 10}{8 - 0.003} = 3.75 \times 10^{- 3} Ω$