A) 150
B) 170
C) 110
D) 220
Correct Answer: A
Solution :
The deflection of one division of galvanometer is given by die current \[=4\times {{10}^{-4}}A\] or the current sensitivity of galvanometer is \[=4\times {{10}^{-4}}A/div\] \[\therefore \] \[{{i}_{g}}=\]full scale deflection = current sensitivity\[\times \]total number of turns \[=4\times {{10}^{-4}}\times 25={{10}^{-2}}A\] Resistance of galvanometer\[=100\Omega \] The value of resistance required, \[R=\frac{V}{{{i}_{g}}}-G=\frac{2.5}{{{10}^{-2}}}-100\] \[=250-100=150\Omega \]You need to login to perform this action.
You will be redirected in
3 sec