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

Question

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be : (A) CE (B) CE (r1/(r2 + r)) (C) CE (r2/(r + r2)) (D) CE (r1/(r1 + r))

Tardigrade Admin · Accepted Answer

In steady state, flow of current through capacitor will be zero. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/490cfbf61ba66fc0257de0c760d7bc1a-.png" /> i= (E/r + r2) VC = i r2 C = (Er2C/r +r2) VC = CE (r2/r +r2)

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

$CE$

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

$CE$ $\frac{r _{2}}{( r + r _{2} )}$

$CE$ $\frac{r _{1}}{( r _{1} + r )}$

In steady state, flow of current through capacitor will be zero.

$i = \frac{E}{r + r _{2}}$
$V_{C} = i r_{2} C = \frac{E r _{2} C}{r + r _{2}}$
$V_{C} = CE \frac{r _{2}}{r + r _{2}}$

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

CE

CE (r2​+r)r1​​

CE (r+r2​)r2​​

CE (r1​+r)r1​​