Question

a resistor has a resistance of 100 ω at 20 °c. when the temperature is increased to 25 °c, the resistance increases to 105 ω. if the initial temperature is doubled, from 20 °c to 40 °c, how much resistance will this object have at the final 40 °c temperature? assume a linear relationship between temperature and resistance. o 220 ω o 120 ω o 200 ω o 240 ω

Explanation:

Step1: Find the temperature - coefficient of resistance

The formula for resistance as a function of temperature is $R = R_0(1+\alpha\Delta T)$. We know that at $T_0 = 20^{\circ}C$, $R_0 = 100\Omega$ and at $T_1=25^{\circ}C$, $R_1 = 105\Omega$.
First, find $\Delta T_1=T_1 - T_0=25 - 20 = 5^{\circ}C$.
Substitute into the formula $R_1 = R_0(1+\alpha\Delta T_1)$:
$105=100(1 + 5\alpha)$
$1+\ 5\alpha=\frac{105}{100}=1.05$
$5\alpha=1.05 - 1=0.05$
$\alpha=\frac{0.05}{5}=0.01/^{\circ}C$

Step2: Calculate the resistance at $T_2 = 40^{\circ}C$

Now, find $\Delta T_2=T_2 - T_0=40 - 20 = 20^{\circ}C$.
Use the formula $R = R_0(1+\alpha\Delta T_2)$
Substitute $R_0 = 100\Omega$, $\alpha = 0.01/^{\circ}C$ and $\Delta T_2 = 20^{\circ}C$
$R=100(1+0.01\times20)$
$R = 100(1 + 0.2)$
$R=120\Omega$

Answer:

$120\Omega$