Kirchhoff's laws for an electric circuits

Kirchhoff's laws: Kirchhoff had given two laws for electric circuits i.e.

Kirchhoff's Current Law or Junction Law

Kirchhoff's Voltage Law or Loop Law

Kirchhoff's Current Law or Junction Law: Kirchhoff's current law state that

The algebraic sum of all the currents at the junction in any electric circuit is always zero.

$\sum_{1}^{n}{i_{n}}=0$

Sign Connection: While applying the KCL, the current moving toward the junction is taken as positive while the current moving away from the junction is taken as negative.
The flow of Current in a junction
So from figure,the current $i_{1}$,$i_{2}$,$i_{5}$ is going toward the junction and the current $i_{3}$,$i_{4}$, So

$\sum{i}= i_{1}+i_{2}+(-i_{3})+(-i_{4})+i_{5}$

According to KCL $\sum{i}= 0$, Now the above equation can be written as

$i_{1}+i_{2}+(-i_{3})+(-i_{4})+i_{5}=0 \qquad$

$i_{1}+i_{2}+i_{5}=i_{3}+i_{4}$

Thus, the sum of current going towards the junction is equal to the sum of current going away from the junction.

In other words, at any junction, neither the charge accumulates nor the charge is removed. So this law represents conservation of charge.

Kirchhoff's Voltage Law or Loop Law: Kirchhoff's voltage law state that:

The algebraic sum of all the voltage or emf in any closed loop of an electric circuit is always zero.

$\sum_{1}^{n}{E_{n}}=0$

This means that the algebraic sum of all the emf applied in any closed loop is always equal to the algebraic sum of the product of current and resistance in the closed loop.

$\sum{E}=\sum {i.R}$

Sign Connection: While applying this law, a product of current and resistance is taken as positive when we traverse in the direction of the conventional current and the emf is taken positively when we traverse from negative to the positive electrode through the electrolyte.
Distribution of Current in a Loop of Circuit
So from the figure:

For Mesh $(1)$

$E_{1}-E_{2}=i_{1}R_{1}-i_{2}R_{2} \qquad(1)$

For Mesh $(2)$

$E_{2}=i_{2}R_{2}+\left( i_{1}+i_{2} \right)R_{3} \qquad(2)$

Comments

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

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Fraunhofer diffraction due to a single slit

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Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

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