Condensers with capacities` C, 2 C, 3C` and `4C` are charged to the voltage, `V, 2 V, 3 V` and `4 V` correspondingly. The circuit is closed. Find the voltage on all condensers in the equilibrium.

Answer

`-(19)/5V,7/5V` and `(14)/(5)V`.

Explanation

Let a charge `q` flow in clockwise direction after closing the awitch.
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Applying loop law in `abcda`, we get
`(q_(1)-q)/C+(q_(2)-q)/(2C)+(q_(3)-q)/(3C)+(q_(4)-q)/(4C)=0`
Substuting values of `q_(1),q_(2),q_(3)`, and `q_(4),`we get
`q=(24)/5CV`
`|DeltaV|_(c)=|(q_(1)-q)/(C)|=(19)/5V`
`|DeltaV|_(2C)=|(q_(2)-q)/(2C)|=2/5V`
`|DeltaV|_(3C)=|(q_(3)-q)/(3C)|=7/5V`
and `|DeltaV|_(4C)=|(q_(4)-q)/(4C)|=(14)/5V`.