How does voltage behave in a parallel circuit? I am studying the series and parallel combinations of capacitors and resistors. I do know that voltage is divided in series whereas it's the same for all circuit components in parallel combination and the vice versa for current, I just want to know the logic behind it. I have read other posts about it, but I don't quite understand them so I'd just like a summarized, to the main point answer. Any help would be appreciated.
 A: I think it helps to remember that normal electronics operates in quasi-static limit so charges in the wires are essentially in equilibrium, any current is a slight perturbation on top of otherwise random motion.
Imagine connecting two or more wires into a vertex. Imagine the vertex-facing end of the first wire is at different potential to other wires you just connected. What will happen? The charges from all the wires will rush either towards or away from different-potential wire-end. As they do so, further charges will be repelled, thus very quickly a new equilibrium will be reached where all vertex-facing wire end are at the same electrical potential.
So we have the following 'law': Wires joined together are at the same electrostatic potential.
Now consider parallel and in-series resistors:

Parallel resistors
Imagine there is a voltage drop of $V_1$ across $R_1$. Let $\phi_{W1}=0V$ (we are free to choose the value of electrostatic potential once, since it's the differences between them that matter). Then we know that $\phi_{U1}=\phi_{W1}+V_1=V_1$. But since the three wires are joined at vertex $A$ we also then know that $\phi_U=\phi_{U1}=\phi_{U2}=V_1$. At the same time, three wires are joined at vertex $B$ so $\phi_W=\phi_{W1}=\phi_{W2}=0$. So then the voltage drop across the second resistor is $V_2=\phi_{U2}-\phi_{W2}=V_1$. Voltage drop across both resistors is $V=\phi_U-\phi_W=V_1$.
Series resistors
Let voltage drops across $R_1$ and $R_2$ be $V_1$ and $V_2$, respectively. Choose $\phi_W=0V$. Then we know that $\phi_L=V_2+\phi_W=V_2$. But two wires join at vertex $A$ so $\phi_K=\phi_L=V_2$. The voltage drop across $R_1$ is $V_1$ so $\phi_U=\phi_K+V_1=V_2+V_1$. Thus the voltage drop across both resistors is $\phi_U-\phi_W=V_1+V_2$.
To figure out what happens with current use the same diagrams but remember that charge is essentially incompressible in this setting, so all the current that goes into a vertex, or into a resistor, or any portion of your circuit, must come out of it. So, for example, in case of in-series resistors, at vertex A. If you know current in two out of three wires, the current in the third wire will simply be whatever is needed to ensure that all the current that goes into $A$, leaves it.

You can think of electric potential here as the pressure applied on all the charges. As long as pressure is the same, there is only random motion, but any difference in pressure leads to compensating re-distribution.
The picture I gave above is for electrostatic regime and zero-resistance wires, things become more complex once you depart from these assumptions
