Assume I have a capacitor with capacity \$C\$ and is charged to a voltage level \$V_a\$ and then discharged.

1) During discharge it dissipates \$E\$ joules of energy. What is the equation to find the final voltage \$V_b\$ of the capacitor after \$E\$ joules have been discharged ?

2) After the same capacitor discharges and reaches \$V_b\$, we re-charge it up to \$V_a\$ again. Assume the charging current is (\$I\$) (Amps). What is the equation to find how much time (\$t\$) it take for the capacitor to charge from \$V_b\$ to \$V_a\$ ?

I am looking for equations without resistance (\$R\$) involved, as I do not know the resistance of the circuit (MCU based), but only the energy consumption of the load.

I googled for these equations, but could not come up with a specific answer. I have learnt these equations in school/university, but have forgotten now.

thank you

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    \$\begingroup\$ These questions sound very much like homework questions, and this is not Chegg...we won't do your homework for you. Can you explain more about why you are asking this question? What is your actual design problem? \$\endgroup\$ – Elliot Alderson May 14 at 19:41
  • \$\begingroup\$ Q = C * V = I * t. E = 0.5 * C * V^2.Work out E1 for initial V, Subtract E to get E2, get V2 from that. \$\endgroup\$ – Brian Drummond May 14 at 19:41
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    \$\begingroup\$ Also, \$i = C \frac{dV}{dt}\$. That should be enough. \$\endgroup\$ – Elliot Alderson May 14 at 19:49
  • \$\begingroup\$ @ElliotAlderson - it is to analyze a MCU connected to an energy harvester with a capacitor. I know the energy consumption of the MCU, need to know the charging duration, if charged from a known voltage \$\endgroup\$ – bd3lk May 14 at 23:25

1) Because the capacitance remains constant $$E= \frac{1}{2} CV^2$$ can be used to compare the energy stored. This can be compared for the states \$a\$ and \$b\$.

2) Again as Elliot Anderson stated $$ i(t) = C \frac{\Delta v(t)}{\Delta t}$$ is the relationship between the voltage and current. Here the same equation is used for the states \$a\$ and \$b\$.

Just as an aside, the same equations for an inductance are $$E= \frac{1}{2} LI^2$$ and $$ v(t) = L \frac{\Delta i(t)}{\Delta t}.$$

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    \$\begingroup\$ just to clarify regarding the second equation: if Va=2.7V and Vb=2.4V and C=1F, then t = 1*(2.7-2.4) = 0.3s ? isn't the charging time exponential ? am i missing something ? \$\endgroup\$ – bd3lk May 14 at 23:23
  • \$\begingroup\$ If the current is constant, such as in your example, the voltage rise is linear. Multimeters use this feature to test the size of large capacitors by timing the voltage rise from -2 V to 2 V. \$\endgroup\$ – skvery May 15 at 10:06

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