A 16-Ω platinum wire at 20°C with a temperature coefficient of resistivity of 3.9 × 10−³ (°C)−¹ will have what resistance at 80°C? A)12 Ω B)13 Ω C)20 Ω D)16 Ω E)3.7 Ω

Question

Quizplus · Accepted Answer

The resistance of a material at a specific temperature can be calculated using the formula $R = R_0[1 + \alpha(T - T_0)]$, where $R_0$ is the initial resistance, $\alpha$ is the temperature coefficient of resistivity, $T$ is the final temperature, and $T_0$ is the initial temperature. Given $R_0 = 16 \Omega$, $\alpha = 3.9 \times 10^{-3} \, \text{°C}^{-1}$, $T = 80°C$, and $T_0 = 20°C$, substituting these values gives $R = 16[1 + 3.9 \times 10^{-3}(80 - 20)] = 16[1 + 3.9 \times 10^{-3} \times 60] = 16[1 + 0.234] = 16 \times 1.234 = 19.744 \Omega$, which rounds to approximately 20 Ω.

A 16-Ω Platinum Wire at 20°C with a Temperature Coefficient