JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 9)
For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :
1$$\Omega$$
5$$\Omega$$
4$$\Omega$$
2$$\Omega$$
Explanation
$${I_{\max }} = {{50} \over 2} = 25$$ mA
$$R = {V \over I} = {{50mV} \over {25mA}} = 2\Omega $$
$$R = {V \over I} = {{50mV} \over {25mA}} = 2\Omega $$
Comments (0)
