Time Constant of a tricky RC circuit

Question

I'm trying to find the stored charge on capacitor1 at time \$t\$

So, I think I have to find the voltage on the capacitor first. In terms of Current, the total current \$I(t)\$ should be the sum of the current on \$C_1\$ which is \$I_1(t)\$ and the current on \$R_2\$ (\$=\mathrm{d}Q_2(t)/\mathrm{d}t\$).
I mean... $$ I(t) = I_1(t)+I_2(t). $$ And when we look at the outer loop, the total voltage \$V\$ should be the sum of the voltage on \$R_1\$, \$R_2\$, and \$C_2\$, so $$ V-R_1\cdot I(t) - R_2\cdot\frac{\mathrm{d}Q_2(t)}{\mathrm{d}t} - \frac{Q_2(t)}{C_2} = 0 $$ Similarly, at the inner loop, the total voltage \$V\$ should be the sum of the voltage on \$R_1\$ and \$C_1\$, so $$ V - R_1\cdot I(t) - \frac{Q_1(t)}{C_1} = 0 $$ But here, I don't know what to do....

Plz help me :(

PS

• I didn't learn about the Thevenin theorem

• when \$t=0\$, there was no charge stored on \$C_1\$ and \$C_2\$

Answer 1

Just use the method of open circuit time constants. If you take output at C1, then you have a system with two poles and a zero.
The time constant for the poles are: $$\tau_1 = R_1C_1$$ $$\tau_2 = (R_1+R_2)C_2$$ $$H^2 = \frac{R_2}{R_1+R_2}$$ $$H^0 = 1$$ Here, \$H^0\$ is the transfer function with all capacitors open circuit and \$H^2\$ is transfer function with \$C_2\$ short circuit and \$C_1\$ open.
The transfer function becomes: $$H(s) = \frac{H^0+s\tau_2H^2}{(1+s\tau_1)(1+s\tau_2)} = \frac{1+sR_2C_2}{(1+sR_1C_1)(1+s(R_1+R_2)C_2)}$$ To find the voltage with time use the step response as follows: $$X(s) = V/s$$ $$V(s) = H(s)V/s$$ Take inverse Laplace to get your answer.