If I have the following circuit... (resistor & capacitors values are made up)


...and I hold down SW3 until the capacitor is charged, then I release SW3 then hold down SW1 why wouldn't the capacitor discharge into the negative terminal of the battery? (according to the comment I posted on this question: Capacitors and simple circuit understanding)

I think I've misunderstand some fundamentals here, as in my mind this circuit would be OK.

  • \$\begingroup\$ Unless BAT1- is connected to the circuit GND, there will be no current flow to charge C2. \$\endgroup\$ – theorifice Aug 6 '16 at 15:10
  • \$\begingroup\$ Ground capacitor lower terminal. Connect Sw3 from top of cap to ground to get hard discharge or top of R1 to get slower discharge. Current path MUST be via a continuous circuit. \$\endgroup\$ – Russell McMahon Aug 6 '16 at 16:37

SW3 doesn't charge the capacitor. I assume you mean SW1 to charge.

Closing SW3 grounds that part of the circuit but there is no ground connection anywhere else on your circuit so there is nowhere for current to flow. i.e., There is no "circuit" back to the battery through SW3 - just a dead-end branch.

Note: When using the built-in schematic editor just hit save rather than taking a screen-grab. That way you or we can edit the schematic or copy and edit in our answers.

  • \$\begingroup\$ Right I see, I think I've got confused about circuits and breadboards. I'm using a breadboard and trying to translate the schematics to the board but its quite confusing. Also thanks for the tip. \$\endgroup\$ – CS Student Aug 6 '16 at 15:14
  • \$\begingroup\$ Also another thing that confused me was that when I had a volt meter on the capacitor it looked as if the voltage was dropping (current flowing) but that was due to the meter itself \$\endgroup\$ – CS Student Aug 6 '16 at 15:25
  • \$\begingroup\$ Correct regarding the multimeter. I mentioned this in the other answer. As an instructive experiment charge up your capacitor to 9 V and let go SW1. Then put the meter across it and time how many seconds it takes to discharge by 63% (to about 3 V). Now you should find the time constant \$ \tau = R \times C \$ where R is the input resistance of the meter. \$\endgroup\$ – Transistor Aug 6 '16 at 15:31

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