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enter image description here

When I charge the circuit as shown where the switch is at A, the p.d. across C1 is 10V, which is correct.

When I discharge the circuit, however, by changing the switch S from A to B, the total capacitance of the circuit is 200μF as C1 and C2 are in parallel.

This must mean the time constant of the discharging circuit is

$$200 \times 10^{-6} \times 100 \times 10^3 = 20s$$

In the solution, however, the time constant of the discharging circuit is quoted as 5 seconds, where it states C1 and C2 are in series when the switch is at B.

Who is right, the solutions or me?

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  • $\begingroup$ I suggest you redraw the circuit with the switch at B. $\endgroup$ – probably_someone Jun 28 '17 at 17:54
  • $\begingroup$ yes I did that but it still doesn't deny that c1 and c2 are parallel @probably_someone $\endgroup$ – vik1245 Jun 28 '17 at 17:56
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    $\begingroup$ Moving the switch to B disconnects the battery and the left resistor from the circuit. What remains is one resistor and two capacitors, which are now in series. $\endgroup$ – probably_someone Jun 28 '17 at 17:58
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    $\begingroup$ physics.stackexchange.com/questions/337737/… I posted a similar question on a different thread with a similar circuit, so why are the two circuits different where in this question it is series and in the other thread, it is parallel? $\endgroup$ – vik1245 Jun 28 '17 at 18:00
  • $\begingroup$ Because the capacitors are arranged differently - in that case, they're on opposite sides of the switches, whereas here, they're on the same side. $\endgroup$ – probably_someone Jun 28 '17 at 18:02
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the total capacitance of the circuit is 200μF as C1 and C2 are in parallel.

They're not in parallel for either switch position.

When the switch is in position A, C2 and the 100k resistor are in series but one end of the resistor is 'dangling' so there is no path for current through the series combination.

When the switch is in position B, C1 is placed in series with the C2 + 100k series RC combination. This should be obvious since there is only one path for current, all three circuit elements have identical current.

If the capacitors were parallel connected, C1 and C2 would 'split' the current through the 100k resistor but clearly, all of the current through the resistor is through either capacitor and so, the capacitors are series connected.

As you already know, the equivalent capacitance of 2 identical series connected capacitors is 1/2 the individual capacitance and thus

$$C_{eq} = 50\mu\mathrm{F}$$

and the time constant is

$$\tau = 100\mathrm{k\Omega}\cdot 50\mu\mathrm{F} = 5\mathrm{s}$$

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    $\begingroup$ physics.stackexchange.com/questions/337737/… how does the circuit in this post then vary from that circuit? $\endgroup$ – vik1245 Jun 28 '17 at 18:54
  • $\begingroup$ @BobSmith, the key difference is that there are resistors in this circuit and so there are three circuit elements connected together on the right hand side when the switch is in position B. In the circuit at the link, when S1 is open and S2 is closed, there are just two circuit elements, the two capacitors, on the right hand side. This is a 'degenerate' configuration that is both series and parallel at the same time, i.e., the two capacitors have identical voltage across as with a parallel connection and the two capacitors have identical current through as with a series connection. $\endgroup$ – Alfred Centauri Jun 28 '17 at 19:10
  • $\begingroup$ By adding the resistor, a third circuit element, the degeneracy is broken and then either the capacitors are series connected or parallel connected. Since, in this circuit, the capacitors have identical current through, they are series connected. $\endgroup$ – Alfred Centauri Jun 28 '17 at 19:11
  • $\begingroup$ @BobSmith, it's possible to connect the resistor differently so that all three are parallel connected. $\endgroup$ – Alfred Centauri Jun 28 '17 at 19:15
  • $\begingroup$ @BobSmith, assume for clarity that, with the switch in the B position, the current is out of the 'top' plate of C1. Since there is only one path and that path is to the resistor, all of the current out of C1 enters the resistor's left-most terminal and exits the resistor's right-most terminal. Again, there is only one path and that is to C2 and so all of the current out of the resistor enters C2 through the top plate and exits the bottom plate. Finally, there is only path and that is to C1 and so all of the current out of C2 enters C1 through the bottom plate. Do you see? $\endgroup$ – Alfred Centauri Jun 28 '17 at 19:25

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