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Q. In the following network, the potential at $O$ is
Question

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
Let the potential at O is $V_{0}.$
Application of Kirchhoff's first law at junction O gives
$\frac{8 - V_{0}}{2}=\frac{V_{0} - 4}{4}+\frac{V_{0} - 2}{2}$
$=\frac{V_{0} - 4 + 2 V_{0} - 4}{4}$
$\frac{4 \left(8 - V_{0}\right)}{2}=V_{0}-4+2V_{0}-4$
$16-2V_{0}=3V_{0}-8$
$16+8=5V_{0}$
$V_{0}=\frac{24}{5}=4.8 \, V$