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I understand that increasing current decreases the time taken for a capacitor to both charge and discharge, and also increasing the potential difference and charge increase the time taken for a capacitor to charge while decreasing the time taken for it to discharge.

However, I am having troubles with deducing what effect resistance will have on it? Is it as simple as V = IR, and increasing resistance with a constant potential difference will decrease current (and thus increase the time taken for a capacitor to both charge and discharge?)

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When a capacitor with capacitance $C$ is charged by applying a voltage source $V$ in series with a resistance $R$, the voltage $V_{cap}$ of the capacitor (and thus charge) increases according $$V_{cap}=V[1-\exp ({-t/RC)}]$$ Thus, as expected, the charging time of the capacitor increases with increasing $R$.

The discharge has the same time constant $RC$ $$V_{cap}=V[exp ({-t/RC)}]$$

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  • $\begingroup$ Thank you. However, shouldn't -t/RC be a power of e (or exp)? $\endgroup$
    – Rowan A.
    Commented Oct 31, 2016 at 22:37
  • $\begingroup$ @Rowan A. - This is the case in the formula given. $\exp{\frac {-t}{RC}}=e^{\frac {-t}{RC}}$ $\endgroup$
    – freecharly
    Commented Oct 31, 2016 at 22:45

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