Discussion Forum



1)

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

  20102019967_capa.JPG


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

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

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

D) CE

Answer:

Option B

Explanation:

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

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