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'

 Kirchhoff's Law for Electrical Networks:

First law :

       It states that the algebraic sum of currents meeting at a point is zero. 

       This law may be called Kirchhoff's current law (KCL). 

       To explain this law consider a number of wires connected at a point P. Currents i₁, i₂, i₃, i₄ and i₅ flow through these wires in the directions as shown in Fig. 1.

Fig. 1. Currents meeting at a point. 

       To determine their algebraic sum of electric currents, we follow the following sign conventions. 

• The currents approaching a given point are taken as positive. 
• The current leaving the given point are taken as negative. 

       Following these sign conversations, we find that i₁ and i₂ are positive while i₃, i₄ and i₅ are negative. 

       According to kirchhoff's first law,

          i₁ + i₂ - i₃ - i₄- i₅ = 0        or   ∑i = 0

       The most evident conclusion of this law is that the current entering a point must be the same as that leaving it. Hence, there cannot be any accumulation of charge at any point in a conductor. This is in accordance with law of convention of charge. 

Second law :

       It states that in a closes electric circuit, the algebraic sum of e.m.f. is equal to the algebraic sum of the products of the resistances and the currents flowing through them. 

       This law may be called kirchhoff's voltage law (KVL). It is in accordance with the law of conservation of energy. 

       Fig. 2. Shows a closed electric circuit.

Fig. 2. A closed electric circuit. 

ABCD containing resistances r₁, r₂, r₃, r₄ and r₅ in the parts AB, BC, CD, DA and AC respectively. Also, let i₁, i₂, i₃, i₄ and i₅, be the respective currents flowing in these parts in the directions shown by arrow heads. Two sources of e.m.f's. E₁ and E₂ are also connected in the mesh. 

       In order to use kirchhoff's voltage we shall follow the following sign conventions. 

(1) If the electric current flows through the electrolyte of the cell from negative to positive terminal [Fig. 3(i)] the e.m.f.of the cell is taken as positive (+E). 

(2) If the electric current flows through the electrolyte of the cell from positive to negative terminal [Fig. 3(ii)] the e.m.f. of the cell is taken as negative (-E).                                                                                        

Fig. 3. Sign conventions for source of e.m.f.

       
(3) If the path taken to traverse the resistance is along the direction of current [below Fig. 4(i)], the final point (B) is at a lower potential than the initial point (A). The product of current and resistance in this case is taken as negative (-ir).

Potential difference = current (i) × resistance (r) 

  = (Potential of final point) - (Potential of initial point) 

  = Vb - Va

Since  Vb < Va,(i × r) is negative. 

Fig. 4. Sign convention for 'ir'. 

(4) If the path taken to traverse the resistance is against the direction of current [above Fig. 4(ii)], the final point (A) is at a potential higher than that of initial point (B). The product of current and resistance in this case is taken as positive (+ir).

Potential difference = current (i) × resistance (r) 

   = (Potential of final point A) - (Potential of initial point B) 

   = Va - Vb

Since Va - Vb , (i × r) is positive.

 

It may be noted here that whenever there is a single cell in a closed circuit and the current is unidirectional, the sum of the products of currents and resistances is taken to be equal to the e.m.f. of the cell, irrespective of the direction of current. 

 

     Apply kirchhoff's second law to the mess ABC. We can write

          i₁r₁ + i₂r₂ − i₅r₅ = E₁ − E₂

     Aɡain, applyinɡ kirchhoff's second law to the mess ACD, we ɡet

         i₅r₅ − i₃r₃ − i₄r₄ = E₁

     

     The above two equations may be written in the followinɡ ɡeneral form

        ∑ir = ∑E

                                                                              

Important notes :

1. From kirchhoff's current law it can be concluded that the potential at any point, in an electric circuit, remains constant whatever may be the number of wires connected at it. 
2. Kirchhoff's voltage law, may be seen as the generalisation of Ohm's law keeping in mind the convention regarding signs of E and i. Ohm's law is also based upon conservation of energy. 

Example 1. A battery of e.m.f. 6 volt and internal resistance 5 ohm is joined in parallel with another of e.m.f. 10 volts and internal resistance 1 ohm and combination is used to send a current through an external resistance of 12 ohm. Calculate the current through each battery. 

Solution. 

Let batteries of e.m.f.s E₁ and E₂ send currents i₁ and i₂ respectively as shown in Fig. 5. The two currents meet at A so that a current 'i' flows through the external resistance of 12 ohm. 

Fig. 5.

