A) 1.25
B) 8
C) 10
D) 20
Correct Answer: B
Solution :
Let the original resistance is \[R\,\Omega \]. \[\therefore \] \[V=I\,R\] \[V=5\times R=5\,R\] ... (i) When \[2\,\Omega \]. resistance is inserted, then total resistance \[=(R+2)\,\Omega \] \[\therefore \] \[V=I\,(R+2)=4\,(R+2)\] ... (ii) From Eqs. (i) and (ii), we get \[5R=4(R+2)\] \[\therefore \] \[R=8\,\,\Omega \]You need to login to perform this action.
You will be redirected in
3 sec