The figure shows a network of four resistances and three batteries Mark the incorrect statement.
Diffcult
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Apply Kirchhoffs law$:$
For left loop $BCFAB,$ $4 I_{1}+2 I_{1}=30-6$ or $6 I_{1}=24 \Rightarrow I_{1}=4 A$
In this equation on $I_{2}$ term so the current flowing in the left loop is independent of the right loop.
For right loop $DCFED,$ $1 I_{2}+2 I_{2}=15-6$ or $3 I_{2}=9 \Rightarrow I_{2}=3 A$
Here also no $I_{1}$ term so The current flowing in the right loop is independent of the left loop.
Net current in through $6 \mathrm{V}$ battery is: $I=I_{1}+I_{2}=4+3=7 \mathrm{A}$
Thus, the $6 \mathrm{V}$ battery is getting charged. Both $15 \mathrm{V}$ and $30 \mathrm{V}$ batteries produce current in $6 V$ battery.
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