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I am working on a problem involving a capacitor in series with a resistor. Basically there is a circuit, with a capacitor of capacitance $200\mu F$, connected in series to a resistor of resistance $470k\Omega$, which is all powered by a cell of $1.5V$. There is also a switch to turn the flow of current on and off.

If I wanted to calculate the maximum energy stored on the capacitor I would assume I can use $E = \frac{1}{2}CV^2$, but I am not sure if I can use $V = 1.5$.

I know voltage splits in a series circuit, but I'm not sure how it would split with the capacitor because technically it does not have resistance in the normal sense. The current value constantly changes. Am I also correct in saying that the resistor will only affect the rate of charge, and not the total charge/energy stored?

Basically I'm asking what the difference in this would be compared to if the resistor wasn't in the circuit.

Diagram: circuit

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  • $\begingroup$ en.wikipedia.org/wiki/RC_circuit, scroll down to series circuit $\endgroup$ – physicopath Mar 2 '17 at 16:51
  • $\begingroup$ So if I use the equation under that section to get the voltage on the capacitor, I compute: $\frac{1}{1 + RC}*V$ where $R = 470*10^{3}, V = 1.5, C = 200*10^{-6}$, and then use this as my $V$ value in the energy equation I will get the maximum energy? $\endgroup$ – Tiernan Watson Mar 2 '17 at 16:57
  • $\begingroup$ Include a circuit diagram. $\endgroup$ – DanielSank Mar 2 '17 at 17:34
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The voltage drop, $V_R$, across a resistor $R$ is simply given by:

$$ V_R = IR $$

When the capacitor is fully charged the current through the resistor is zero so the voltage drop $V_R = 0$. That means the full 1.5V is across the capacitor.

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  • $\begingroup$ Thanks! So then say, after OPENING the switch in the circuit so it is not complete, I added another capacitor of $400\mu F$ in parallel with that one, so that the resistor is then in series with the two in parallel. Would the new voltage across the capacitors be $0.5V$? Since the total capacitance would be $600\mu F$ and $Q=0.0003$ from the previous arrangement, so $V = \frac{0.0003}{0.0006}$? $\endgroup$ – Tiernan Watson Mar 2 '17 at 17:50
  • $\begingroup$ @TiernanWatson: yes $\endgroup$ – John Rennie Mar 2 '17 at 18:08

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