Kirchhoff’s Voltage Law (KVL,) or Kirchhoff’s Loop Rule. This law is based on the conservation of energy and may be stated as under:
In any closed electrical circuit or loop, the algebraic sum of all the electromotive force (e.m.f s) and voltage drops in resistors is equal to zero, i.e., in any closed circuit or loop.
The algebraic sum of e.m.f s + Algebraic sum of voltage drops = 0
The validity of Kirchhoff’s voltage law can be easily established by referring to the loop ABCDA shown in Fig.
If we start from any point (say point A) in this closed circuit and go back to this point (i.e., point A) after going around the circuit, then there is no increase or decrease in potential. This means that algebraic sum of the e.m.f.s of all the sources (here only one e.m.f. source is considered) met on the way plus the algebraic sum of the voltage drops in the resistances must be zero. Kirchhoff’s voltage law is based on the law of conservation of energy, i.e., the net change in the energy of a charge alter completing the closed path is zero.
V1 + V2 − V = 0
Kirchhoff’s voltage law is also called as loop rule.
Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).
RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5
RN = 7.66
The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.
According to voltage Division Rule
VAB = V1R3 ⁄ (R1 + R3)
= 24 × 8 ⁄ (4 + 8)
VAB = 16 V
Now Norton Current IN is
IN = VAB ⁄ RN
IN = 16 ⁄ 7.66
IN = 2.08
According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.
From the above Diagram
Current Flowing towards the Point: I2, I6, I4
Current Flowing Away from the Point: I1, I3, I5
Hence I2 + I6 + I4 = I1 + I3+ I5
Putting the value of the current
2A + 7A + I4 = 4A + 3A + 8A
I4 = 15A − 9A = 6A
I4 = 6A
The resistance of the conductor is determined by the
R = ρL/A
ρ = Resistivity of the conductor
L = Length of the conductor = 10 m
A = Area of the conductor = πR2 = 3.14 × 0.22 = 0.125 m2
R = Resistance = 2 Ω
Therefore the resistivity is
ρ = A × R ⁄ L
= 0.125 × 2 ⁄ 10
ρ = 0.025 Ω-m