Wheatstone Bridge

It consists of four resistors connected as shown in figure, one of them is to be measured. Let us assume that R₂ is unknown and R₃ can be varied and it is adjusted until no current flows in the galvanometer. In this condition, Wheatstone bridge is said to be balanced, and the resistances satisfy the ratio.

Important Points

Equivalent resistance
1. Between E and C i.e. across the diagonal of the cube
2. Between A and B i.e. across one side of the cube
3. Between A and C i.e. across the diagonal of one face of the cube
If a wire of resistance R, cut in n equal parts and then these parts are collected to form a bundle then equivalent resistance of combination will be
Wheatstone bridge is most sensitive if all the arms of bridge have equal resistances i.e. P = Q = R = S
In Wheatstone bridge to avoid inductive effects the battery key should be pressed first and the galvanometer key afterwards.
The measurement of resistance by Wheatstone bridge is not affected by the internal resistance of the cell.

Kirchoff's Laws

To analyze the circuit Kirchoff gave two laws: Junction Law and Loop Law

The Junction law

At any junction the sum of the currents entering the junction must equal the sum of the currents leaving the junction.

In figure, the currents directed towards the junction point are and .

Thus,

Note: This law is based on "conservation of charge".

Loop Law

The algebraic sum of all the potential differences along a closed loop in a circuit is zero.

While using this rule, one starts from a point on the loop and goes along the loop, either clockwise or anticlockwise, to reach the same point again. Any potential drop encountered is taken to be positive and any potential rise is taken to be negative. The net sum of all these potential differences should be zero.

Note: The loop law follows directly from the fact that electrostatic force is a conservative force and the work done by it in any closed path is zero.

Sign Convention

While applying Kirchhoff's voltage law to a closed circuit, algebraic sums are considered. Therefore, it is very important to assign proper signs to e.m.fs and voltage drops in the closed circuit. The following sign conventions may be followed:

A rise in potential should be considered positive and fall in potential should be considered negative.

(i) Thus in Fig. (i), as we go from A to B (i.e., from negative terminal of the cell to the positive terminal), there is a rise in potential. In Fig. (ii), as we go from A to B, there is also a rise in potential.

(ii) In Fig. (i), as we go from C to D, there is a fall in potential. In Fig. (ii), as we go from C to D, there is again a fall in potential.

Illustration of Kirchhoff's Laws

Kirchhoff's law can be beautifully explained by referring to Fig. The directions in which currents are assumed to flow is unimportant, since if wrong direction is chosen, it will be indicated by the negative sign in the final result.

(i) The magnitude of current in any branch of the circuit can be found by applying Kirchhoff's current law. Thus at point C in Fig., the incoming currents to the junction C are I₁ and I₂. Obviously, the current in branch CF will be I₁ + I₂

(ii) There are three loops in Fig. viz., ABCFA, CDEFC and ABCDEFA. Since there are only two unknown quantities (i.e., I₁ and I₂), we need only two equations in terms of I₁ and I₂. This can be achieved by applying Kirchhoff's voltage law to any two loops.

Loop ABCFA: In this loop, e.m.f. E₁ will be given positive sign. It is because as we consider the loop in the order ABCFA, we go from the negative terminal to the positive terminal of the battery in the branch AB and, hence, there is rise in potential. The voltage drop in branch CF is (I₁ + I₂) R₁ and shall bear negative sing. It is because as we consider the loop in the order ABCFA, we go with the current in the branch CF and there is a fall in potential. Applying Kirchhoff's voltage law to the loop ABCFA, we have

or ... (i)

Loop CDEFC: As we go round the loop in the order CDEFC, drop I₂R₂ is positive, e.m.f. E₂ is negative and drop (I₁ + I₂) R₁ is positive. Therefore, applying Kirchhoff's voltage law to this loop, we get,

I₂R₂ _ E₂ + (I₁ + I₂) R₁ = 0

or, I₂R₂ +(I₁ + I₂) R₁ = E₂

Since E₁, E₂, R₁ and R₂ are known, we can find the values of I₁ or I₂ from the above two equations.

Steps to Solve Circuits by Kirchhoff's Laws

Assume unknown currents in the given circuit and show their directions by arrows.
Choose any loop and find the algebraic sum of voltage drops plus the algebraic sum of e.m.fs in that loop and put it equal to zero.
Write equations for as many loops as the number of unknown quantities. Solve the equations to find the unknown quantities.
If the value of assumed current comes to be negative, it means that actual direction of current is opposite to that of assumed direction.

Wheatstone Bridge and Kirchhoff's Laws

This figure shows unbalanced Wheatstone bridge. Under such conditions, the galvanometer will carry current I_g Let the total current delivered by the battery be I. At junction A, total current is divided into two parts : I₁ flowing through P and (I _ I₁) flowing through R. At junction B, current I₁ is divided into two parts, I_g flowing through galvanometer and (I₁ _ I_g) passes through Q. At junction D, I_g and (I _ I₁) add up so that (I _ I₁ + I_g) flows through X. Let G be the resistance of the galvanometer.

Loop ABDA: Applying Kirchhoff's voltage law to the closed circuit ABDA, we have,

(_I₁P) + (_I_gG) +[(I _ I₁)R] = 0

or, I₁P + I_gG _ (I _ I₁) R = 0

Loop BCDB: Applying Kirchhoff's voltage law to the closed circuit BCDB, we have,

[_(I₁ _ I_g) Q] + [(I _ I₁ + I_g) X] + [I_gG] = 0

or (I₁ _ I_g) Q _ (I _ I₁ + I_g) X _ I_gG = 0

Under Balanced Conditions

Under balanced conditions of the bridge, I_g = 0. Putting I_g = 0 in eqs. (i) and (ii) above, we get,

I₁P = (I_I₁) R ...(iii)

I₁Q = (I_I₁) X ... (iv)

Dividing [(iii)/(iv)] we get

P/Q = R/X

or, PX = QR

Measurement of Temperature Using Wheatstone Bridge

We know that resistance of a conductor increases with increase in temperature. A platinum wire wound non-inductively on a mica former is connected in the arm CD of Wheatstone bridge as shown in Fig.

The platinum wire is immeresed in ice at 0ºC. Making P = Q, the value of R is adjusted so that on closing keys K₁ and K₂, the galvanometer gives zero reading. Since the values of resistances in the three arms AB, BC and AD are known, the value R₀ of the platinum wire at 0ºC can be found using Wheatstone bridge formula.
The platinum wire is then maintained at 100ºC by placing the wire in steam. Making P = Q, the value of R is so adjusted that on closing the keys K₁ and K₂, the galvanometer gives zero reading. Using Wheatstone bridge formula, we can find the resistance R₁₀₀ of the platinum wire at 100ºC.
The platinum wire is then placed in the bath whose temperature tºC is to be determined. Following the above procedure, the resistance R_t of the platinum wire at tºC can be found.

Calculation: If a is the temperature coefficient of resistance of platinum wire, then,

or, ...(i)

Also

or, ...(ii)

From eqs. (i) and (ii), we have,

or,

Since the values of R₀, R_t and R₁₀₀ are known, the unknown temperature t can be determined.