Home
Class 12
PHYSICS
The resistance of each arm of Wheatstone...

The resistance of each arm of Wheatstone Bridge is `10 Omega`. A resistance of `10 Omega` is connected in series with the galvanometer then equivalent resistance across battery will be

A

`10 Omega`

B

`15 Omega`

C

`20 Omega`

D

`40 Omega`.

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

A battery of internal resistance 3Omega is connected to 20Omega resistor and the potential difference across the resistor is 10V. If another resistor 30Omega is connected in series with the first resistor and battery is again connectecl to the combination, then calculate the e.m.f and terminal potential difference across the combination..

With a resistance R connected in series with a galvanometer of resistance 100 Omega , it acts as a voltmeter of range 0-V. To double the range a resistance of 1000 Omega is to be connected in series with R. Then the value of R is ( Omega )

A galvanometer of resistance 50 Omega gives a full scale deflection for a current 5xx10^(-4) A. A. The resistance that should be connected in series with the galvanometer to read 3 V is

A battery of e.m.f. 12V and internal resistance 2Omega is connected in series with a tangent galvanometer of resistance 4Omega . The deflection is 60^@ when the plane of the coil is along the magnetic meridian. To get a deflection of 30^@ , the resistance to be connected in series with the tangent, galvanometer is :

Three resistances of 4 Omega,5 Omega and 20 Omega are connected in parallel. Their combined resistance is :