The current $i_1$ and $i_2$ through the resistors $R_1(=10\,\Omega )$ and ${R_2}\left( { = 30\,\Omega } \right)$ in the circuit diagram with $E_1 = 3\,V$, $E_2 = 3\,V$ and $E_3 = 2\,V$ are respectively
Medium
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$I_{1}=\frac{3-1}{10}=0.2 \,A$
$I_{2}=\frac{3-0}{30}=0.1 \,A$
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