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

$CE \frac{r_{1}}{(r_{2}+r)}$

$CE \frac{r_{2}}{(r_{}+r_{2})}$

$CE \frac{r_{1}}{(r_{1}+r_{})}$

D) CE

Option B

In steady-state , no current flow through the capacitor, So, resistance becomes ineffective.

So, the current in circuit

$I= \frac{E}{r+r_{2}(Total Resistance)}$

$\because$ Potential drop across capacitor

= Potential drop across r₂ =I r₂ = $\frac{Er_{2}}{r+r_{2}}$

$\therefore$ Stored charge of capacitor . Q= CV

= $CE \frac{r_{2}}{(r_{}+r_{2})}$

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