question_answer

Two resistances \[{{R}_{1}}\] and \[{{R}_{2}}\] are joined as shown in the figure to two batteries of e.m.f. \[{{E}_{1}}\] and \[{{E}_{2}}\]. If \[{{E}_{2}}\] is short-circuited, the current through \[{{R}_{1}}\] is           [NDA 1995]

A)                    \[{{E}_{1}}/{{R}_{1}}\]

B)                    \[{{E}_{2}}/{{R}_{1}}\]

C)                    \[{{E}_{2}}/{{R}_{2}}\]

D)                    \[{{E}_{1}}/({{R}_{2}}+{{R}_{1}})\]