How can we understand Kirchhoff's voltage law from the principle of conservation of energy? Kirchoff's voltage law states that the algebraic sum of the voltage drops in a closed circuit is zero. It is mentioned in a textbook that it is a statement of the conservation of energy but doesn't explain it in any detail. How can we understand this?
 A: A vector field $\boldsymbol{A}$ is conservative if the integral
$$\int_{\Gamma}\boldsymbol{A}\cdot d\boldsymbol{l}\tag{1}$$
is path independent i.e. it does not depend on $\Gamma$.
$(1)$ is equivalent to
$$\oint\boldsymbol{A}\cdot d\boldsymbol{l}=0\tag{2}$$
for any closed path.
Now, electrostatic energy difference between to points $A$ and $B$ is given by
$$\Delta\mathcal{E}=qV=-q\int_{\Gamma_0}\boldsymbol{E}\cdot d\boldsymbol{l} \tag{3}$$
As you can see, $\Delta\mathcal{E}=0$ if $\Gamma_0$ is a closed path because the electrostatic field is conservative. $(2)$ applied to electric field is none other than KVL; now, if the energy difference were to depend on the path, in general $(3)$ would be non-zero even for closed path, yielding a energy difference even though $A=B$, thus energy would be path dependent.
A: In simple terms, going around a closed loop Kirchhoff tells you that the sum of the potential differences is zero.
For example, suppose a simple series circuit with three nodes $a,\,b $ and $c$, then  $V_{\rm a \to b}+V_{\rm b \to c}+V_{\rm c \to a}=0$.
Multiply each term by the current $I\Rightarrow IV_{\rm a \to b}+IV_{\rm b \to c}+IV_{\rm c \to a}=0$ which you can interpret as the sum of sources of electrical power plus the sum sinks of electrical power all adding up to zero power, ie a restatement of the law of conservation of energy useful for electrical circuits.
A: Kirchhoff's voltage law is a consequence of the scalar nature of Coulombs law and the conservative nature of the Coulomb force. The Coulomb potential only depends on position, not on the path connecting positions. It is valid only if the current is stationary. And yes it is also consistent with energy conservation.
A: The short explanation is the gain in electrical potential energy per unit charge that a voltage source supplies a circuit equals the sum of the losses in electrical potential energy per unit charge in the circuit due to the work required to move the charge through a circuit. This is equivalent to saying the voltage rise of the voltage source connected to a circuit equals the sum of the voltage drops in the circuit, which is a statement of Kirchhoff's voltage law.
Keep in mind that the voltage difference between two points is defined as the change in electrical potential energy per unit charge (Joules per Coulomb, or J/C) moving the charge between the two points. That change can be positive or negative.
For example, for conventional current, positive charge moving inside a 10 V battery from the negative terminal to positive terminal gains electrical potential energy of 10 Joules per Coulomb (10 J/C), from the conversion of chemical potential energy to electrical potential energy.
Let that Coulomb of charge move through three resistors in series of $2\Omega$, $3\Omega$, and $5\Omega$. From Ohms law the current is 1 Ampere and the resulting voltage drops in the three resistors are 2V, 3V and 5V, or a loss of 2 J/C, 3 J/C and 5J/C of potential energy, respectively. The total loss of 10 J/C of potential energy (ultimately dissipated as heat), equals the gain of 10 J/C of potential energy from the battery, for conservation of energy.
Hope this helps.
A: In the approximation of electrical circuits, we can consider that an electrical circuit is made up of capacitive elements (Electroquasistatics), for which the energy $\frac{1}{2}\frac{1}{C}Q^2$ is mainly stored in the electric field, and inductive elements (Magnetoquasistatics) for which the energy $\frac{1}{2}Li^2$ is stored in the magnetic field.
The total energy is of the form $E_{em}=\frac{1}{2}Li^2+\frac{1}{2}\frac{1}{C}Q^2$
If there are no dissipative elements, we have $\frac{dE_{em}}{dt}=0$ which gives: $Li\frac{di}{dt}+\frac{Q}{C}\frac{dQ}{dt}=0$
Limiting ourselves to the simple circuit $(L,C)$, we have $i=\frac{dQ}{dt}$ and we obtain Kirchoff's law: $L\frac{di}{dt}+\frac{Q}{C}=0$
If there is also a resistive element that dissipates electromagnetic energy, we must write $\frac{dE_{em}}{dt}=-Ri^2$ and we obtain Kirchoff's law: $L\frac{di}{dt}+\frac{Q}{C}+Ri=0$
