The value of current in the $6 \,\Omega$ resistance is $....\,A$
JEE MAIN 2021, Diffcult
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Applying $KCL$ at point $P,$
$\frac{V-0}{6}+\frac{V-90}{5}+\frac{V-140}{20}=0$
$\Rightarrow 10 {V}+12 {V}-1080+3 {V}-420=0$
$\Rightarrow {V}=60$
$\therefore \text { Current in } 6 \Omega=\frac{{V}-0}{6}=10 {A}$
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