Home
Class 12
PHYSICS
We have a galvanometer of resistance 25 ...

We have a galvanometer of resistance `25 Omega`. It is shunted by a `2.5 Omega` wire. The part of total current that flows through the galvanometer is given as

A

`(I)/(I_0)=(1)/(11)`

B

`(I)/(I_0)=(2)/(11)`

C

`(I)/(I_0)=(3)/(11)`

D

`(I)/(I_0)=(4)/(11)`

Text Solution

Verified by Experts

The correct Answer is:
A
`G=25Omega, S=2.5 Omega , (I_g)/(I)=?`
`I_g = ((S)/(S+G))I`
`therefore , (I_g)/(I)=(S)/(S+G)=(2.5)/(2.5+25)=(2.5)/(27.5)=(1)/(11)`
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

A galvanometer of resistance 20Omega is shunted by a 2 Omega resistor. What part of the main current flows through the galvanometer ?

A galvanometer has a resistance 50 Omega and it shunted by a 0.5Omega resistor. The fraction of the main current that flows throught the galvanometer is

A galvanometer has a resistance of 98 Omega .it is shunted by 2Omega resistance. The fraction of the total current passing through the galvanometer is

A galvanometer of resistance 90 Omega is shunted by a resistance of 10 Omega . What fraction of main current passes through the shunt ?

A galvanometer of resistance 36 Omega is changed into an ammeter by using a shunt of 4 Omega . The fraction f of total current passing through the galvanometer is

A galvanometer has resistance 36Omega if a shunt of 4Omega is added with this, then fraction of current that passes through galvanometer is:

A galvanomter of resistance 50Omega is shunted by a resistance of 5Omega . What fraction of the main current passes through the galvanometer? Through the shunt?

When a galvanometer of resistance G is shunted with a low resistance S, then the effective resistance R_(eff) of galvanometer becomes R_(eff)=(GS)/(G+S) If current is passed through such a galvanometer, then the major amount of current flows through shunt and the rest through galvanometer, i.e., the current divides itself in the inverse ratio of resistance. Read the above passage and answer the following questions: (i) Why is the resistance of shunted galvanometer lower than that of shunt? (ii) A galvanometer of resistance 30Omega is shunted by a resistance of 3Omega . What fraction of the main current passes (i) through the galvanometer and (ii) through the shunt? (iii) What are the basic values you learn from the above study?

A 36 Omega galvanometer is shunted by resistance of 4Omega . The percentage of the total current, which passes through the galvanometer is