∴                i = i₁ + i₂

Applying Kirchhoff's second law to the mesh containing battery of e.m.f. E₁ and external resistance R, we get

             i₁ × 5 + i × 12 = 6

or                 5i₁ + 12i = 6          .... (1) 

Again, applying Kirchhoff's second law to the mesh containing battery of e.m.f. E₂ and external resistance R, we get

     

            i₂ × 1 + i × 12 = 0

or                   i₂ + 12i = 0          ....(2) 

Multiplying equation (2) by 5 and adding to equation (1), we get

              5(i₁ + i₂) + 72i = 56

or                   5i₁ + 72i = 56      [∵ i₁ + i₂ = i]

or                            77i = 56

or                                i = 56/77 A. 

Substituting this value of i in equation (1), we get 

             5i₂ + 12 × 56/77 = 6

or         5i₁ = [6 - 672/77]A = -210/77 A

or           i₁ = - 210/385 A = - 6/11 A. 

Again      i = i₁ + i₂

or           i₂ = i - i₁ = 56/77 A - (-6/11) A

                  = 98/77 A

                  = 14/11 A. 

Example 2. A cell of e.m.f. 2 volt and internal resistance 1 ohm is connected with two wires of resistances 2 and 5 ohm in parallel. Find the current through each wire. 

Solution. 

Suppose the cell sends a current 'i' through the circuit (Fig. 6). At the point C, the current is divided into two parts i₁ and i₂.

Fig. 6. 

                i₁ + i₂ = i

The currents i₁ and i₂ flow through resistances of 2 ohm and 5 ohm respectively. Applying Kirchhoff's second law to the mesh 'FECDE', we get

  

          i × 1 + i₁ × 2 = 2     or      i + 2 i₁ = 2

or      (i₁ + i₂) + 2i₁ = 2      [∵ i = i₁ + i₂]

or                      3i₁ = 2            .... (1) 

Applying Kirchhoff's second law to the mesh DBAC, we get

or          2i₁ - 5i₂ = 2,    2i₁ = 5i₂

or                    i₁ = 5/2 i₂

Substituting this value of i₁ in equation (1), we get

                 3 × 5/2 i₂ + i₂ = 2      

or                                 i₂ = 4/17 A

Again,                         i₁ = 5/2 × 4/17 A

                                        = 10/17 A. 

Electromotive force (E.M.F) and Potential Difference

 Electromotive Force (E.M.F.): 

Electric current flows through a conductor due to the motion of charge. We know that motion cannot take place without the application of force. So we can safely, assume that the flow of current is the result of application of a force on the charge carriers. This force comes into existence only when a source of electricity (a cell, battery, etc.) is connected across the conductor. Due to some process in source of electricity, energy is liberated which crates a potential difference across the two ends of conductor. As a result of this an electric field is set up in the conductor which exerts a force on the charge, thus, producing motion on it. As the charge 'q' is taken round the circuit, some work 'W' is done. 

     Work done per unit charge to take the charge around the circuit is called electromotive force. 

        E.M.F = W/q

     Unit of E.M.F. in S.I. is joule/coulomb.

     Since joule/coulomb represents the potential difference so, potential difference and E.M.F. are dimensionally similar.   

Electro-motive force (e.m.f.)

It is the difference of potential across the two terminals of a source of current when there is  no electric current flowing through the external resistance. 

Electromotive Force and Potential Difference:

Fig. 1. shows a source of current i.e., a cell connected to a conductor of resistance R through a key K. A voltmeter V is connected across the two terminals of the cell. 

     When the key K is open [Fig. 1(i)], the reading of voltmeter gives the electromotive force (e.m.f.) E of the cell. 

     The electromotive force E of a cell is defined as the difference of potential between its terminals when there is no current in the external circuit, i.e., when the cell is in open circuit. 

     When the key K is open [Fig. 1(ii)], an electric current i flies from positive to negative terminal in the external circuit and from negative terminal to positive terminal within the cell. Thus, a potential drop 'r' takes place across the internal resistance 'r' (shown separately) of the cell. The direction of 'ir' is opposite to the e.m.f. 'E' of the cell. The reading of the voltmeter now decreases and gives the potential difference V between the terminals of the cell. 

Fig. 1. Distinction between e.m.f. and potential difference. 

     The potential difference of a cell is the difference of potential between two terminals when it is in closed circuit. 

∴       V = E - ir       or        E = V + ir

     Hence, E is greater than V by an amount equal to the drop of potential within the cell due to its internal resistance. 

Distinction Between E.M.F. (E) and Potential Difference (V):

E.M.F. (E) and potential difference 'V' can be distinguished from each other in following respects. 

1. E.M.F. is the cause while potential difference is its effect. 

2. It is associated only with cells while potential difference can exist across a resistor. 

3. E.M.F. is measured by the motion of a charge around a circuit while potential difference is measured by motion of charge between two points. 

