A circuit has a section `ABC` if the potential at point `A,B and C` are `V_(1) , V_(2) and V_(3)` respectively , calculate the potential at point `O`

Answer

`[(V_(1))/(R_(1)) + ( V_(2))/(R_(2)) + V_(3))/(R_(3))][(1)/(R_(1)) +(1)/(R_(2)) + (1)/(R_(3))]^(-1)

Explanation

According to kirchhoff's first rule at `O`
`- l_(1) - l_(2) - l_(3) = 0 or l_(1) + l_(2) + l_(3) = 0 `…..(i) If `V_(0)` is the electric potential at `O`
`V_(1) ,V_(2) and V_(3)` be the potential at `A,B and C` respectively then `V_(0) - V_(1) = l_(1) R_(1)`
`V_(0) - V_(2) = l_(2) R_(2) and V_(0) - V_(3) = l_(3) R_(3)`
This , `l_(1) = ((V_(0) - V_(1))/(R_(1)) , l_(2) = ((V_(0) - V_(2))/(R_(2))`
and ` l_(3) = ((V_(0) - V_(3))/(R_(3))`
putting valume of `l_(1) ,l_(2) and l_(3) `in (i)( we get
`(V_(0) - V_(1))/(R_(1)) + (V_(0) - V_(2))/(R_(2)) + V_(V_(0) - V_(3))/(R_(3)) = 0`
or `V_(0) [(1)/(R_(1)) +(1)/(R_(2)) + (1)/(R_(3))]- [(V_(1))/(R_(1)) + ( V_(2))/(R_(2)) + V_(3))/(R_(3))] = 0`
or `V_(0) = [(V_(1))/(R_(1)) + ( V_(2))/(R_(2)) + V_(3))/(R_(3))][(1)/(R_(1)) +(1)/(R_(2)) + (1)/(R_(3))]^(-1)`