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Hi,
It will be 6V across C1(20µ) and 4V across C2(30µ).
This is because the capacitors act as containers of electrical charges and the voltage shows the level of the charges accumulated, almost similar to cylindrical containers of liquids where the level of the liquid corresponds to the voltage.If two cylinders of different capacity contains same volume of liquid, the level of the liquid in the cylinder with less capacity will be higher than the other. Similarly, when two series connected capacitors of differing capacities are charged from a common supply, the charging current flowing through both remains same, the charges accumulated on both capacitors will be same, but the potential developed across the lower capacitor will be higher than the other.
To be precise, the voltage developed across a capacitor is inversely proportional to its capacity as given by the equation v = ∫idt/c. For the circuit shown, voltage across C1 will be V•C2/(C1+C2) and viceversa.
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