Question

Write the statement of Junction and Loop Law of Kirchhoff s.

Answer 1

“In an electric circuit, the algebraic sum of current at any junction point must be zero.”
i. e., ΣI = 0
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kirchhoff s First Law or Junction Law.

It is based on the law of Conservation of Charge. When currents in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.

In figure at junction O,
I₁ + I₂ – I₃ – I₄ + I₅ = 0
or I₁ + I₂ + I₅ = I₃ + I₄ ………………. (1)
In figure the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.

Kirchhoff’s Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., ΣV = 0 …………… (1)
The algebraic sum of the emf’s in any loop of a circuit is equal to the sum of the voltages across the resistance.
ΣE = ΣIR ………….. (2)
Sign convention for applying Loop Rule
1. The product of current and resistance (IR) is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance (IR) is taken as negative if the resistor is traversed in the opposite direction of assumed current.

In loop we can move from A to B then the product of IR take as positive V_AB = +IR. [In the direction of current]
In a loop we can move from B to A then the product of IR take as negative.
∴ V_BA = -IR [Opposite direction of current]

2. When we move from B to A then E is taken positive (+). When we move from A to B, then E is taken negative (-).

Loop law can be understood by the figure
In the given figure for point d by the Junction Law, current flowing through the resistance R% will be
I₃ = I₁ + I₂ …………… (3)
In the loop adcba by the Kirchhoff s Loop Law.
I₁R₁ – I₂R₂ = ε₁ – ε₂
or I₂R₂ = I₂R₂ = ε₂ – ε₁ ………… (4)
In the loop defcd by the Loop Law
I₃R₃ + I₂R₂ = ε₂ …………….. (5)
By solving equations (3), (4) and (5) we can find out electric currents in different branches as well as we can find potential differences across the resistance.