Home
Class 12
PHYSICS
A galvanometer of resistance 90 Omega 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

`1//10`

B

9/100

C

1/100

D

`9//10`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

A galvanometer of resistance 95 Omega shunted by a resistance of 5 Omega gives deflection of 50 divisions when joined in series with resistance of 20 k Omega and 2.0 V accumulator. The current sensitivity of the galvanometer in division per mu A is

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

A resistance of 2Omega is connected in parallel to a galvanometer of resistance 48Omega . Find the percentage of fraction of total current passing through the resistance of 2Omega .

If a galvanometer is shunted by one third of its resistance, find the fraction of total current passing through the galvanometer.

If galvanometer is shunted by (1/n)^ th resistance, then the the fraction of current passing through the galvanometer will be

If galvanometer is shunted by 1//n^(th) of its value, then what fraction of the current passes through the galvanometer ?

A galvanometer has a resistance of 3663Omega . A shunt Sis connected across it such that (1/34) of the total current passes through the galvanometer. Then the value of the shunt is :

A galvanometer has a resistance of 16Omega . It shows full scale deflection when a current of 20mA is passed through it. The only shunt resistance available is 0.06W which is not appropriate to convert galvanometer into ammeters. How much resistance should be connected in series with the coil of galvanometer so that the range of ammeter is 8A?

A cell of e.m.f. 3 V and internal resistance 4 Omega is connected to two resistances of 10 Omega and 24 Omega joined in parallel. Find the current through each resistance using Kirchhoff's laws