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The switched circuit is above. I am trying to understand how the switch causes the capacitor to charge and discharge. Most of the examples I can find involve using a switch to connect a capacitor to a voltage source, charging it up until the voltage across the capacitor equals the source. Then a switch is used to isolate the capacitor from the source, discharging the capacitor.

This circuit is much more complicated than that. Does closing the switch cause the current to bypass the capacitor, discharging it?

  • \$\begingroup\$ When S1 is closed, it shorts the capacitor and it should discharge (not instantly though because it's discharging through a resistor R3). The voltage on R2 will also drop to zero. \$\endgroup\$ – Havenard Feb 6 '18 at 20:50
  • \$\begingroup\$ can you explain the reason for the voltage divider? Why not just remove R2? \$\endgroup\$ – Stone Preston Feb 6 '18 at 21:00
  • \$\begingroup\$ R2 sets the maximum voltage C can be charged to, with the switch open. Maximum voltage is equal to the ratio of R1/R2. \$\endgroup\$ – Sparky256 Feb 6 '18 at 23:00

Closing the switch drops the voltage above R3 causing the voltage across it and the cap C1 to drop to zero, when the switch opens the cap begins charging again. The resistor R3 simply limits the current that can flow in and out of the capacitor for a slower discharge/charge.

The R1 and R2 form a resistive divider to divide the voltage of V1 at point D, but R1 also limits the current from the source when the switch is closed so it's not a direct short across the source when the switch is closed

  • \$\begingroup\$ ah ok so since the switch is connected to ground, when its closed point D voltage is zero (and the voltage above R3 since its a single node). So the capacitor discharges once the voltage across is becomes zero. Is that what you are saying? \$\endgroup\$ – Stone Preston Feb 6 '18 at 20:53
  • \$\begingroup\$ Yep, that is what I'm saying \$\endgroup\$ – Voltage Spike Feb 6 '18 at 20:53
  • \$\begingroup\$ Ok thanks for the explanation. I get that part now. I guess im still confused on the point of the voltage divider. What is it accomplishing \$\endgroup\$ – Stone Preston Feb 6 '18 at 20:55
  • \$\begingroup\$ @StonePreston The divider reduces the maximum charged voltage for the capacitor when the switch is open and limits the current taken from the voltage source and going through the switch when the switch is closed (though the upper resistor of the divider could also achieve this last part.) \$\endgroup\$ – jonk Feb 6 '18 at 21:03

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