Say there is a circuit connected in series

  • Capacitor (120 V - 10000 µF)
  • Load (8 ohm)
  • DC power supply (100 V - 5 A)

After charging the capacitor to 100 V from the power supply, how much current will be in the circuit while discharging?

Will it be the maximum current of power supply (5 A) or will it be according to Ohm's law 100/8= 12.5 A?

Will the capacitor act as separate circuit with load or does the maximum current of circuit comes from the power supply?

enter image description here

  • 2
    \$\begingroup\$ Once the capacitor is charged, no current flows. (Assuming an ideal capacitor, with no leakage). \$\endgroup\$ Aug 16 at 12:34
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    \$\begingroup\$ In the circuit you've shown the capacitor is not discharging. It has 100 volts across it, and the power supply is pushing 100 volts with the opposite polarity, and nothing is moving. \$\endgroup\$ Aug 16 at 12:46
  • 1
    \$\begingroup\$ Please draw the two different states you are proposing, charging and discharging. Then you will see that In your present discharge is not possible. \$\endgroup\$
    – winny
    Aug 16 at 13:13
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    \$\begingroup\$ You've changed the circuit again. Please unaccept my answer so I can delete it. \$\endgroup\$
    – JRE
    Aug 16 at 14:18
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    \$\begingroup\$ @Tito Changing your circuit like that undermines all the effort everyone's put in trying to answer your question. \$\endgroup\$
    – James
    Aug 16 at 14:21

4 Answers 4


Once the capacitor is charged in your circuit, no current will flow.

If the capacitor is fully discharged, then the current at the start will be 100 V/8 Ω = 12.5 A, but since the power supply can only deliver 5 A you will only get 5 A during the charge phase. As the capacitor charges, the current flow will go to zero.

  • \$\begingroup\$ So if i close swtich between supply and capacitor (charge) then close switch between capacitor and load (discharge) the current to load will be 12.5A and decreasing with capacitor voltage? \$\endgroup\$
    – Tito
    Aug 16 at 12:53
  • \$\begingroup\$ Yes. It is so :) \$\endgroup\$
    – Arseniy
    Aug 16 at 12:55
  • \$\begingroup\$ Assuming the power supply can deliver 12.5A and stay at 100V doing so. OP mentioned a 5A PSU. \$\endgroup\$
    – marcelm
    Aug 16 at 14:00

In your circuit, both switches must be closed to charge the capacitor. If either or both switches are opened the capacitor will not discharge but will retain the voltage it has when the switch is opened. Closing the both switches again will allow charging to continue until the capacitor voltage reaches 100V. Your circuit will not allow discharge.

The following circuit will allow both charging and discharging


simulate this circuit – Schematic created using CircuitLab

Assuming that the respective switch stays closed for at least $$t=5RC$$

SW1 closed, SW2 open: C1 will charge according to $$v_{C}=100(1-e^{-t/RC}) V, i=12.5e^{-t/RC} A$$

SW2 closed, SW1 open: C1 will discharge according to $$v_{C}=100e^{-t/RC} V, i=-12.5e^{-t/RC} A$$

The 5 A current limit will maintain the current at 5 A for charging, but the discharging will be the same. SW1 must remain closed for longer to achieve full charge.

SW1 closed, SW2 open: C1 will charge according to $$v_{C}=\frac{5}{C}t V, v_{C}<60 V$$ $$v_{C}=100(1-e^{-(t-t_{0})/RC})+60e^{-(t-t_{0})/RC} V, v_{C}>60 V$$

$$i=5 A, v_{C}<60 V$$

$$i=5e^{-(t-t_{0})/RC} A, v_{C}>60 V$$

  • \$\begingroup\$ If both were closed, how many current? \$\endgroup\$
    – Tito
    Aug 16 at 14:12
  • 2
    \$\begingroup\$ @Tito As you have drawn it with zero resistance, infinite current flows between the shorted out power supply terminals, or at least as much as the power supply can provide before it shuts off or burns up. No current flows into the capacitor. \$\endgroup\$ Aug 16 at 14:22
  • \$\begingroup\$ SW1 closed, SW2 open : current couldn't reach 12.5A as the powersupply max current is 5A \$\endgroup\$
    – Tito
    Aug 16 at 14:44
  • \$\begingroup\$ Thanks Tito. I added the current limiting to my answer \$\endgroup\$
    – RussellH
    Aug 16 at 15:49

If the capacitor is initially totally discharged and then the switch between the psu and capacitor is closed (other switch open) then this will not cause the capacitor to charge but both plates of the capacitor will go to 100 V.

Now if you close the switch between the capacitor and the load the power supply voltage will be instantly pulled down to 40 V (5 A * 8 ohms = 40 V), if the psu has current limiting. If there is no current limiting the psu might blow a fuse or possibly be damaged. Assuming the psu has current limiting, then the next thing to happen will be that the capacitor starts to charge in a linear manner, the voltage across the load stays constant at 40 V and the psu output voltage rises until it reaches a maximum of 100 V where it stabalizes. From this point the capacitor continues to charge and the voltage across the load and current through it exponentially reduce to 0 V and 0 A respectively.

So, the maximum current through the load is equal to the maximum current that the psu can supply which is 5 A.

This all happens because the currents in the two leads of a capacitor must always be equal to each other and so the psu must supply the same current to the capacitor as the capacitor is supplying to the load. It is a series loop, the current at any point in the loop must be identical to the current at every other point.


I guess there must be a key between the source and the capacitor or between the capasitor and the load.

If this is the first case there will be a 5A current at the close key moment. If this is the second case there will be a 12.5A current at the close key moment.

If you tell that the capacitor will be charged first I guess that you mean the second case. So it is 12.5A.


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