### Shortcut Methods

**Shortcut Methods and Tricks**

**1. Temperature Dependence of Resistivity**

- Remember the approximate value of the temperature coefficient of resistivity for copper: 0.0039/°C.
- For small temperature changes, the change in resistivity can be approximated as:

$$\Delta \rho = \rho_0 * \alpha * \Delta T$$ where:

- $\Delta \rho$ is the change in resistivity
- $\rho_0$ is the resistivity at the reference temperature
- $\alpha$ is the temperature coefficient of resistivity
- $\Delta T$ is the change in temperature.

**2. Mobility and Resistivity**

- The mobility of charge carriers is inversely proportional to their mass.
- The resistivity of a material is inversely related to the mobility of its charge carriers. This means that a material with high mobility will have a low resistivity, and vice versa.

**Sample Problems**

**1. Temperature Dependence of Resistivity**

A copper wire has a resistance of 1 Ω at 25°C. What will its resistance be at 50°C?

**Solution:**

Using the formula for the change in resistivity:

$$ \Delta R=R_0\alpha (T_2-T_1) $$

where $$R_0=1 \ \Omega,$$ $$\alpha = 0.0039 / °C,$$ $$T_1=25 \ °C, \text{ and }T_2=50 \ °C$$

Substituting these values into the equation, we get:

$$\Delta R=1 \Omega * 0.0039 / °C * (50 °C-25 °C) = 0.0975 \Omega$$

Therefore, the resistance of the copper wire at 50°C will be:

$$R = R_0 + \Delta R = 1 \Omega + 0.0975 \Omega = 1.0975 \Omega$$

**2. Mobility and Resistivity**

A semiconductor material has a mobility of 0.1 m^2/V-s and a resistivity of 10^-3 ohm-m. What is the concentration of charge carriers in the material?

**Solution:**

The conductivity of the material can be calculated using the formula:

$$\sigma = n e \mu$$

where:

- $\sigma$ is the conductivity in S/m
- $n$ is the concentration of charge carriers in m^-3
- $e$ is the elementary charge in coulombs
- $\mu$ is the mobility in m^2/V-s

Substituting the given values into the equation, we get:

$$10^{-3} S/m = n \times 1.602 \times 10^{-19} C \times 0.1 m^2/V-s$$ $$n = \frac{10^{-3} S/m}{1.602 \times 10^{-19} C \times 0.1 m^2/V-s} = 6.24 \times 10^{15} m^{-3}$$

Therefore, the concentration of charge carriers in the semiconductor material is 6.24 x 10^15 m^-3.