I would like to know what will happen in the circuit diagram below when the switch is on( connected to a voltage source and R1) and switch off(connected to the R2)

My thoughts... For switch on only the R1 and VT will be left? for switch off VT and R1 will be removed? the capacitor will act as a source, am I right?


simulate this circuit – Schematic created using CircuitLab

  • 2
    \$\begingroup\$ Why not just simulate it? \$\endgroup\$ – Justin Trzeciak Mar 11 '14 at 1:48
  • \$\begingroup\$ ^ jrtzeciak is right...if you already drew it in CircuitLab simply simulate it. \$\endgroup\$ – EasyOhm Mar 11 '14 at 1:56

The way this is drawn R2 does nothing. When the switch is off (as drawn) nothing happens the circuit is incomplete.

When the switch is on you will charge the capacitor until it is fully charged and current will stop flowing. R1 and R2 simply sum to 1 equivalent resistor R. And the charge profile will look like this:

enter image description here

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html


The way the circuit is drawn does not match the wording of the question, so my answer assumes the wording is correct and the circuit switch is wrong as that makes more logical sense from a learning circuits standpoint.

When on, the resistors R1 and R3 act as series equivalent resistors and are therefore just summed together. If the switch is closed long enough, the capacitor will be charged to Vt. When in the off position, where R3 and R2 are now connected, they will discharge the capacitor through the new equivalent series resistor which may also be summed together to find the equivalent resistance.


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