A) R
B) 2R
C) \[\frac{R}{4}\]
D) \[\frac{R}{2}\]
Correct Answer: A
Solution :
Key Idea: The balanced condition of Wheatstone's bridge is \[\frac{{{R}_{AB}}}{{{R}_{BC}}}=\frac{{{R}_{AD}}}{{{R}_{DC}}}\] As bridge is in balanced condition, no current will flow through BD. \[{{R}_{1}}={{R}_{AB}}={{R}_{BC}}\] \[=R+R=2R\] \[{{R}_{2}}={{R}_{AD}}+{{R}_{CD}}+R+R=2R\] Also,\[{{R}_{1}}\]and\[{{R}_{2}}\]are in parallel combination. Hence, equivalent resistance between A and C will be \[\therefore \] \[{{R}_{eq}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{4{{R}^{2}}}{4R}=R\]You need to login to perform this action.
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