Home
Class 12
PHYSICS
A galvanometer of resistance 100ohms 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

A

1 ohm

B

11 ohms

C

10 ohms

D

9 ohms

Text Solution

Verified by Experts

The correct Answer is:
C
`S = (G)/( (i)/(i_g) - 1) `
Promotional Banner

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(H.W) (AMPERE.S CIRCUITAL LAW BIOT-SAVART LAW AND ITS APPLICATIONS|7 Videos

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(H.W) (SOLENOID AND TOROID)|1 Videos

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(C.W) (MOVING COIL GALVANOMETER)|2 Videos

  • MAGNETISM AND MATTER

    NARAYNA|Exercise EXERCISE - 4 (SINGLE ANSWER TYPE QUESTION)|17 Videos

  • NUCLEAR PHYSICS

    NARAYNA|Exercise LEVEL-II-(H.W)|9 Videos

Similar Questions

Explore conceptually related problems

If a shunt is to be applied to a galvanometer of resistance 50Omega so that only 5% of total current passes through the galvanometer. The resistance of shunt should be

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

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

A galvanometer of resistance 20Omega is to be shunted so that only 1% of the current passes through it. Shunt connected is 99//20Omega

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

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 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

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