Find current in a nontrivial circuit

Question

Let us assume that we have this circuit:

We know that:

$$I_1 = I_2 + I_3$$ $$I_1R_1 + I_3R_3 = V$$ $$I_1R_1 + I_2R_2 + \dfrac{q_c}{C_1} = V$$

Therefore we can write:

$$I_1 = k_1 - k_2*q_c$$

with:

$$k_1 = \dfrac{V(1/R_2+1/R_3)}{1+R_1/R_2+R_1/R_3}$$ $$k_2 = (R_2*C_1(1+R_1/R_2+R_1/R_3))^(-1)$$

The question is, Can we write the equation:

$$\dfrac{dq(t)}{dt}={k_1-k_2q(t)}$$

I doubt if $q_c=q(t)$ because $I_3$ does not flow through the branch which contains the capacitor, therefore:

$$I_1 = I_2 + I_3 => q_1 = q_2 + q_3$$

I think $q(t)=q_1$ But if it is so $q_2 = q_c \neq q(t)$. How can I write down the differential equation in order to find $q(t)$?

Answer 1

How can I write down the differential equation in order to find q(t)?

There are generally several ways to solve a circuit but some may be far more straightforward than others. The most straightforward approach to find $q_C(t)$ in this case is to transform this circuit into a simple series RC circuit.

By finding the Thevenin equivalent circuit seen from the capacitor terminals, we get by inspection:

$$V_{TH} = V\dfrac{R_3}{R_1 + R_3}$$

$$R_{TH} = R_2 + R_1||R_3$$

Since we now have a simple series RC circuit, the solution for the capacitor current is well known:

$$i_C(t) = i_C(0)e^{-t/\tau} $$

$$\tau = R_{TH}C_1 $$

The solution for the capacitor charge follows by integration:

$$q_C(t) = \tau i_C(0)(1 - e^{-t/\tau}) + q_C(0) = C_1V_{TH}(1 - e^{-t/\tau}) + q_C(0)e^{-t/\tau} $$