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In the circuit shown in fig, the energy ...

In the circuit shown in fig, the energy stored in both capacitors is `U_(1)`. If swich `S` is opened and a dielectric slab of constant 5 is put in free spaces of the capacitors, the energy stored is found to be `U_(2)`. Calcualte `U_(1)//U_(2).

Text Solution

Verified by Experts

The correct Answer is:
`5//13`
In Fig, when `S` is closed,
`U_(1) = (1)/(2) CV^(2) = CV^(2)`
when `S` is open and electric is introduced.
Capacity of each condenser `= KC = 5 C`.
Condenser to left is connected to battery. Its potential remains `V`. But condenser on right is not connected to battery. So on introducing dielectric its potential becomes `(V)/(5)`
`:.` Total energy,
`U_(2) = (1)/(2) (5 C) V^(2) + (1)/(2) (5C) ((V)/(5))^(2)`
`= (1)/(2) (5 C) V^(2) (1 + (1)/(25))`
`U_(2) = (5 CV^(2))/(2) xx (26)/(25) = (13)/(5) CV^(2)`
`:. (U_(1))/(U_(2)) = (CV^(2))/((13)/(5) CV^(2)) = (5)/(13)`
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