I find it difficult to understand as to why current behaves differently in an RL and an RC circuit. At time $t=0$, in an RC circuit the current is maximum, whereas in an RL circuit its $0$. Why is it so? The current should rise from $0$ to some value in both the circuits, so where am I going wrong?
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[@Floris' answer](https://physics.stackexchange.com/a/143737/44176) being good, i'll give another view on the matter.

A **capacitor is equivalent to an open circuit** (since simply put, a capacitor is an element consisting of two plates which do not actually touch but through another medium, the *dielectric*, the circuit is **not connected** at that point where the capacitor is located), whereas an **inductor is equivalent to a short circuit** (since simply put, an inductor is an element which is a wire folded around a magnetic material, the circuit is **connected** at that point)

**For Direct current (DC):**

In DC how would these behave (as a first approximation)? 

The capacitor would be an open circuit while the inductor would be a closed circuit.

Now a capacitor (and an inductor) **are not immediately open or closed circuits**, for example the capacitor needs to be charged first, then it acts as an open circuit. Similarly the inductor needs to overcome any magnetic effects before acting as a closed circuit.

**For Alternating current (AC):**

So lets assume a frequency of operation where the capacitor does not have time to charge completely, then it will **not act as an open circuit** but maybe close to it (depending on frequency of operation)

Similarly an inductor will not act as closed circuit (depending on frequency of operation), but close to it.

The above provide basic explanation (and mnemonic rules) to infer the behaviour of capacitors/inductors in AC circuits

The images are from different sites (just for illustration), represent the resistor voltage in RC and RL circuits:

**RC in series (Resistor voltage in time):**

![RC Resistor voltage in time](https://i.stack.imgur.com/yyK3e.jpg)

**RL in series (Resistor voltage in time):**

![RL Resistor voltage in time](https://i.stack.imgur.com/Tq8kG.gif)
> 
> Why is it so?
> 
> 
> 

Well, it isn't actually *always* so. It will depend on the actual circuit configuration and whether the switch opens or closes when $t=0$.

But first, here are a couple of *crucial* results to *always* keep in mind when solving these type of switched circuits:

* the current through an inductor must be continuous
* the voltage across a capacitor must be continuous

In this context, this means that the current through the inductor the instant *after* the switch changes position must be the same as the current through the inductor *before* the switch changes position.

Similarly, the voltage across the capacitor the instant *after* the switch changes position must be the same as the voltage across the capacitor *before* the switch changes position.

So, for example, if we have a resistor $R$, inductor $L$, and switch in series with a voltage source $V$ and the switch is *closed* at $t=0$, it is indeed true that the inductor current is initially zero and asymptotically approaches a constant value

$$i\_L(0-) = i\_L(0+) = 0$$
$$i\_L(t) \rightarrow \frac{V}{R}\,,\quad t\rightarrow \infty $$

This must be the case since the inductor current is zero before the switch is closed and so, must be zero the instant after the switch is closed.

If the inductor is replaced with a capacitor, the voltage across the capacitor $C$ is initially zero and so, must be zero the instant after the switch is closed.

This means that the entire voltage $V$ is across the resistor and so, the series current must 'jump' from zero to $\frac{V}{R}$ at $t=0$ and then decay asymptotically to zero.

$$v\_C(0-) =v\_C(0+) = 0 \Rightarrow i\_C(0+) = \frac{V}{R}$$

$$i\_C(t) \rightarrow 0\,,\quad t \rightarrow \infty $$