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

Answer

`6.00`

Explanation

Current will flow only through the branch containing resistance `R_(2)`
`therefore I= E/(R_(2) +r) = 5/(4+1)` = 1 amp.
Potential difference across `R_(2)= 1 xx 4 = 4` volt
If q be the charge on each plate of the capacitor, then
`q/C + q/C=4`or `(2q)/C=4`
or `(2q)/(3 xx 10^(-6))=4` or `q = 6 muC`