Kirchhoff's law

Kirchhoff's Law plays a crucial role in physics, especially in circuit analysis. It enables engineers and physicists to design and analyze electrical circuits by assessing current distribution and voltage levels. Kirchhoff's Voltage Law (KVL) is particularly useful for calculating voltages in loops, facilitating the development of complex systems like power distribution networks and electronic devices. Moreover, these laws are essential for tackling challenges in various fields, including telecommunications, robotics, and renewable energy systems.

1.0Kirchhoff’s First Law Statement

• This law is also known as junction law or current law (KCL).According to this-In an electric circuit, the net sum of the currents gathering at any junction in the circuit is zero.
• Sum of the currents arriving the junction is equal to sum of the currents departing the junction

$\sum i=0$

$i_2+i_3+i_4-i_1-i_5=0$

$\sum i=0$

$i_2+i_3+i_4=i_1+i_5$  ………..(1)

$\text{ Sum of Current Entering }=\text{ Sum of Current Leaving }$

KCL is Based On Law Of Conservation Of Charge

• Charges do not accumulate at a junction in a circuit. The total charge entering a junction during a given time interval is equal to the total charge leaving it. This means that a junction cannot act as a source or sink of charge; rather, it maintains a balance where the incoming charge matches the outgoing charge over the same period.
• Multiplying equation (1) by small time interval t, then

$i_2+i_3+i_4=i_1+i_5$ ………..(1)

$i_{2}t+i_{3}t+i_{4}t=i_{1}t+i_{5}t\left(\therefore I=\frac{Q}{t}\right)$

$Q_2+Q_3+Q_4=Q_1+Q_5$

$\sum Q=0$

2.0Examples of Kirchhoff’s First Law

Example: Figure shows three resistances are connected with switch S. Initially the switch is open. When the switch S is closed, find the current passing through it.

Solution: Let V be the potential of the junction as shown in figure.

Applying junction law, we have 

$i_1+i_2=i_3$

$\frac{20-V}{2}+\frac{5-V}{4}=\frac{V-0}{2}$

40-2 V+5-V=2 V

5 V=45

V=9 V

$i_3=\frac{V}{2}=\frac{9}{2}=4.5\mathrm{A}$

3.0Kirchhoff’s Second Law Statement

It is also known as loop rule or voltage law (KVL) and according to it in any closed circuit the algebraic sum of e.m.f. and algebraic sum of potential drops is zero. 

$\sum IR+\sum E=0$

Sign convention for the application of Kirchhoff’s law 

1.   The change in potential in traversing a resistance in the direction of the current is –iR, while in the opposite direction it is +iR.

2. The change in potential in traversing an emf source from negative to positive terminal is +e, while in the opposite direction it is -e regardless of the direction of current in the circuit.

3. The change in potential in traversing a capacitor from the -ve terminal to the +ve terminal is +q/C, while in the opposite direction it is -q/C.

In the following closed loop, ABCDEA

$-I_1{R}_1+I_2{R}_2-E_1-I_3{R}_3+E_2+E_3-I_4{R}_4=0$

4.0Examples of Kirchhoff’s Second Law

Example: Find the current through 12 resistor in figure.

Solution: Let V be the potential at P; then applying KCL at junction P, we get

$I=I_1+I_2+I_3$

$\frac{15-V}{12}=\frac{V-2}{6}+\frac{V-3}{4}+\frac{V-4}{8}$

$V=\frac{68}{15}V$

$I=\frac{15-V}{12}=\frac{15-\left(\frac{68}{15}\right)}{12}=\frac{157}{180}A$

5.0Sample Questions on Kirchhoff’s Law

Q-1.Differentiate between Kirchhoff’s First Law and Kirchhoff’s Second Law.

Solution:

Q-3.Find the current in each Wire?

Solution:

Let potential at point B=0,Let potential at E=x. Applying Kirchhoff’s Law at junction E,

$\frac{x-10}{1}+\frac{x-30}{2}+\frac{x+14}{2}=0$

4 x=36 

x=9 V

Current in EF = $\frac{10-9}{1}=1A\text{ from }F\text{ to }$

$\frac{10-9}{1}=1\mathrm{A}\text{ from }F\text{ to }E$

Current in BE= $\frac{30-9}{2}=10.5\mathrm{A}\text{ from }B\text{ to }E$

Current in DE= $\frac{9-(-14)}{2}=11.5\text{ A from }E\text{ to }D$

Q-4.Find the potential at point A.

Solution:

Let potential at A=x ,Applying Kirchhoff’s Current Law at junction A,

$\frac{x-20-10}{1}+\frac{x-15-20}{2}+\frac{x+45}{2}+\frac{x+30}{1}=0$

$\frac{2x-60+x-35+x+45+2x+60}{2}=0$

6 x+10=0

$x=-\frac{5}{3}V$

Potential at point A= $-\frac{5}{3}V$