Home
Class 12
PHYSICS
To send 10% of the main current through ...

To send 10% of the main current through a moving coil galvanometer of resistance `99omega`, the shunt required is –

A

`9.9 Omega`

B

`10 Omega`

C

`11 Omega`

D

`9 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
Required resistance which should be added in series so that , the current in the circuit is given by Initial resistance = Final resistane
i.e., `G=((Gs)/(G+s))+R`
`therefore R=G-((GS)/(G+S))=(G^2)/(G+S)`.
Promotional Banner

Topper's Solved these Questions

  • MAGNETIC EFFECT OF ELECTRIC CURRENT

    NIKITA PUBLICATION|Exercise MCQs (Sensitivity and accuracy of M.C.G.)|1 Videos

  • KINETIC THEORY OF GASES & RADIATION

    NIKITA PUBLICATION|Exercise MCQs (Question Given in MHT-CET)|79 Videos

  • MAGNETISM

    NIKITA PUBLICATION|Exercise MCQs|249 Videos

Similar Questions

Explore conceptually related problems

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

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 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

The shunt required for 10% of main current to be sent through the moving coil galvanometer of resistance 99 Omega will be-

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

If 2% of the main current is to be passed through the galvanometer of resistance G , the resistance of shunt required is

A galvanometer of resistance 100ohms is shunted so that only 1//11 of the main current flows through the galvanometer. The resistance of the shunt is

If only 2% of the main current is to be passed through a galvanometer of resistance G , then the resistance of shunt will be

If only 2% of the main current is to be passed through a galvanometer of resistance G , then the resistance of the shunt will be