0
\$\begingroup\$

How is the (dis)charging process of capacitors in series calculated?

I know that $$C_{total}=\frac1{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...}$$ $$Q_{total}=Q_{1}=Q_{2}=Q_{3}=...$$ $$V_{1/2/3}=\frac{Q_{total}}{C_{1/2/3}}$$ I also know that $$V(t)=V_0*(1-e^{\frac{-t}{R*C}})$$ $$I(t)=\frac{V_0}{R}*e^{\frac{-t}{R*C}}$$ My goal is to calculate the voltage progression while (dis)charging two capacitors in series. Is that even possible or is this process too random to be calculated?

Note that i do not want to calculate the behaviour of the capacitive electronic component the two capacitors form together. Instead I want to know the voltage drop at each capacitor (, their charge and the electric current flowing through them) at any point in time.

Hopefully this is an achievable task and not connected to excessive testing and measuring.

Thank You in Advance

\$\endgroup\$
3
  • 3
    \$\begingroup\$ This is difficult, if not impossible, since you can't know the initial voltage on each cap when you start as all capacitors have leakage and this will dominate at DC. But if you assume they all start with identical charge so you can get the initial voltage on each. Now if you know the current then you know how the charge changes in each capacitor with time and can calculate the voltage. Since they are in series the charge will be changing at the same rate. \$\endgroup\$ – Warren Hill Feb 19 '20 at 18:28
  • \$\begingroup\$ Charge = current * time. Voltage = charge / capacitance. \$\endgroup\$ – user253751 Feb 19 '20 at 19:06
  • \$\begingroup\$ As current is changing \$ Q = \int I \text{ dt} \$, \$ V = \frac{Q}{C} \$. \$\endgroup\$ – Warren Hill Feb 19 '20 at 19:45
1
\$\begingroup\$

You can replace the series capacitors with their equivalent total capacitance and then determine the voltage across and current through this equivalent capacitance as a function of time, just as you would if the capacitors where in fact the single equivalent capacitance.

Since the capacitors are in series you know that the current through them must be exactly the same. This means that the change in the total charge must be the same for each capacitor. So the change in voltage for each capacitor must be equal to that change in total charge divided by the individual capacitances. Basically, if you know the current and the capacitance values then you know the voltage.

\$\endgroup\$
0
\$\begingroup\$

Assuming ideal caps all around:

If \$V_0\$ is the total voltage across the caps, and \$C_t\$ is your total capacitance, the voltage between the caps could be anything. It might be 100V across \$C_1\$ and -99V across \$C_2\$ for a \$V_0\$ of 1V. The overall system will work the same.

The homework may be asking you to assume the capacitors were also charged in series, so that \$V_n\$ is proportional to \$1/C_n\$. If not, assume a center voltage \$V_1\$ and use it as a term in your solution. The change in voltage across each cap over time won't be affected.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.