4. E.M.F. is measured in an open electric circuit while potential difference measured in a closed electric circuit. 

5. Refer Fig. 2(i), cell is getting discharged through an external resistance R. 

Fig. 2. Discharging and charging of a cell. 

          i = E/R + r        or       E = iR + ir

Where 'r' is 5he internal resistance of the cell. 

Since, iR = potential difference across external resistance = V

∴      E = V + ir       or      E - V = ir      .... (i) 

During charging Fig. 2(ii), the charger sends a current through the cell. It can be noted that direction of current through internal resistance 'r' is in direction opposite to that in first case. 

Charge i to -i in equation

               E - V = -ir                             .... (ii) 

From equations (i) and (ii) we can conclude that :

• E > V during discharging of cell
• E > V during charging of cell
• E = V in open circuit (i = 0).

Examples-

1) 36 cells each of internal resistance 0.5 ohm and e.m.f. 1.5 V each are used to send current through an external circuit of 2 ohm resistance. Find the best mode of grouping them and the current through the external circuit ? 

Solution-

Given                 E = 1.5 V,  r = 0.5 Ω 

and                     R = 2 Ω

Total number of cells,      mn = 36             ... (i) 

For maximum current,   

 

                       R = nr/m

                       2 = n × 0.5/m

or                   n = 4 m                                    ... (ii) 

Substituting for n in equation (i), we get

                         m × 4m = 36

or        m² = 9         or            m = 3               

 Using equation (ii), we get        n = 12

Thus, the best mode of grouping will be to have 12 cells connected in series with each other and 3 cells such rows in parallel. 

Current        i = mnE/mR + nr

                        = 36 × 1.5/3 × 2 +12 × 0.5  . A

                        = 36 × 1.5/12 

                        = 4.5 A. 

2) 12 cells each having the same e.m.f. are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is in series with an ammeter and two cells identical with other. The current is 3 A when the external cells aid this battery and is 2 A when the cells oppose the battery. How many cells in the battery are wrongly connected ? 

Solution- 

Let 'E' be the e.m.f. of each cell and 'n' be number of cells wrongly connected. 

Number of cells in correct order = 12 - n

Net e.m.f. of the battery

           = (12 - n) E - nE

           = (12 - 2n) E

(i) When the two external cells aid this battery

    Net e.m.f.  = (12 - 2n) E +2E

∴             (12 - 2n) E +2E/R = 3

where R = external resistance. 

or                 (14 - 2n) E = 3R                     ... (i) 

(ii) When the two external cells oppose this battery

     Net e.m.f. = (12 - 2n) E - 2E 

∴               (12 - 2n) E - 2E/R = 2

or                        (10 - 2n) E = 2R            ... (ii) 

Dividing equation (i) by equation (ii), 

             (14 - 2n) E/(10 - 2n) E = 3R/2R

                                      28 - 4n = 30 - 6n

or                                          2n = 2

or                                            n = 1

Thus, there is only one cell which is wrongly connected in the battery. 

Resistivity (specific resistance)

Resistivity (ρ) :     

                                                     

As from the equation R = ρ × l/A  

                                                    

It is clear that, resistance 'R' of a conductor depends on the following factors:

    (i) Length 'l' of the conductor;     R ∝ l

i.e., resistance of a conductor varies directly as its length. 

    As potential difference is applied across the two ends, free electrons move from the end at lower potential to the end at a higher potential. In this process they collide against each other and undergo retardation. A greater length of conductor results in greater number of collisions, thereby producing greater retardation and hence greater resistance. 

    (ii) Area of cross-section 'a';     R ∝ 1/a

    Resistance of a conductor varies inversely as its area of cross-section. 

    For a conductor having greater area of cross-section more free electrons cross that section of conductor in one second, thereby giving large current. A large current means a lesser resistance. 

Combining the two factors together, 

      R ∝ l/a        or        R = ρ l/a

Where ρ is the resistivity or the specific resistance of the material. 

   

  If        l = 1, A = 1, R = ρ

Definition of Resistivity:

The resistivity of a material is defined as the resistance of a conductor made up of the material having unit length and area of cross section. 

Alternatively, it is also defined as the resistance in between the two opposite faces of a unit cube made up of that material. 

Unit of resistivity (ρ) :

in S.I= ohm × m²/m

         = ohm × m

n C.G.S= stat ohm×c.m²/c.m

             = stat ohm×c.m

Dimension of resistivity

[ρ] = [R] × [a]/[l]

[ρ] = [M1 L2 T-3 A-2] × [L2]/[L1]

[ρ] = [M1 L3 T-3 A-2]

So, the dimensions of resistivity are 1, 3, -3 and -2 in mass, length, time and electric current respectively. 

