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Current Electricity - Temperature Dependence of Resistivity

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The resistivity of a material changes with temperature. For a limited range of temperatures, the relation is approximately linear: ρT=ρ0[1+α(T−T0)]\rho_T = \rho_0 [1 + \alpha(T - T_0)].

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Temperature Coefficient of Resistivity (α\alpha): It is defined as the fractional change in resistivity per unit rise in temperature. Its SI unit is K−1K^{-1} or ∘C−1^{\circ}C^{-1}.

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For Conductors (Metals): α\alpha is positive. As temperature increases, the amplitude of vibrations of lattice ions increases, leading to more frequent collisions. This reduces the relaxation time τ\tau. Since ρ=mne2τ\rho = \frac{m}{ne^2\tau}, resistivity increases.

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For Semiconductors and Insulators: α\alpha is negative. As temperature increases, more electrons jump to the conduction band, significantly increasing the charge carrier density nn. This increase in nn dominates over the decrease in τ\tau, so resistivity decreases exponentially.

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For Alloys (e.g., Manganin, Constantan, Nichrome): These materials exhibit a very weak dependence of resistivity on temperature (very small α\alpha). This property makes them ideal for constructing standard resistors and resistance boxes.

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The resistance of a conductor also follows a similar relation: RT=R0[1+α(T−T0)]R_T = R_0 [1 + \alpha(T - T_0)], assuming the dimensions of the conductor do not change significantly with temperature.

📐Formulae

ρT=ρ0[1+α(T−T0)]\rho_T = \rho_0 [1 + \alpha(T - T_0)]

RT=R0[1+α(T−T0)]R_T = R_0 [1 + \alpha(T - T_0)]

α=RT−R0R0(T−T0)\alpha = \frac{R_T - R_0}{R_0(T - T_0)}

ρ=mne2τ\rho = \frac{m}{ne^2\tau}

ρ(T)=ρ0e−Eg/kBT (For semiconductors)\rho(T) = \rho_0 e^{-E_g / k_B T} \text{ (For semiconductors)}

💡Examples

Problem 1:

The resistance of a platinum wire of a platinum resistance thermometer at the ice point is 5Ω5 \Omega and at steam point is 5.39Ω5.39 \Omega. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795Ω5.795 \Omega. Calculate the temperature of the bath.

Solution:

Given: R0=5ΩR_0 = 5 \Omega (at 0∘C0^{\circ}C), R100=5.39ΩR_{100} = 5.39 \Omega (at 100∘C100^{\circ}C), and Rt=5.795ΩR_t = 5.795 \Omega. Using the formula: t=Rt−R0R100−R0×100t = \frac{R_t - R_0}{R_{100} - R_0} \times 100 Substituting the values: t=5.795−55.39−5×100t = \frac{5.795 - 5}{5.39 - 5} \times 100 t=0.7950.39×100t = \frac{0.795}{0.39} \times 100 t=2.03846×100≈203.85∘Ct = 2.03846 \times 100 \approx 203.85^{\circ}C

Explanation:

The temperature of a resistance thermometer is calculated using the linear variation of resistance with temperature. The ratio of the change in resistance at the unknown temperature to the change in resistance over a known interval (100∘C100^{\circ}C) gives the temperature value.

Problem 2:

A heating element using nichrome connected to a 230V230 V supply draws an initial current of 3.2A3.2 A which settles after a few seconds to a steady value of 2.8A2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0∘C27.0^{\circ}C? Temperature coefficient of resistance of nichrome is 1.70×10−4 ∘C−11.70 \times 10^{-4} \, ^{\circ}C^{-1}.

Solution:

Initial resistance R1R_1 at T1=27∘CT_1 = 27^{\circ}C: R1=VI1=2303.2=71.875ΩR_1 = \frac{V}{I_1} = \frac{230}{3.2} = 71.875 \Omega Final resistance R2R_2 at steady temperature T2T_2: R2=VI2=2302.8=82.143ΩR_2 = \frac{V}{I_2} = \frac{230}{2.8} = 82.143 \Omega Using R2=R1[1+α(T2−T1)]R_2 = R_1 [1 + \alpha(T_2 - T_1)], we get: T2−T1=R2−R1R1αT_2 - T_1 = \frac{R_2 - R_1}{R_1 \alpha} 82.143−71.87510.268\begin{array}{r} 82.143 \\ - 71.875 \\ \hline 10.268 \end{array} T2−27=10.26871.875×1.70×10−4T_2 - 27 = \frac{10.268}{71.875 \times 1.70 \times 10^{-4}} T2−27≈840.35T_2 - 27 \approx 840.35 T2=840.35+27=867.35∘CT_2 = 840.35 + 27 = 867.35^{\circ}C

Explanation:

The current decreases as the temperature of the heating element increases because the resistance of the nichrome wire increases with temperature. We calculate the resistance at both states and use the temperature coefficient formula to find the final temperature.

Temperature Dependence of Resistivity Class 12 Notes & Examples