Question

(a) Use Kirhhoffs rules to obtain the balance condition in a Wheatstone bridge.

(b) Draw a circuit diagram for determining the unknown resistance R using meter bridge. Explain brefly its working, giving the necessary formula used.

Answer 1

Wheatstone Bridge: Wheatstone bridge is an electrical arrangement which forms the basis of the instruments used to determine an unknown resistance.

Principle of Wheatstone bridge: If four resistances P, Q, R and S are arranged as shown in figure, if the galvanometer does not show any defection, then

\(\frac{P}{Q} = \frac{R}{S}\)

Derivation

Wheatstone bridge consists of four resistances P, Q, R and S arranged as shown in Fig. with a cell E between the points A and C and a galvanometer G between B and D.

Let I be the total current given out by the cell which on reaching the point A divides itself into two parts : (a) I₁ flowing through P and (b) (I - I₁) through R. The current I₁, on reaching the point B, divides itself into two parts : (a) I_g through the galvanometer G and (b) (I₁ - I_g) through Q. The current Ig through the arm BD and (I - I₁) through AD combine at D to send a current (I - I₁ + I_g) through S. On reaching the point C, the current (I₁ - I_g) through BC and (I - I₁ + I_g) through DC combine to give the total current I, thus completing the circuit. (The values of the currents at a junction can be verified by applying Kirchhoffs first law at the junction).

Applying Kirchhoffs second law to the closed mesh ABDA, we get

I₁P + I_gG - (I - I₁) R = 0. ...................(1)

where G is the resistance of the galvanometer.
Again applying Kirchhoffs second law to the closed mesh BCDB we get,

(I - I_g) Q - (I - I₁ + I_g) S - I_gG = 0 ...........................(2)

The value of R is so adjusted that the points B and D are at the same potential and as such no current flows in the arm BD i.e. I_g = 0. This is indicated by zero deflection in the galvanometer G. In this position, the bridge is said to be balanced. Putting I_g = 0 in equations (1) and (2), we get 

I₁P  - (I - I₁) R = 0

or  I₁P = (I - I₁)R .......................(3)

and I₁ Q - (I - I₁) S = 0

i.e. I₁ Q = (I - I₁) S ........................(4) 

Dividing (3) by (4), we get

\(\frac{I_1P}{I_1Q} = \frac{(I-I_I)R}{(I_1-I_1)S}\)

or \(\frac{P}{Q} = \frac{R}{S}\)

The practical form of Wheatstone’s bridge is slide wire bridge.

Slide wire bridge: It consists of uniform wire AC, usually one metre long, soldered to the ends of two thick rectangular strips of copper A and C, fixed on a wooden base.

A sliding contact B, called the jockey, can be moved along a graduated scale. By pressing a small key mounted on the jockey, contact can be made at any point along the length of the wire.

If the length of the wire is 1 metre, it is called meter bridge. The unknown resistance is placed across the gap DC and a known resistor R is placed across the gap AD. The two resistances P and Q are obtained from the wire AC by moving the jockey B connceted to the strip D through a galvanometer G.

Working: The key K in the cell circuit is inserted and the jockey is moved along the wire till, for a certain position B, that galvanometer shows no deflection. The bridge is then said to be balanced, if P and Q are the resistances of the parts AB and BC of the wire, we have

\(\frac{P}{Q} = \frac{R}{S}\)

If the slide wire is of uniform cross-section, the resistance, of AB and BC shall be proportional to their lengths, say l and (100 - l) respectively. If r is the resistance per cm length of the wire, we have

\(\frac{P}{Q} = \frac{lr}{(100-l)r} = \frac{l}{100-l}\)

Substituting in the above equation, we have

\(\frac{l}{100-l}\) = \(\frac{R}{S}\)

or S = \(\frac{100-l}{l}R\) 

Knowing l and R; S can be calculated.

This arrangement cannot be used for determining very low resistance because this apparatus is most sensitive only when all the resistances of all arms [P, Q, R and S] are comparable i.e. neither too large, nor too small.

Advantage: It is a null method and therefore it is accurate.