Wheatstone bridge and its applications

Wheatstone Bridɡe:

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

      It consists of four resistances P, Q, R and S connected in the four arms of a square ABCD [Fig. 1]. A cell of e.m.f. E, is connected between the points A and C through a one way key K₁. A sensitive galvanometer of resistance G is connected between the terminals B and D through another one way key K₂. After closing the keys K₁ and K₂, the resistance P, Q, R and S are so adjusted that the galvanometer shows no deflection. In this position the wheatstone bridge is said to be balanced. 

Fig. 1. Wheatstone bridge

      Using kirchhoff's current law, the distribution of current and their directions through various resistances are shown in Fig. 1.

      Now, giving positive sign to the currents flowing in clockwise direction, and negative sign to the currents flowing in anti-clockwise direction and applying kirchhoff's voltage law to the mesh ABD, we can write, 

         i₁ P + ig G - (i - i₁) R =0                     ..... (1) 

      Similarly applying kirchhoff's second law to the mesh BCD, we can write, 

        (i₁ -.ig) Q - (i - i₁ + ig) S - ig . G = 0    ..... (2) 

      The right hand sides of both the equations (1) and (2) are zero because there is no source of e.m.f. in both the closed circuits ABD and BCD. 

      Since the bridge is balanced, therefore, the current 'ig' flowing through the arm BD is zero. Putting ig = 0 in equations (1) and (2), we get

           i₁P - (i − i₁) R = 0

or                         i₁P = (i - i₁) R             .... (3) 

and     i₁Q - (i - i₁) S = 0

or                        i₁Q = (i - i₁) S             .... (4) 

      Dividing equation (3) by equation (4), we get

    i₁P/i₁Q = (i₁ - i₁) R/(i₁ - i₁) S  or  P/Q = R/S

      This is the required condition for the bridge to be balanced and gives the principle of Wheatstone bridge. 

Special Case. Wheatstone bridge can also be balanced by connecting it in fourth arm CD (Fig. 2). In this arm BD may be connected through a key K₂ only. 

Fig. 2. A wheatstone bridge. 

      Keeping K₂ open press K₁ and note the deflection, say, θ. Current through CD is (i - i₁). Now press K₂ also. If the bridge is unbalanced, a current will flow through arm BD in either direction. This will result in a change in current through arm CD which is turn will change the deflection. If the bridge is balanced, there is no current through arm BD. So, on pressing key K₂ there will be no change in deflection in galvanometer. In such a case

          P/Q = R/G           or          G = R. Q/P

      This gives as a method for determination of resistance of galvanometer. This method is called Kelvin's method. 

Important notes :

1. A balanced wheatstone bridge shall remain balanced if the positions of source of e.m.f. and galvanometer is interchanged. 
2. When the galvanometer is between B and D, the indication for a balanced bridge is zero deflection in galvanometer. 
3. When the galvanometer in the fourth arm (CD), the indication of balanced bridge is same deflection in galvanometer whether the connection between B and D is open or closed. 

Applications of Wheatstone Bridge

1) Determination of unknown resistance. The unknown resistance 'X' to be determined by the Wheatstone bridge is connected in arms CD while three resistance boxes P, Q and R connected as shown in Fig. 3.

Fig. 3. Wheatstone bridge. 

 The values of resistances P, Q and R are so adjusted that the bridge is balanced and thus no current flows through the galvanometer. In this position, 

          P/Q = R/X         or         X = R × Q/P

Since the values of P, Q and R are known, therefore, the value of 'X' can be calculated. 

2) Slide wire bridge or meter bridge. It is an instrument to determine the unknown resistance of a conductor. It is a practical form of the Wheatstone bridge. 

    It consists of a thin uniform wire AC made of manganin or eureka. It is one meter long. It is stretched and fixed to two thick copper strips at A and C and is placed along a meter scale over a wooden board. A jockey J is connected to the terminal D through a galvanometer G as shown in Fig. 3.

Fig. 4. Slide wire bridge. 

 A resistance box is inserted in the gap between A and D while n known resistances R can be introduced between A and D. A cell is connected across A and C through a key K. 

   To find the value of the unknown resistance, the key K is closed and a suitable resistance R is introduced in the resistance box. Now the jockey is moved along the wire till, for a certain position, the galvanometer shows zero deflection. Let B be this position. The points A, B, C and D correspond to the same points as on the Wheatstone bridge. The resistances of wires of lengths AB (= l₁) 

and BC (= l₂) correspond to resistances P and Q respectively. Applying the principle of Wheatstone bridge, 

              P/Q = R/S                ..... (i) 

Since the wire has a uniform area of cross-section, the resistances P and Q should be proportional to their respective lengths l₁ and l₂.

∴          l₁/l₂ = P/Q                ..... (ii) 

Using equations (i) and (ii), we get, 

             

                 S = l₂/l₁ R

Knowing l₁, l₂ and R, the unknown resistances can be calculated. 

3) Post office box. It is a practical form of Wheatstone bridge. It measures resistance with greater accuracy. 

  

     Originally, it was designed to measure resistance of telegraph wires and for other similar work in the post offices. For this reason, it was named as 'post office box'. 

