Question

Give statements of Kirchhoff s junction and loop laws. Find out the balance condition of Wheatstone’s bridge with the help of these laws.

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 emfs in any loop of a circuit is equal to the sum of the voltages across the resistances.
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 resistances.

Construction: The construction of Wheatstone’s bridge is shown in following diagram (figure) in which a quadrilateral ABCD is prepared by

connecting four resistances P, Q, R and S. A cell is connected between the points A and C through key K₁ which is known as battery key and a galvanometer between the point B and D through another key K₂ which is known as galvanometer key. If on pressing the key K₁ and K₂, there is no deflection in galvanometer, then the bridge is said to be balanced condition. In this situation the relation between all the resistances is given by:
\(\frac{P}{Q}=\frac{R}{S}\)

Principle of Wheatstone Bridge and Condition of Balance
When battery key K₁ is pressed, then main current I starts flowing in the circuit. At junction A this current splits in two parts I₁ and I₂ as shown in figure. These currents I₂ and I₂ again obtain two paths at junctions B and D respectively. Now following three situations are possible:

(i) When V_B > V_D. then current I₂ passes as such through resistance S but current I₁ splits in two parts at junction B, one part I_g passes through galvanometer showing deflection in one direction while rest current (I₁ – I_g) passes through Q.

(ii) When V_B < V_D. then opposite situation to previous situation (i) is observed i.e. .current I₂ splits in two parts at D. Now the current I_g produces deflection in galvanometer in opposite direction to previous deflection.

(iii) When V_B = V_D , then no current passes through galvanometer i.e.. galvanometer shows no deflection because now I _g = 0. This situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called ‘Bridge circuit’ Thus in balanced condition.

Therefore, “it is clear that in balanced situation of the bridge, ratio of resistances of any two corresponding arms of quadrilateral ABCD is equal to ratio of two remaining corresponding arms”.
From equation (3).
S = \(\frac{Q}{P} \times R\)
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below :
(i) P and Q are called ratio arms.
(ii) R is called variable resistance arm.
(iii) S is called unknown resistance arm.