Variation of Resistivity with Temperature:

(i) Conductors. The resistance R of a material is given by

                     R = ρ. l/a

                                                   

As from the equation R = ρ × l/A

                                                  

                            R = m/ne²τ . l/a

Comparing these two equations 

                            ρ = m/ne²τ

     'τ' is the 'relaxation time' which depends upon the temperature of the conductor. An increase in the temperature results in an increase in amplitude of atomic vibrations which in turn decreases τ. Therefore, resistivity ρ, of the conductor, increases with an increase in temperature. Vibration of ρ with temperature for copper is shown [in Fig. 1]. The curve is a straight lines around room temperature. It deviates beyond 800 K and also in the low temperature region. At absolute zero the material possesses some minimum finite resistivity. 

Fig. 1. Variation of resistivity with temperature for a conductor. 

If ρ₀ and ρt are resistivity of the conductor at 0⁰ at t⁰C respectively, then

         ρt = ρ₀ [ 1 + ∝ (T - T₀)

or      ∝ = ρt - ρ₀/ ρ₀(T - T₀)

'∝' is called the temperature coefficient of the resistivity. 

          ρ₀ = 1, T - T₀ = 1, ∝ =  ρt -  ρ₀

Temperature coefficient of resistivity of a conductor is defined as the change in resistivity per unit resistivity per degree celsius rise of temperature. 

(ii) Insulators. In case of insulators the resistivity, almost, increases exponentially with a decrease in temperature and tends to infinity as the temperature approaches absolute zero. Resistivity  ρ₀ and  ρt at 0⁰C and t⁰C, respectively, are connected by the relation

          ρt =  ρ₀  e-Eg/kT      ..... (1) 

Where 'Eg' is called the positive energy of the insulator and  'k' is Boltzmann constant. Eg represents the difference between conductor band and valence band and has value ranging from zero to several electron volt. Depending upon the value of Eg, the substance behaves like an insulator or as a semi-conductor.  

For Eg < 1 eV, value of  'ρ' at room temperature is not very high. Therefore, the substance acts as a semi-conductor. For Eg > 1 eV, the value of ρ at room temperature is high (≈ 103 ohm). So, the substance acts as an insulator. 

Since number of electrons per cm³ of the material varies inversely as the resistivity, 

              n ∝ 1/ρ

∴     From equation (1),

              nt = n₀ e-EgkT

Where  n₀ has a value of the order of 10²⁸.

Values of resistivity and temperature coefficient of some conductors, semi-conductors and insulators are given in table. 

Table:

Material	Resistivity at 0⁰ C (ohmmeter)	Temperature co-eff. (∝)
Conductors         Aluminium         Copper         Silver         Gold         Iron         Lead         Platinum Semi-conductor         Germanium         Silicon         Carbon Insulators         Glass         Rubber	2.7 × 10-8 1.7 × 10-8 1.6 × 10-8 2.42 × 10-8 10 × 10-8 22 × 10-8 11 × 10-8 0.46 2300 3.5 × 10-5 10¹⁰ - 10¹⁴ 10¹³ - 10¹⁶	3.9 × 10^-3 3.9 × 10^-3 3.8 × 10^-3 3.4 ×10^-3 5.0 × 10^-3 3.90 × 10^-3 3.92 × 10^-3 -48 × 10^-3 -75 × 10^-4 -0.5 × 10^-3     —     —

Important Notes:

1. Resistivity is a constant of material. Two wires having different lengths and thickness but made up of same material will have same resistivity. 
2. Resistivity of a conductor is much small while that of insulator is large. 
3. Resistivity of a conductor increases with an increase in temperature. 
4. Resistivity of a insulator increases with a decrease in temperature. 
5. Resistivity and conductivity are inverse of each other. 

Example-

A hollow copper tube of 5 m length has got external diameter equal to 0.1 m and the walls are 5 mm thick. If the resistivity of copper is 1.7 × 10-8 ohm metre, calculate the resistance of the tube. 

Solution. A hollow tube of radius r, thickness d and length l [Fig. 2(i)] can be considered to be equivalent to a metallic sheet of breadth 2πr, thickness d and length l [Fig. 2(ii)].

Fig. 2.

 Area of Cross-section of the sheet,                  

                    A = 2πr × d

∴                  R = ρ × l/A = ρ × l/2πr × d

Here            r = 5 × 10-2 m,

d = 5 mm = 5 × 10-3 m, l = 5 m, 

ρ = 1.7 × 10-8 ohm metre 

Substituting, we get

     R = 1.7 × 10-8 × 5 / 2 × π × 5 × 10-2 × 5 × 10-3 ohm

     R = 5.4 × 10-5 ohm.