A) 1A
B) \[2\sqrt{3}\text{A}\]
C) 4A
D) 6 A
Correct Answer: D
Solution :
According to the principle of tangent, galvanometer \[I=k\,\tan \theta \] or \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\tan \,{{\theta }_{1}}}{\tan \,{{\theta }_{2}}}\] or \[{{I}_{2}}=\left( \frac{\tan {{\theta }_{2}}}{\tan {{\theta }_{1}}} \right){{I}_{1}}\] Given \[{{I}_{1}}=2A,\] \[{{\theta }_{1}}={{30}^{o}},\] \[{{\theta }_{2}}={{60}^{o}}\] \[\therefore \] \[{{I}_{2}}=\left( \frac{\tan {{60}^{o}}}{\tan {{30}^{o}}} \right)2=6A\]
You need to login to perform this action.
You will be redirected in
3 sec
