Two cells of emf $2\, E$ and $E$ with internal resistance $r _{1}$ and $r _{2}$ respectively are connected in series to an external resistor $R$ (see $figure$). The value of $R ,$ at which the potential difference across the terminals of the first cell becomes zero is
JEE MAIN 2021, Diffcult
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$i =\frac{3 E }{ R + r _{1}+ r _{2}}$
$TPD =2 E - ir _{1}=0$
$2 E = ir _{1}$
$2 E =\frac{3 E \times r _{1}}{ R + r _{1}+ r _{2}}$
$2 R +2 r _{1}+2 r _{2}=3 r _{1}$
$R =\frac{ r _{1}}{2}- r _{2}$
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