# Charging and discharging a capacitor [closed]

So from what I understand, in the case for the charging capacitor, I understand the voltage across the capacitor as well as the charge can be given as the function of time:

$$V(t)=\epsilon(1-e^{-t/RC})$$ $$Q(t)=C\epsilon(1-e^{-t/RC})$$

But I do not understand why the function of time expression given by the charging capacitor is given in my book: $$I(t)=I_0(1-e^{-t/RC})$$

The reason why I do not understand the current equation is by the expression of the current above, the function reaches an asymptotic value of $$I_0$$ as time goes to $$\infty$$, but shouldn't there be no current passing through the capacitor when $$t\rightarrow \infty$$, because there is no potential difference across the whole circuit when time is very large? Furthermore, by the above equation, if $$t =0,$$ then it would imply that the current through the capacitor is $$0$$ initially?

I am confused to seeing why the current in the charging capacitor would increase to the maximum value of $$I_0$$, but I do understand why the charge and voltage increase to an asymptotic value.

I would appreciate the clarification to my confusion.

## closed as unclear what you're asking by Emilio Pisanty, user191954, John Rennie, Kyle Kanos, Jon CusterJan 11 at 15:08

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• It’s hard to know how to help without more info on the circuit. Are the R and C perhaps in parallel? – Bob Jacobsen Jan 11 at 4:47
• They are in series. I dont get it cause some texts say that current is both $I =I_0e^{-t/RC}$ for charging and discharging. – Aurora Borealis Jan 11 at 6:40
• Which book? It's likely that your book is wrong, but it's also possible that you've misinterpreted the context, and it's impossible to tell the difference (i.e. your question is unanswerable) without a full reference to the text where you found this (and, if it's hard to find, a full quotation of the relevant passage). – Emilio Pisanty Jan 11 at 8:56
• bowlesphysics.com/images/AP_Physics_C_-_RC_Circuits.pdf page 14, slides my teacher gave me. – Aurora Borealis Jan 11 at 9:22

## 2 Answers

I have edited the slide to which you are referring.

The voltage $$V(t)$$ referred to in the slide is the voltage across the capacitor.

The voltage across the resistor is $$\mathcal E - V(t) = \mathcal E - \mathcal E\left ( 1-e^{-\frac{t}{RC}}\right )=\mathcal E \,e^{-\frac{t}{RC}}$$ and dividing this voltage by the resistance $$R$$ gives the correct current relationship.

No what you said was definitely right . The formulas for the charge and potential difference are also correct . But in your book the mistake they have made is that they have applied ohm's law to get the expression for current. Well that is wrong because Ohm's law can be applied only for purely resistive circuits i.e without any capacitance or inductance. The actual formula can be derived by differentiating the expression for charge with respect to time.