# Will a capacitor draw charge/current if it's not feeding a load?

I have a number of capacitors connected in parallel, as in the following circuit: simulate this circuit – Schematic created using CircuitLab

1. If the switch is open, will the caps still draw charge/current?
2. Will this only happen when the load draws current from/through them? (Is the load required for the caps to charge?)

I have a basic understanding of how caps work and don't have an LCR meter or scope with which to test this.

• After the initial charge, there is a small leakage in each that the power supply needs to feed them with. Usually uA of current. Feb 20, 2020 at 7:23
• Yes, that's fine. (It's not an issue that needs attention in the current application of the circuit, which is short-term. The aim is to charge the capacitors, disconnect the power source and use the circuit for about a minute (or as long as the charge in the caps lasts), possibly cutting flow to the load during that period, and then recharge the caps if need be.) Feb 20, 2020 at 7:35 simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's circuit redrawn.

If we redraw the circuit to lay it out in a conventional fashion (reading left to right with positive supply at top it becomes a little more obvious what is going on.

1. If the switch is open, will the caps still draw charge/current?

It should be clear now that there will be an initial current to charge up the capacitors. As they charge the current will fall reaching zero when fully charged. Closing the switch will not affect the capacitor current.

2, Will this only happen when the load draws current from/through them? (Is the load required for the caps to charge?)

With the redrawn schematic you can see that the capacitors will charge whether the load is connected or not.

Note 500 kΩ on a 9 V supply will limit the LED current to $$\ \frac {9}{500k} = 0.018 \ \text {mA} \$$ which isn't enough to light an LED. 1 kΩ would give about 7 mA to split between the two LEDs but usually we would use one resistor per LED to balance the currents or connect the two LEDs in series to reduce the current required by the whole circuit.

• It's possible I measured the resistor's value incorrectly (by an order of magnitude), but I have a functioning circuit with a resistor that I'm sure has that value. Feb 20, 2020 at 7:48
• If you can see the LEDs you must have great eyesight and be sitting in a very dark room! Feb 20, 2020 at 8:30

By them selves the capacitors will have slight self discharge and with the battery connected there will also be a small leakage current.

If the switch is open, will the caps still draw charge/current?

Yes, it will be very small current. See below for an example. Will this only happen when the load draws current from/through them? (Is the load required for the caps to charge?)

When you close the switch the load gets connected, depending on the amount of current drawn by the load, the capacitor might contribute to the load current too. If the 9V supply is very far (inductance of the cable or with a higher internal resistance). Soon after load current is reduced, the capacitors will charge back to the supply voltage. And then, only leakage current seen above will be there.

1. Theoretically, zero current will flow when the switch is open. This is because once the capacitors fully charges up, they will not conduct any further. Or you can say, Capacitor blocks DC. Practically, a negligible amount of leakage current flows through capacitors, which may be insignificantly low for most applications.

2. Capacitor don't need a Resistor to charge. They charge theoretically instantly as soon as you apply a voltage to them. Resistors are mostly included, and that too in series, to delay the charging time of capacitor. Because you resistor is not in series but in parallel to the capacitor, it will charge instantly.

Yes, since real capacitors do have a finite resistor parallel to the capacitance. The value of this resistor depends on the type and quality of the capacitor. Electrolytic capacitors as shown in the circuit usually do have a low resistance compared to film capacitors thus drawing more leakage current.