Question

Find the charge $Q$ on the capacitor at time $t$ .

Answer 1

We can use the formula of charge on a capacitor as product of capacitance and voltage across it, and then using the equation of potential balance across each element of circuit, we will get a differential equation, and by solving it, we can find charge as a function of time.

Complete step by step answer:
Equivalent resistance with series in capacitor is $\dfrac{{R \times R}}{{R + R}} + R$
On solving this, we can equivalent resistance $\operatorname{Re}$ as
${\text{Re = }}\dfrac{{\text{R}}}{2} + {\text{R}}$
On further simplification, we get,
$\operatorname{Re} = \dfrac{{3R}}{2}$
Now, equation of potential balance, across each element of circuit will be,
$V - {V_C} - {V_R} = 0$
Where,
${V_C}$ is voltage across capacitance, that is, $\dfrac{q}{C}$ ( $q$ is charge at time $t$ and $C$ is capacitance )
${V_R}$ is voltage across resistance, that is, $\operatorname{Re} I$ ( $\operatorname{Re}$ is equivalent resistance, $I$ is current at time $t$ )
On further simplification we get,
$V = \dfrac{q}{C} + \operatorname{Re} I$
Now current is rate of change of charge, that is, $\dfrac{{dq}}{{dt}}$, so we get,
$V = \dfrac{q}{C} + \operatorname{Re} \dfrac{{dq}}{{dt}}$
On further simplification, we get,
$\dfrac{{dt}}{{\operatorname{Re} C}} = \dfrac{{dq}}{{CV - q}}$
On integrating both sides and applying limits in time from $0$ to $t$ and on charge from $0$ to $q$, we get,
$\int_0^t {\dfrac{{dt}}{{\operatorname{Re} C}}} = \int_0^q {\dfrac{{dq}}{{CV - q}}}$
On integration we get,
$\left( {\dfrac{t}{{\operatorname{Re} C}}} \right)_0^t = \left( { - \ln (CV - q)} \right)_0^q$
On solving limits, we get,
$\left( { - \dfrac{t}{{\operatorname{Re} C}}} \right) = \left( {\ln (CV - q) - \ln (CV)} \right)$
So we get,
$\dfrac{{ - t}}{{\operatorname{Re} C}} = \ln \left( {\dfrac{{CV - q}}{{CV}}} \right)$
On further simplifications, we get,
$q = CV\left( {1 - {e^{\dfrac{{ - t}}{{\operatorname{Re} C}}}}} \right)$
Where $\operatorname{Re} = \dfrac{{3R}}{2}$, $CV$ is initial charge (if we denote it by ${q_0}$ ) then we get,
$q = {q_0}\left( {1 - {e^{\dfrac{{ - 2t}}{{3RC}}}}} \right)$

Note: We got $\operatorname{Re} C$ in the expression of charge, this is known as Time Constant of the RC Circuit. This time constant has a role of opposing the change in charge of the capacitor. This can be denoted as the inertia of the circuit. It means, greater the time constant, slower will be the charging rate of the capacitor.