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(A) Let the current through the $30\,V$ battery be $I_1$ and the current through the $15\,V$ battery be $I_2$. The current through the central $3\,\Om

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$3$

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(A) Let the current through the $30\,V$ battery be $I_1$ and the current through the $15\,V$ battery be $I_2$. The current through the central $3\,\Omega$ resistor is $I = I_1 + I_2$ (flowing downwards).Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop $(ABCF)$:$-30 + 4I_1 + 3(I_1 + I_2) + 2I_1 = 0$$9I_1 + 3I_2 = 30 \implies 3I_1 + I_2 = 10$  --- $(1)$Applying $KVL$ to the right loop $(CDEF)$:$-15 + 1I_2 + 3(I_1 + I_2) + 2I_2 = 0$$3I_1 + 6I_2 = 15 \implies I_1 + 2I_2 = 5$  --- $(2)$From $(1)$,$I_2 = 10 - 3I_1$. Substituting into $(2)$:$I_1 + 2(10 - 3I_1) = 5$$I_1 + 20 - 6I_1 = 5$$-5I_1 = -15 \implies I_1 = 3\,A$.The current through the $30\,V$ battery is $3\,A$.

The figure shows a network of five resistances and two batteries. The current through the $30\,V$ battery is ............... $A$.

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