Determine the potential at point `1` of the circuit shown in Fig., assuming the potential at the point `O` to be equal to zero.

using the symmetry of the formula obtained, write the expressions for the potentials at points 2 and 3.

Explanation

In the loop `ABDEA`, using `-Delta varphi = 0`
`-xi_(3) + (q_(1))/(C_(3)) + (q_(1) + q_(2))/(C_(1)) + xi_(1) = 0` ....(1)
similarly in the loop `ODEF, O`
`(q_(1) + q_(2))/(C_(1)) + xi_(1) - xi_(2) + (q_(2))/(C_(2)) = 0` ....(2)
Solving Eqs. (1) and (2) we get,
`q_(1) + q_(2) = (xi_(2) C_(2) - xi_(1) C_(2) - xi_(1) C_(3) + xi_(3) C_(3))/((C_(3))/(C_(1)) + (C_(2))/(C_(1)) + 1)`
Now, `varphi_(1) - varphi_(0) = varphi_(1) = - ((q_(1) - q_(2)))/(C_(1))`, as `(varphi_(0) = 0)`
`= (xi_(1) (C_(2) + C_(3)) - xi_(0) C_(2) - xi_(3) C_(3))/(C_(1) + C_(2) + C_(3))`
And using the symmetry, `varphi_(2) = (xi_(2) (C_(1) + C_(3)) - xi_(1) C_(1) - xi_(3) C_(3))/(C_(1) + C_(2) + C_(3))`
and `varphi_(3) = (xi_(3) (C_(1) + C_(2)) - xi_(1) C_(1) - xi_(2) C_(2))/(C_(1) + C_(2) + C_(3))`
The answers have wrong sign in the book.