On solving an RC circuit Given the following RC circuit:

it's required to calculate how the electric current varies and how the other electrical quantities vary.
Based on that request, I was wondering if it was correct to consider this other RC circuit:

therefore, as usual, distinguish between:

*

*charge process: $i(t) = \frac{\Delta V}{R_{eq}}e^{-\frac{t}{R_{eq}\,C}}$ and $q(t) = C\,\Delta V\left(1 - e^{-\frac{t}{R_{eq}\,C}}\right)$;


*discharge process: $i(t) = -\frac{\Delta V}{R_{eq}}e^{-\frac{t}{R_{eq}\,C}}$ and $q(t) = C\,\Delta V\,e^{-\frac{t}{R_{eq}\,C}}$;
or is this simplification not correct and it's necessary to solve the initial circuit through differential equations?
In the latter case if I was given some help on how to do it I would be happy, thanks!

EDIT:
Thanks to the clear answers of @Ritam_Dasgupta and @Señor O, I managed to write the associated Cauchy problem:
$$
\begin{cases}
\epsilon = R_1\,\dot{q}_1(t) + \frac{q(t)}{C} \\
\epsilon = R_2\,\dot{q}_2(t) + \frac{q(t)}{C} \\
\epsilon = R_3\,\dot{q}_3(t) + \frac{q(t)}{C} \\
\dot{q}(t) = \dot{q}_1(t) + \dot{q}_2(t) + \dot{q}_3(t) \\
q(0) = q_1(0) = q_2(0) = q_3(0) = \overline{q}
\end{cases}
$$
which can be declined in the two classic cases:

*

*charge process: $\epsilon = \Delta V$, $\overline{q} = 0$, then:
$$
\begin{aligned}
& q(t) = C\,\Delta V\left(1 - e^{-\frac{t}{R_{eq}\,C}}\right),
\quad \quad \quad \,
i(t) = \frac{\Delta V}{R_{eq}}\,e^{-\frac{t}{R_{eq}\,C}}\,; \\
& q_k(t) = r_k\,q(t),
\quad \quad \quad \quad \quad \quad \quad \quad \quad
i_k(t) = r_k\,i(t)\,; \\ 
& U_{vs} = \frac{q^2(t)}{C}\,, 
\quad \quad
U_C = \frac{q^2(t)}{2\,C}\,,
\quad \quad
U_R = \frac{q^2(t)}{2\,C}\,; \\
\end{aligned}
$$


*discharge process: $\epsilon = 0$, $\overline{q} = C\,\Delta V$, then:
$$
\begin{aligned}
& q(t) = C\,\Delta V\,e^{-\frac{t}{R_{eq}\,C}},
\quad \quad \quad \quad \quad \quad \quad \quad
i(t) = -\frac{\Delta V}{R_{eq}}\,e^{-\frac{t}{R_{eq}\,C}}\,; \\
& q_k(t) = (1 - r_k)\,C\,\Delta V + r_k\,q(t),
\quad \quad \quad \,
i_k(t) = r_k\,i(t)\,; \\
& U_{vs} = 0\,, 
\quad \quad \quad \quad
U_C = \frac{q^2(t)}{2\,C}\,,
\quad \quad \quad \quad
U_R = \frac{q^2(t)}{2\,C}\,; \\
\end{aligned}
$$
where $r_k = \frac{R_{eq}}{R_k}$ and $R_{eq} = \left(\begin{aligned}\sum_{k = 1}^3\end{aligned} \frac{1}{R_k}\right)^{-1}$.
I hope I haven't made any mistakes, in case I will correct.
 A: Yes, what you did is most certainly correct. Now if you have to find out currents in individual branches you just have to notice that since potential difference across each resistor is the same the current passing through a resistor would be inversely proportional to its resistance.
You can verify this by writing the differential equation itself. Consider the current in the branches to be $i_1$, $i_2$ and $i_3$ respectively such that $i_{eq}=i_1+i_2+i_3$. Now the differential equations for different paths would be:
$$ V-\frac {q_{eq}} {C}- i_1R_1=0$$
$$V-\frac {q_{eq}}{C}-i_2R_2=0$$
$$V-\frac {q_{eq}}{C}-i_3R_3=0$$
To proceed you should divide these $3$ equations with $R_1$, $R_2$, and $R_3$ respectively and add.
We have, $\frac {1}{R_{eq}}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$
Hence, we get:
$$\frac {V}{R_{eq}}-\frac {q}{R_{eq}C}-i_{eq}=0$$
Since we have $\frac {dq}{dt}=i_{eq}$, all that remains is to integrate, and you'll obtain the required result.
A: Yes you can do all the operations as if it were a circuit with the equivalent resistor.
The reason for this can be seen if you break down what's happening across each of the 3 split wires. The currents all add up while the voltage drops are the same:
$ I_1 + I_2 + I_3 = I \\
 V_1 = V_2 = V_3 = V$
The differential equation is based off of using $I = \frac{\partial Q}{\partial t}$. For each wire you get a separate differential equation:
$  V = R_i \frac{\partial Q_i}{\partial t} + \frac{Q}{C}$
Where $\frac{\partial Q_1}{\partial t} = \frac{\partial Q}{\partial t} - \frac{\partial Q_2}{\partial t} - \frac{\partial Q_3}{\partial t}$ and so on. Tediously, you can do algebra this system of equations to get to a differential equation only in terms of the total current$\frac{\partial Q}{\partial t}$ without reference to the current on the individual wires $\frac{\partial Q_i}{\partial t}$.
You'll then see that you get an easily solveable diff eq where the coefficient in the exponent is exactly the same as $R_{eq}$.
