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

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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\$

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    \$\begingroup\$ This is a second order circuit. It really wouldn’t have a time constant. You’d have to get the second order differential equation and determine the characteristic roots. This would tell you if you got a underdamped, critically damped or underdamped response. \$\endgroup\$
    – Leoman12
    Apr 22, 2020 at 3:17
  • \$\begingroup\$ Thanks. Is there another way to find the stored charge without using the second order differential equation? Because I don't know how to calculate the second order diff eq. \$\endgroup\$
    – Robert
    Apr 22, 2020 at 3:31
  • \$\begingroup\$ If you want the stored charge on C1 then you can use the formula Q=CV. Since at steady state both capacitors equal V supply. \$\endgroup\$
    – Leoman12
    Apr 22, 2020 at 3:41
  • \$\begingroup\$ Also DC flat line circuits do not have a time constant. Your capacitors will just get charged with current equal to voltage over the resistor divided by the resistance. In order to have a time constant you need impulse current. The time constant is the time necessary to charge the capacitor to 63.2%, after 5 time constants, the charge is 100%. There is an app for android called EveryCircuit, the older versions are not paid. Its not good for creating your own circuits, but the examples are good and will give you a good idea of the basic circuits. \$\endgroup\$ Apr 22, 2020 at 4:04
  • \$\begingroup\$ "Is there another way to find the stored charge without using the second order differential equation?" Check answers from @VerbalKint like electronics.stackexchange.com/a/492981/200815 \$\endgroup\$
    – Huisman
    Apr 22, 2020 at 5:06

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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.

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