As shown in the figure, a network of resistors is connected to a battery of $2\,V$ with an internal resistance of $3\,\Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are :
JEE MAIN 2023, Diffcult
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Equivalent resistance of circuit
$R _{ eq } =3+1+2+4+2$
$=12 \Omega$
Current through battery $i =\frac{24}{12}=2 A$
$I _4=\frac{ R _5}{ R _4+ R _5} \times 2=\frac{5}{20+5} \times 2=\frac{2}{5} A$
$I _5=2-\frac{2}{5}=\frac{8}{5} A$
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