A) 100 \[\Omega \]
B) 50\[\Omega \]
C) 25 \[\Omega \]
D) 75\[\Omega \]
Correct Answer: C
Solution :
The given circuit can be drawn as follows : Since, ratio of resistances in the opposite arms of the bridge is same, hence, it is a balanced Wheatstone bridge. Therefore, arm BD is ineffective. The circuit now reduces as follows: We have two resistors of\[2R\,\Omega \]. Each connected in parallel. Hence, resultant resistance is \[\frac{1}{R}=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}\]\[\Rightarrow \] \[R-R=25\,\Omega \]You need to login to perform this action.
You will be redirected in
3 sec