### JEE Main On 16 April 2018 Question 22

##### Question: A galvanometer with its coil resistance $ 25\Omega $ requires a current of 1mA for its full deflection. In order to construct an ammeter to read up to a current of 2A, the approximate value of the shunt resistance should be [JEE Main 16-4-2018]

#### Options:

A) $ 2.5\times {10^{-2}}\Omega $

B) $ 1.25\times {10^{-3}}\Omega $

C) $ 2.5\times {10^{-3}}\Omega $

D) $ 1.25\times {10^{-2}}\Omega $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

Formula: $ S={G^*}I/(I-Ig) $

where S= shunt resistance ,

(I-Ig)= current through the shunt resistance ,

I=Current in the circuit.

Here, galvanometer resistance, G= 25 ohm ;

Ig= 0.001 A , I= 2A $ s=({25^*}0.0001)/(2-0.001) $

$ =1.25\times {10^{-3}}\Omega $

A shunt resistance of $ 1.25\times {10^{-2}}\Omega $ must be connected to make the galvanometer in the range of 0 to 2A