To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at 25 and resistance of the semiconductor arm is 3 mΩ. Arm BC is cooled at a constant rate of 2/s. If the galvanometer G shows no deflection after 10 s, then α is: (a) -1 × 10⁻² ⁻¹ (b) -2 × 10⁻² ⁻¹ (c) -2.5 × 10⁻² ⁻¹ (d) -1.5 × 10⁻² ⁻¹

Answer: (a). At balance (no deflection): RABRBC = RADRDC ⇒ 0.8RBC = 1/3, so RBC = 2.4 mΩ. Temperature change in 10 s: Δ T = -2 × 10 = -20. R = R₀(1 + α Δ T) ⇒ 2.4 = 3(1 - 20α) ⇒ |α| = (0.6)/(3 × 20) = 1 × 10⁻² ⁻¹ A semiconductor has a negative temperature coefficient. So the answer is (a) -1 × 10⁻² ⁻¹.

To measure the temperature coefficient of resistivity α of a… | JEE Main 2024 (1 Feb, Shift 2)

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Intrinsic & Extrinsic Semiconductors

To measure the temperature coefficient of resistivity $\alpha$ of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at $25^\circ\text{C}$ and resistance of the semiconductor arm is $3\text{ m}\Omega$ . Arm $\text{BC}$ is cooled at a constant rate of $2^\circ\text{C/s}$ . If the galvanometer $\text{G}$ shows no deflection after $10\text{ s}$ , then $\alpha$ is:

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Answer: (a)

At balance (no deflection): $\dfrac{R_{AB}}{R_{BC}} = \dfrac{R_{AD}}{R_{DC}} \Rightarrow \dfrac{0.8}{R_{BC}} = \dfrac{1}{3}$ , so $R_{BC} = 2.4\ \text{m}\Omega$ .
Temperature change in 10 s: $\Delta T = -2 \times 10 = -20\ ^\circ\text{C}$ .
$R = R_0(1 + \alpha\,\Delta T) \Rightarrow 2.4 = 3(1 - 20\alpha) \Rightarrow |\alpha| = \dfrac{0.6}{3 \times 20} = 1 \times 10^{-2}\ ^\circ\text{C}^{-1}$
A semiconductor has a negative temperature coefficient.
So the answer is (a) $-1 \times 10^{-2}\ ^\circ\text{C}^{-1}$ .

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