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In Fig, E =5 volt , r = 1 Omega, R(2) =...

In Fig, `E =5` volt , `r = 1 Omega, R_(2) = 4 Omega, R_(1) = R_(3) = 1 Omega` and `C = 3 muF`. Then the numbercal value of the charge on each plate of the capacitor is

A

3`muC`

B

6`muC`

C

12`muC`

D

24`muC`

Text Solution

Verified by Experts

The correct Answer is:
2
At steady state, No current will flow through capacitor, So I = `E/(R_(2)+r)=5/(4+1)=1A`
Potential difference a cross capacitor = `ixxR_(2)`
`=1xx4=4v`
Charge on each capacitor =`(C/2)xx4`
`=(3/2)xx4=6muC`
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