question_answer

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are\[{{R}_{1}}\]and\[{{R}_{2}}({{R}_{2}}>{{R}_{1}})\].   If   the   potential difference across the source having internal resistance\[{{R}_{2}}\]is zero, then          [AIEEE 2005]

A) \[R=\frac{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}{({{R}_{2}}-{{R}_{1}})}\]

B) \[R={{R}_{2}}-{{R}_{1}}\]

C) \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\]

D) \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{2}}-{{R}_{1}})}\]

Correct Answer: B

Solution :

[b] The equivalent resistance of the circuit

\[l=\frac{2E}{{{R}_{1}}+{{R}_{2}}+R}\]

According to the question,

\[-({{V}_{1}}-{{V}_{B}})\,=E-l\,{{R}_{2}}\]

\[0=E-l\,{{R}_{2}}\]

\[E=\,l{{R}_{2}}\]