Why use Kirchhoff's law for capacitors? [closed]

My teacher stated that we could use the Kirchhoff law for capacitors but the law is applicable for the moving charge how do we apply it on stationary charges. And he mentioned we take the product of current and resistance in the law according to the direction in which we are looking if the charge is seen to go in direction of current we take negative sign but if opposite positive sign? Also how, $$\sum IR=\sum V$$And he stated that the current at the junction of the parallel connected capacitors is "0".

closed as unclear what you're asking by Jon Custer, Yashas, Bill N, David Hammen, Michael SeifertMay 12 '17 at 15:04

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

For resistors when we write $V=IR$, we actually mean that if $I$ current is passing through a resistor of resistance $R$, then a unit charge would lose $V$ amount of energy.

The total energy (including energy lost due to internal resistance of the voltage supplier) lost by the unit charge in the whole circuit, is equal to the emf of the battery.

This is applicable for moving charges because, when a charge starts and ends its motion at a same point under electric field, net work done in $0$. This is because, electric force is a conservative force.

This is the working principle of Kirchhoff's law.

For capacitors,

$C=\frac{Q}{V}$, capacitance can be defined as the amount of charge required to raise the potential of a body by $1 volt$.

$V=\frac{Q}{C}$.

This can be interpreted as that when the charge on the capacitor is $Q$ of capacitance $C$ then V volts of energy will be lost across it. In a capacitor an electron from one plate goes to the other plate and these plates are at the same potential. So the net work is 0.

Regarding the sign problem,

The basic idea is when a unit charge goes from High potential to low potential it loses energy and thenceforth comes the -ve sign. Current flows from high potential to low potential.

If ,while solving a circuit problem by Kirchhoff's law, you select any arbitrary direction of current (which may or may not be correct) and use the sign conventions properly (according to your assumed direction irrespective of the actual direction) then you will get the correct answer. This is the beauty of this law.

It's true because after getting the resultant equation from Kirchhoff's law, you equate it to $0$. So it won't matter if you multiply both side by -1, you would still get the same answer, as long as you form the equation correctly.

If you get +ve value of current, your initial assumption of the current's direction was correct and if you get a -ve value the direction of current is opposite to the one you initially assumed but it has the same absolute value.