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Q.
Find the current through $10\, \Omega$ resistor:-
Solution:
$KCL$ at point $A$
$I _{1}+ I _{2}+ I _{3}=0$
$\Rightarrow \frac{ X -3}{10}+\frac{ X -0}{6}+\frac{ X -4.5}{3}=0$
$\Rightarrow \frac{3 x -9+5 x +10 x -45}{30}=0$
$\Rightarrow 18 x =54$
$\Rightarrow X =3$ volt
$I _{1}=\frac{ X -3}{10}=0 A$
