In the given circuit diagram when the current reaches steady stte in the circuit, the charge on the capacitor of capacitance C will be:-

Options

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

Answer

B

Explanation

In steady state, no current flows through the capacitor So, resistor `r_(P)` becomes ineffective. So, the current in circuit,
`I=(E)/(r+r_(2)("Total resistance"))because ` Potential drop across capacitor=Potential drop across `r_(2)=Ir_(2)=(E_(2))/(r+r_(2))`
`therefore`Stored charge of capacitor `Q=CV=CE(r_(2))/(r+r_(2))`.