In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be .....

Options

A. `CE (r_(2))/((r + r_(2)))`
B. `CE(r_(2))/((r + r_(2)))`
C. CE
D. `CE (r_(1))/((r_(2) + r))`

Answer

A

Explanation

`CE (r_(2))/((r + r_(2)))`

Current flowing in circuit, I = `(E)/(r + r_(2)) `
[ `because` Current through `r_(2)` will be 0 ]
now , V = E - IR
V = E - `(Er)/(r + r_(2))`
` therefore V = (Er + Er_(2) - Er)/(r + r_(2))`
`therefore V = (Er_(2))/(r + r_(2))`
Change on capacitor
Q = CV
= ` (CEr_(2))/(r + r_(2))`
` = CE (r_(2))/(r + r_(2))`