Variations of Electrical Resistivity with Temperature

Not only resistance but specific resistance or resistivity of metallic conductors also increases with rise in temperature and vice versa. As seen from Fig. the resistivities of metals vary linearly with temperature over a significant range of temperature-the variation becoming non-linear both at very high and at very low temperatures. Let, for any metallic conductor.

ρ1 = resistivity at t1 °C
ρ2 = resistivity at t2 °C
m = Slope of the linear part of the curve

then ,$m=\frac{\rho 2-\rho 1}{t2-t1}$

The ratio of m/ρ1 is called the temperature coefficient of resistivity at temperature t1 °C. It may be defined as numerically equal to the fractional change in ρ1 per °C change in the temperature from t1 °C. It is almost equal to the temperature-coefficient of resistance ${\alpha }_1$ . Hence, putting ${\alpha }_1$= m/ρ1 we get

$\rho 2=\rho 1\left[1+\alpha 1(t2-t1)\right] or \rho t=\rho 0(1+\alpha 0t)$

Note: It has been found that although temperature is the most significant factor influencing the resistivity of metals. other factors like pressure and tension also affect resistivity to some extent For most metals except lithium and calcium. increase in pressure leads to decrease in resistivity. However, resistivity increases with increase in tension.

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Examples based on theory 1) Electrical Conductance and Conductivity 2) Effect of Temperature on Electrical Resistance  3) Temperature Coefficient of Electrical Resistance  4) Value of α at Different Temperatures ( In terms of...
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Example A copper conductor has its specific resistance of 1.6×10-6 ohm-cm at O°C and a resistance temperature coefficient of 1/254.5 per °C at 20°C. Find (1) the specific resistance and (2) the resistance - temperature coefficient at 60°C. 

Solution :

$\alpha 20=\frac{\alpha 0}{1+\alpha 0\times 20} or \frac{1}{254.5}=\frac{\alpha 0}{1+\alpha 0\times 20}$$$     so ,$\alpha 0=\frac{1}{234.5}per °C$
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(1)       ρ60 = ρ0 (1 +$\alpha$0 x 60) =  2×10-6 Ω-cm

(2)      $\alpha$60= $\alpha$0 / (1+$\alpha$0x60) = 1/294.5 per °C.

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Example: A platinum coil has a resistance of 3.146 ohm at 40°C and 3.767 ohm at 100°C. Find the resistance at O°C and the temperature-coefficient of resistance at 40°C. 

Solution:

R100 = R0 (1+100 $\mathrm{\alpha <span\; style="font-family:\; Arial,Helvetica,sans-serif;">0\; )</span>}$  _ _ _ _ _ _ _ (1)

R40 = R0 (1+ 40$\mathrm{<span\; style="font-family:\; Arial,Helvetica,sans-serif;"> \; </span>\alpha 0) \; <span\; style="font-family:\; Arial,Helvetica,sans-serif;">\_\; \_\; \_\; \_\; \_\; \_ \; \_\; (2)</span>}$

$\frac{3.767}{3.146}=\frac{1+100 \alpha 0}{1+40 \alpha 0} or \alpha 0=\frac{1}{264} per °C$

From eq (1)       3.767 = R0 (1+100x0.00379 )  so R0=2.732 ohm

and  α40 = α0 / (1+40 α0 ) = 1/304 per °C

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Example: A potential difference of 250 V is applied to afield winding at 15°C and the current is 5 A. What will be the mean temperature of the winding when current has fallen to 3.91 A. applied voltage being constant. Assume $\mathrm{\alpha 15}$ =1/254.5.

Solution:

Let R1 =winding resistance at 15°C; R2 = winding resistance at unknown mean temperature t2°c.

R1 = 250/5 = 50 ohm; R2= 250/3.91 = 63.94 ohm.

R2 = R1 [1 + α15 (t2 - t1)]   so t2=86 °C

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Example: Two coils connected in series have resistances of 600 ohm and 300 ohm with temp coeff. of 0.1% and 0.4% respectively at 20°C. Find the resistance of the combination at a tempt. of 50°C. What is the effective tempt. coeff. of combination?

Solution:

Resistance of 600 ohm resistor at 50°C is = 600 [1 + 0.001 (50 - 20)] = 618 Ohm
Similarly, resistance of 300 ohm resistor at 50°C is = 300 [1 + 0.004 (50 - 20)] = 336 Ohm

So, total resistance of combination at 50°C is = 618 + 336 =954 ohm

Let β = resistance-temperature coefficient at 20°C

Now, combination resistance at 20°C = 900 ohm
Combination resistance at 50°C = 954 ohm

954 = 900[ ] + β (50 - 20)]   so, β = 0.002

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Example: Two wires A and B are connected in series at O°C and resistance of B is 3.5 times that of A. The resistance temperature coefficient of A is 0.4% and that of the combination is 0.1%. Find the resistance temperature coefficient of B.

Solution:

A simple technique which gives quick results in such questions is illustrated by the diagram of Fig. It is seen that. RB/RA =0.003 / ( 0.001- $\alpha$)

$$or 3.5 = 0.003/(0.001- $\alpha$)$$

or a = 0.000

143 °C^-1   or 0.0143%
 

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