Two cells are connected between points $A$ and $B$ as shown. Cell $1$ has emf of $12\,V$ and internal resistance of $3\,\Omega$. Cell $2$ has emf of $6\,V$ and internal resistance of $6\,\Omega$. An external resistor $R$ of $4\,\Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be $.............A$.
JEE MAIN 2023, Medium
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$E _{ eq }=\frac{\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\frac{1}{6}}$
$E _{ eq }=6\,V$
$r _{ eq }=2\,\Omega$
$R =4\,\Omega$
So, $i=\frac{6}{2+4}=1\,A$
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