question_answer

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is:                            [AIPMT 2003]

A)  R        

B)                   2R                  

C)  \[\frac{R}{4}\]             

D)       \[\frac{R}{2}\]

Correct Answer: A

Solution :

[a] Key Idea: The balanced condition of Wheat-stone’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.

\[{{R}_{1}}={{R}_{AB}}+{{R}_{BC}}\]

\[=R+R=2R\]

\[{{R}_{2}}={{R}_{AD}}+{{R}_{CD}}=R+R=2R\]

\[{{R}_{1}}\] and \[{{R}_{2}}\] are in parallel combination.

Hence, equivalent resistance between A and C will be.

\[\therefore \]            \[{{R}_{eq}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{4{{R}^{2}}}{4R}=R\]