A) 1.25
B) 8
C) 10
D) 20
Correct Answer: B
Solution :
Let the original resistance is \[R\Omega .\] \[\therefore \]\[V=IR\] \[V=5\times R=5R\] ...(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.
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