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A circuit shown in the figure consists o...

A circuit shown in the figure consists of a battery of emf `10V` and two capacitance `C_(1)` and `C_(2)` of capacitances `1.0muF` and `2.0muF` respectively. The potential difference `V_(A) - V_(B)` is `5V`

A

Charge on capacitor `C_(1)` is equal to charge on capacitor `C_(2)`

B

Voltage across capacitor `C_(1)` is `5V`.

C

Voltage across capacitor `C_(2)` is `10V`

D

Enegry stored in capacitor `C_(1)` is two times the enegry stored in capacitor `C_(2)`.

Text Solution

Verified by Experts

The correct Answer is:
A, D
`V_(A) - (q)/(1) +10 -(q)/(2) = V_(B)`
`V_(A) - V_(B) +10 = (q)/(1) +(q)/(2)`
`5+10 =(3q)/(2) rArr q = 10muC`

`V_(1) = (10muC)/(1muF) = 10, &V_(2) = (10muC)/(2muF) = 5V`
`U_(1) = (q^(2))/(2xxC_(1)), & U_(2) = (q^(2))/(2C_(2))`
Since `C_(2) = 2C_(1)` so `U_(1) = 2U_(2)`
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