The figure shows a network of four resistors | vedclass

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(A) Let the potential at point $F$ be $V_F = 0\,V$.Applying Kirchhoff's voltage law to the left loop $(ABCF)$:$30 - 4I_1 - 6 - 2I_1 = 0$$24 = 6I_1 \im

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Only $6\,V$

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(A) Let the potential at point $F$ be $V_F = 0\,V$.Applying Kirchhoff's voltage law to the left loop $(ABCF)$:$30 - 4I_1 - 6 - 2I_1 = 0$$24 = 6I_1 \implies I_1 = 4\,A$.Applying Kirchhoff's voltage law to the right loop $(CDEF)$:$15 - 1I_2 - 6 - 2I_2 = 0$$9 = 3I_2 \implies I_2 = 3\,A$.The total current flowing through the $6\,V$ battery is $I = I_1 + I_2 = 4 + 3 = 7\,A$.Since the current $I$ enters the positive terminal of the $6\,V$ battery,it is getting charged.For the $30\,V$ battery,current $I_1$ leaves the positive terminal (discharging).For the $15\,V$ battery,current $I_2$ leaves the positive terminal (discharging).Therefore,only the $6\,V$ battery is getting charged.

The figure shows a network of four resistors and three batteries. Which of the batteries is getting charged?

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