The answer depends on when you look

The rules look like the resistor rules with the connections swapped, and they are:

$$C_{\text{parallel}} = C_1 + C_2 + \cdots, \qquad \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$$

So a series combination lands below the smallest capacitor in it: $2\ \mu$F and $4\ \mu$F in series give $1.33\ \mu$F, not $6\ \mu$F.

Check the direction against the plate picture rather than memorising which rule goes where. Connecting capacitors in parallel is effectively adding plate area, which raises $C$. Connecting them in series stacks the gaps, which is a larger effective separation, and that lowers $C$.

One more series fact that gets replaced by a guess: capacitors in series carry the same magnitude of charge on every plate, because the charge on the inner plates is separated out of a region that was neutral. It does not divide in proportion to $C$.

"Capacitors block direct current" is the long-time behaviour, and applying it at $t = 0$ gets the initial current exactly backward.

An uncharged capacitor has $Q = 0$, so $\Delta V = Q/C = 0$ across it. An element with no potential difference across it is behaving like a plain wire. So:

1. Replace the uncharged capacitor with a wire.
2. Solve the resulting resistor circuit.
3. That is the current at $t = 0$, and it is the largest current that branch ever carries.

With a $10$ V battery and a single $100\ \Omega$ resistor, the initial current is $0.10$ A, not zero. Charge has to arrive before any potential difference can build, and only as it builds does the branch stop conducting.

Once the transient has died away, no more charge is arriving, so the current in the capacitor's branch is zero. Redraw with that branch open and solve the resistor circuit that remains.

Then read the capacitor's potential difference off the two points it actually bridges. That is often not the emf. A capacitor across the $4\ \Omega$ resistor of an $8\ \Omega$ and $4\ \Omega$ divider on a $12$ V source ends at

$$\Delta V_C = 12 \times \frac{4}{12} = 4\ \text{V}, \qquad Q = C \times 4\ \text{V}.$$

One consequence catches people out: with no current in that branch, a resistor placed in series with the capacitor has no potential difference across it at all in the steady state. It matters only during the transient, where it sets how fast the crossing happens.

The crossing between the two easy states is exponential, with time constant

$$\tau = RC.$$

In one time constant a charging capacitor reaches about 63% of the charge it will finish with, and a discharging one falls to about 37% of what it started at. The approach continues from there, and after about five time constants it is over for practical purposes.

Two things to keep straight. $\tau$ is not a finishing time: the capacitor is not full at $t = \tau$. And the scaling runs the way the physics does, not the way "faster" sounds: a larger $R$ or a larger $C$ makes the process slower. More resistance means less current available to move charge onto the plates; more capacitance means more charge to move.