The figure shows a network of four resistances and three batteries Which of the battery is getting charged.
Diffcult
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Left loop$:$
$V_{F}+6+4 I_{1}-30+2 I_{1}=V_{F}$
$I_{4}=4 A$
Right loop$:$
$V_{F}+6+1 I_{2}-15+2 I_{2}=V_{F} 30 V$
or $I_{2}=3 A$ Net current through $6 \mathrm{V}$ battery is
$I=I_{1}+I_{2}=4+3=7 A$
The $6 V$ battery is getting charged. Both $15 \mathrm{V}$ and $30 \mathrm{V}$ batteries produce current in $6 \mathrm{V}$ battery.
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