circuit diagram

Consider a circuit as shown with a battery, $$ V_b $$

and two identical capacitors

$$ C_1, C_2 $$

(with C_2 being the one that's off to the side). Both have the same capacitance C. The resistor has resistance R.

Before adding the new capacitor C_2, the current in the ammeter is, say, I. The reed switch flips back and forth such that the capacitor charges fully and discharges fully, (only one capacitor at this moment). Thus when the switch is to the right, the capacitor has voltage V and current is V/R = I.

Now when I add the second capacitor, $$ C_2 $$ can I treat them as a single capacitor with capacitance 2C? If I do, then the current should decrease because:

$$ V = Q/2C $$ $$ I = V/R $$

Even if I consider them to be separate capacitors, Each capacitor will store charge

$$ Q $$

and thus total charge will be, $$2Q$$

Thus, voltage should be

$$ 2Q/2C = V $$

and current should stay the same. Why is this incorrect?

Note that I understand that current should double because you have twice the charge being stored and at each discharge cycle, the rate of charge flow should double and hence current should double as well.

  • \$\begingroup\$ You can. In a steady state. When you change the configuration in a run-time there are transients. \$\endgroup\$
    – Eugene Sh.
    Commented Dec 6, 2017 at 17:35
  • \$\begingroup\$ Why do I get weird answers for current though? What am I doing wrong? \$\endgroup\$ Commented Dec 6, 2017 at 17:41
  • \$\begingroup\$ For what current? There is no current in the steady state with the switch to the right.. \$\endgroup\$
    – Eugene Sh.
    Commented Dec 6, 2017 at 17:43
  • 2
    \$\begingroup\$ I = V/R applies to resistors, not capacitors. For capacitors, I = C dV/dt. \$\endgroup\$ Commented Dec 6, 2017 at 17:46
  • \$\begingroup\$ The current in the SuperCap or battery can use Ohm's Law but just for ESR of the C, so I=V/ESR for short circuit or initial charge. \$\endgroup\$ Commented Dec 6, 2017 at 18:11

5 Answers 5


In reality this circuit will weld the reed to the left contact the first time it charges if the capacitor is of any sensible size for this task. In theory, charges and discharges "fully" take forever. But it will mostly charge in a multiple of R*C that is large enough, say 5 or 10. So it still makes sense to try to solve this.

One way (maybe the best way) to handle this problem is to consider energy. Each time the capacitor is charged from 0V to the battery voltage it gains energy

E = \$\frac {CV_B^2}{2}\$. Each time it discharges "fully" it loses that energy.

If the cycle repeats at a periodic frequency f we know that the power transfer is f*E. So the RMS current through the ammeter must be:

\$I_{RMS} = V_B\sqrt{\frac{fC}{2R}} \$

Since capacitors in parallel add, doubling the capacitance will increase the RMS current by \$\sqrt{2}\$, all other things being equal. You can show that by using the energy equation for each capacitor.

On the other hand, the average current will double.. and it has a particularly simple equation Iavg = f * C * Vb. (R drops out). Proof left to the student.

Note that the current in the resistor will follow an exponential discharge curve from Vb towards zero with time constant RC each cycle, then drop from a very small number to exactly zero when the switch opens.

Can I treat two identical capacitors with capacitance C in parallel as one capacitor with capacitance 2C?


Capacitances in parallel add. There really isn't much more to say.

  • \$\begingroup\$ So why do I get weird answers for the current? Where am I going wrong? \$\endgroup\$ Commented Dec 6, 2017 at 17:40
  • \$\begingroup\$ @NeevParikh - See my comment on the OP. \$\endgroup\$ Commented Dec 6, 2017 at 17:47

When you turn the switch to output current from the capacitors to the resistor, the voltage in the capacitor is the same (=battery voltage), no matter have you the other cap assembled or not. Ohm's law states that the current will be the same (I=U/R). If you have both capacitors assembled, the durrent decays 50% slower due the double charging capacity of the double cap. Two identical capacitors in parallel have exactly double capacitace when compared to a single capacitor.


Let me add this to your question.


simulate this circuit – Schematic created using CircuitLab

Let's say a 18650 LiPo cell charges to 4.1V then operates from 3.7 to 3.3V over 20 hour period and gives 1000 mAh between those voltages.

  1. How many mAh would you get with two perfectly matched cells in parallel?
  2. What would be the circulating discharge current if each ESR was 10 mOhms and ΔVoc open circuit voltage was 100 mV say 3.7, 3.8V between two cells to be connected in parallel?

  3. Now if the Energy of a cap at some initial voltage Vi, E=½CVi² and lower for final voltage at say 10% SoC for Vf , say at 3.3V, what is this LiPo batteries equivalent capacitance if for 20 hours = 1C rate, let's assume Vbat average = 3.6V and Energy [Watt-sec or J ] E = P * t = Vavg * Ah * 3600 [s/h]

    • Now, what is C [kF] of 1 LiPo cell?
    • and what is C [kF] if 2 cells in parallel?
    • and what happens if one is relatively low %SoC and the other is full %SoC then put in parallel for State of Charge (SoC) and both very low ESR ? ( sizzle poof? or ? )

seems quite possible theoretically but needs the minutest of finishing while in practical. the reason being the attainability of stable state. charging-discharging should go simultaneously on both the capacitors. if not, fluctuations will occur.


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