      Fig. 5 shows a post office box in its simplest form. The arms AB and BC contain resistances P and Q respectively. The arms are known as ratio arms. Each of these arms  contains resistances of 10, 100 and 1000 ohm. The third arm AD corresponds to resistance R. In this arm, the resistances are arranged in such a way that a resistance of 1 to 10,000 ohm can be introduced in steps of one ohm. The unknown resistance S is connected between the terminals C and D. This forms the fourth arm S of the Wheatstone bridge. 

      The terminals A and B are connected to the tapping keys K₁ and K₂ respectively. The terminal C is connected to the tapping key K₁ through a cell E. When the tapping key K₁ is pressed, A and C are connected through the cell. The key K₁ is referred to as the battery key. The terminal D is connected to the tapping key K₂ through a galvanometer G. When the tapping key K₂ is pressed, B and D are connected through the galvanometer. Key K₂ is referred to as the galvanometer key. 

Fig. 5. Post office box. 

       The battery circuit is always closed before the galvanometer circuit is closed. If the galvanometer circuit is closed first, then an opposing current (due to induction) will flow through the galvanometer which is not desirable. However, when the circuit is to be broken, it is always the galvanometer circuit which is broken first. 

     In order to determine the unknown resistance S, make P and Q equal to 10 ohm each. In this way the ratio arms will be made equal. Adjust the resistance R in the third arm AB so that on pressing first the battery key K₁ and then the galvanometer key K₂, the galvanometer gives no deflection. 

  

Since,              P/Q = R/S      

or                         S = Q/P R = 10/10 R = R

     So the resistance R in the third arm AB directly determines the unknown resistance S. In actual practice, the unknown resistance S is usually not a whole number, this, the galvanometer will not give zero deflection for any value of resistance R after the ratio arms are made equal. The post office box can measure resistance accurately upto the second place of  decimal, i.e., upto 1/1000th of an ohm. The method of measurement is described below :

     Let the unknown resistance S be 21.34 ohm. 

     Keeping P = 10 ohm and Q = 10 ohm, the resistance R in the third arm AB is adjusted so that R = 21 ohm deflects the galvanometer to one side and R = 22 ohm deflects the galvanometer to the other side. So, the unknown resistance S lies between 21 ohm and 22 ohm. 

Increase the value of P to 100 ohm and keep Q to 10 ohm. 

Now,       S = Q/P R = 10/100 R = 1/10 R

     It will be observed that R = 213 ohm deflects the galvanometer to one side and R = 214 ohm deflects the galvanometer to the other side. So, the unknown resistance S lies between 21.3 ohm and 21.4 ohm. 

Increase the value of P to 1000 ohm and keep Q equal to 10 ohm. 

Now,       S = Q/P . R = 10/1000 . R = 1/10 . R

      It will be observed that for a resistance of 2134 ohm, the galvanometer does not give any deflection. So, the unknown resistance is equal to 21.34 ohm. 

4) Carry-Foster's bridge. It is an improved form of Wheatstone bridge specially designed for measurement of flow resistances. It consists of a wire 1 meter long and having a uniform area of cross-section stretched between two copper strips [Fig. 6(i) ]. There are four gaps in the copper strips. Two resistance boxes P and Q are connected in between gap 2 and 4. A functional resistance box  is connected in gap 4 while the unknown resistance X whose value is to be determined is connected in gap 1. A battery is connected across A and C while a sensitive galvanometer is connected across B and a jockey D which can be slided to and fro on the wire. 

Fig. 6. Carry-Foster's bridge. 

     Introduce some resistance from P and Q. Introduce a small resistance from Y and adjust the jockey along the wire so as to obtain a balance point. Let 'l₁' be the length of wire giving balance point with X in gap 1 and Y in gap 4

        P/Q = resistance between A and D/resistance between C and D

or    P/Q = X + l₁r + ∝/Y + (100 -  l₁) r + β    ... (5) 

Where ∝ and β are the resistances of thick copper strips in between AD and in between CD  respectively and 'r' is the resistance per centimeter of the wire. 

     Interchanging X and Y and again find the length 'l₂' of the wire giving balance point [Fig. 6(ii)].

         P/Q = resistance between A and D/resistance between C and D

or    P/Q = Y + l₂r + ∝/X + (100 - l₂) r + β    ... (6) 

From equation (5) and (6) 

 

       X + l₁r + ∝/Y + (100 - l₁) r + β/Y + l₂r + ∝/X + (100 - l₂) r + β

Adding 1 to both sides

    X + l₁r + ∝/Y + (100 - l₁) r + β +1/Y + l₂r + ∝/X + (100 - l₂) r + β +1

or      X + Y + ∝ + β + 100r/Y + (100 - l₁) r + β = X + Y + ∝ + β + 100r/X + (100 - l₂) r + β

∴       Y + (100 - l₁) r + β = X + (100 - l₂) r + β

or                                 X = Y - (l₁ - l₂) r      ... (7) 

If 'r' is known, value if X can be calculated. 

Determination of 'r'

      To determine value of resistance per centimeter of the wire, X is placed by a thick copper strip (zero resistance). Let l₁' and l₂' be the lengths of wire giving balance point with strip in gap 1, Y in gap 4, and then interchanging the two. Using equation (7) with X = 0.

               0 = Y - (l₁' - l₂') r

or           Y = (l₁' - l₂') r

or            r = Y/l₁' - l₂'