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
  • A${E_1}/{R_1}$
  • B${E_2}/{R_1}$
  • C${E_2}/{R_2}$
  • D${E_1}/({R_2} + {R_1})$
Medium
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