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To send 10% of the main current through ...

To send `10%` of the main current through a moving coil galvanometer of resistance `9Omega`, shunt S required is

A

`81Omega`

B

`1Omega`

C

`10Omega`

D

`9Omega`

Text Solution

Verified by Experts

The correct Answer is:
B

For Parallel connection, potential difference across the node is same, `i_g xx G = (i- i_g) xx S,` where `i_g`, G, i&S represent current in galvanometer, galvanometer resistance, total current and shunt resistance respectively.
`rArr S = (0.1)/(0.9) xx G { i_g = 0.1i}`
`rArr S = 1/9 xx 9 = 1 Omega`
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Knowledge Check

  • If 10% of the main current is to be passsed through a maving coil galvanometer of resistance 99 Omega , then the ratio of its resistance and the shunt resistance will be

    A
    `1:9`
    B
    `9:1`
    C
    `1:11`
    D
    `11:1`

  • A moving coil galvanometer has a resistance of 990Omega. in order to send only 10% of the main currect through this galvanometer, the resistance of the required shunt is

    A
    `0.9 Omega`
    B
    `110 Omega`
    C
    `405 Omega`
    D
    `90 Omega`

  • The resistance of the shunt required to allow 2% of the main current through the galvanometer of resistance 49Omega is

    A
    `1Omega`
    B
    `2Omega`
    C
    `0.2Omega`
    D
    `0.1Omega`

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