RC Circuits (in Capacitor and Resistor in Series vs Capacitors and Resistors in Parallel and beyond)

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JogoBonito

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I've just finished reading most of the threads on this topic here and studying them via TBR. I am still pretty confused on the topic and was wondering if you guys can check my understanding and also answer some questions

So this is what I believe to be correct
First for RC circuits in series.
1. So my understanding here is that when you have a resistor and a capacitor in series, the maximum charge on the capacitor is not affected by the presence of the resistor. Only that it slows down the charging/discharging of the capacitor. The voltage of the capacitor starts at zero, while the the voltage of the resistor starts at a value equaling the total emf of the battery source because current I=VR in the beginning (is this correct?). And over time, the current that flows to the resistor decreases as the capacitor is charged more and more. At the end, the current stops all together in the circuit when the capacitor is fully charged. At this point the voltage of the capacitor= voltage of the battery. Because the current =0, the voltage across the resistor= 0.

  • Does the order of the capacitor and resistor matter? If it went battery-> capacitor-> resistor -> battery vs Battery-> resistor-> capacitor -> battery?
  • How does it make sense that maximum charge on the capacitor is not affected by the presence of a resistor? Does a resistor not dissipate heat (converting electrochemical potential energy into heat) and shouldn't that account for a decrease in total charge that can be accumulated by the capacitor?

For RC Circuits in Parallel
1. Say you started with a scenario with a circuit with only capacitors and resistors that are in parallel but with a switch open. Immediately after the switch is closed, the voltage across the capacitor AND the resistor is 0. When the switch is closed, assuming the capacitor is initially uncharged, all of current goes to exclusively to the capacitor (path of least resistance). But over time, the capacitor will build up charge and when the charge builds up on the capacitor, so doe the voltage across it : because Q= CV. The current through the resistor will start at 0, and build up to some maximum value given by V=IR, where Voltage drop for resistor = VOltage drop for the capacitor. The capacitor and the resistors in parallel end up with the same voltage at the end. The final current through the capacitor is zero. But the final current through the parallel resistor is not zero (is this correct?)
  • I have also read that when the capacitor is charged fully at the end, there will be NO current in that branch of the circuit. Does this refer exclusively to the Capacitor, but not the resistor that is parallel to it? It wouldn't make sense for the resistor to have no current because V=IR and it now has a voltage drop equalling emf of the battery (because no other resistors exist in this scenario)
Having said all that. Can you clarify this question?
circuit-for-31-jpg.16335

If both switches were closed, and a new resistor were inserted into the circuit before S1, then the maximum charge that could be stored on the capacitor would:
A. Increase, because the voltage across the cap. would increase
B. Decrease, because the voltage across the cap. would increase
C. Increase, because the voltage across the cap. would decrease
D. Decrease, because the voltage across the cap. would decrease

Answer is D.

Explanation by TBR Verbatim Being:
If another resistor were inserted into the circuit before switch Sq, then it would be in series with the capacitor. The change this would produce in the maximum charge that could be stored on the capacitor depends on the change in voltage across the capacitor. Voltage and charge for a capacitor are related through Q= CV. If the voltage increases, the charge will increase: and if the voltage decreases, the charge will decrease. Based on that eliminate B and C. What we now have to decide is what effect the new resistor has on the voltage across the capacitor. Without the new resistor, the voltage across the capacitor is 10 mV. With the new resistor in the circuit some of the 10 mV has to go on the resistor and some on the capacitor. Since the whole 10mV is spread across two circuit elements instead of just the capacitor, we can conclude the voltage must decrease across the capacitor.

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so based on my understanding. initially I reasoned adding a resistor in series shouldn't decrease the maximum charge that can be stored on the capacitor. But then I thought, could it be because parallel resistors also? Is it because in a typical resistor and capacitor in series ONLY circuits, voltage is not spit, but voltage is split here? Sorry I've been out of undergrad for quite some time, and this concept has boggled me down quite a bit

Any help would be greatly appreciated

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A short answer is that every series resistor in a circuit will have it's own voltage drop and the capacitance will be based on the voltage difference on each of the plates (measured before and after the capacitor). If you have many resistors in series, and a capacitor placed in line with all the resistors, then the voltage drop between the two resistors that are just before and just after the capacitor will be less than the battery voltage.

Also it is not because of the parallel right half of the diagram.
If you remove R1, R2, and S2 from the diagram and ask the same question, you get the same answer. The charge differential that can be established on each plate is directly dependent to the voltage differential so adding resistance that will decrease the voltage difference (compared to the internal resistance of the battery) will lower the capacitance.
 
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