In the given figure, the value of $V_{0}$ will be $.....V$.
A$3$
B$4$
C$40$
D$8$
JEE MAIN 2022, Diffcult
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B$4$
b By nodal analysis $\frac{V_{0}-2}{1 k \Omega}+\frac{V_{0}-4}{1 k \Omega}+\frac{V_{0}-6}{1 k \Omega}=0$
$3 V _{0}-12=0$
$V _{0}=4$
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