Physics - Current Electricity Question with Solution | TestHub

Applying $KCL$ at point $P,$

$\frac{V-0}{6}+\frac{V-90}{5}+\frac{V-140}{20}=0$
$\Rightarrow 10 V+12 V-1080+3 V-420=0$
$\Rightarrow V=60$
$\therefore$ Current in $6 \mathrm{\Omega }=\frac{\mathrm{V}-0}{6}=10 \mathrm{A}$
Hence, option $3.$