The figure shows a network of five resistances and two batteries The current through the $30\,V$ battery is ............... $A$
Diffcult
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$I=I_1+I_2$
in loop $(1)$ and $(2)$
$-30+4 I_1+3 I+2 I_1=0$
$2 I_1+3\left(I_1+I_2\right)=3 0^{10}$
$3 I_1+I_2=10-(1)$
$-15+I_1+3 I+2 I_2=0$
$I_1+3 I_1+3 I_2+2 I_2=15$
$4 I_1+5 I_2=15$
From equation $(1)$ and $(2)$
$I_1=\frac{5 V -15}{15-4}=\frac{35}{11}=3.02$
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