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Most textbooks use this circuit as an example of the most basic RC circuit.

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The solution for V(t) for this circuit is (if capacitor is initially uncharged):

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I've had an exam recently where the resistor/or capacitor was to the left of the switch. My question is will the general solution to this circuit change in either of the two cases?

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  • \$\begingroup\$ The circuit was a resistor in series with a capacitor with a switch that shorted the top of the cap to ground when switched on? \$\endgroup\$ – sherrellbc Jul 13 '14 at 15:03
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No, it will have no effect.

For the purpose of circuit analysis, when the switch is off, no matter where it is, as long as it is in series with the voltage source, no charge will be able to move through the circuit.

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  • \$\begingroup\$ The circuit above has no moving charge when the switch is open, true. However, the equation presented is for a discharging RC circuit with initial voltage Vs - so the schematic and governing equation here are miss-matched. Also, if the R and C are on the left of the switch then C is initially charged before the switch closes. Upon closing the switch it would discharge, which is governed by the equation above. These two configurations do not have the same representative equations. Or am I missing something here? \$\endgroup\$ – sherrellbc Jul 13 '14 at 15:08
  • \$\begingroup\$ @sherrellbc: Yes and no. It is all about charge density. If the cap is connected to V-, charge will flow from V- up to the point that part of the circuit has a uniform charge distribution. The electrons on the other side of the cap will be repelled. However, the cap won't end up with a Vs difference since as these electrons, instead of being pulled into the V+ terminal, end up squeezing in that isolated part of the circuit. When the switch is then closed, the charging process will resume as normal. I could say the same thing and assume holes are the charge carriers and nothing would change... \$\endgroup\$ – Evan Jul 13 '14 at 17:50
  • \$\begingroup\$ @sherrellbc: ...However, the question refered to standard textbooks. These textbooks normally state "at t=0 the switch is closed" so this assumes all I described in the previous comment don't really matter and the statement could very well be "at t=0 the wires are connected to the voltage source". \$\endgroup\$ – Evan Jul 13 '14 at 17:52
  • \$\begingroup\$ @sherrellbc: The equation is the one for charging. You can use some mathematical software with a plot function to see its output. \$\endgroup\$ – Evan Jul 13 '14 at 17:56
  • \$\begingroup\$ @sherrellbc: Lastly, don't think of ground as having zero charge. Zero volts is different from zero charge. We might use the surface of the earth as 0V reference but it does have a negative charge. In particular, the volt, when first conceived, corresponded to the charge difference between two infinite planes one meter apart that create an electric field of 1N/C which meant if there was one particle with a charge of one coulomb within the field, a force of 1Newton would be exerted on the particle by the field. \$\endgroup\$ – Evan Jul 13 '14 at 18:10

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