NTSE – Physics
Resistance & Resistivity
Topic for NTSE
- Basic definition of resistance
- Ohm’s law
- Resistivity of conductors & insulators
- Temperature dependence of resistance
- Practical applications
1. Simple Explanation
| Concept | What it means |
|---|---|
| Resistance (R) | The ability of a material to oppose the flow of electric current. The larger the resistance, the harder it is for electrons to move through. |
| Resistivity (ρ) | An intrinsic property of a material that tells how strongly it resists current flow per unit length and cross‑section. It is independent of the shape and size of the conductor. |
Why we care
• Determines how much voltage is needed to push a given current.
• Governs power dissipation (P = I^2R).
• Essential in designing electrical circuits, wires, heating elements, sensors, etc.
2. Key Points & Formulae
2.1 Ohm’s Law
[ V = IR \quad\Longrightarrow\quad R = \frac{V}{I} ]
- (V) – voltage (V)
- (I) – current (A)
- (R) – resistance (Ω)
2.2 Resistance of a uniform conductor
[ R = \rho,\frac{L}{A} ]
- (L) – length (m)
- (A) – cross‑sectional area (m²)
- (\rho) – resistivity (Ω·m)
2.3 Resistivity of a cylindrical wire
For a wire of diameter (d) (radius (r = d/2)): [ A = \pi r^2 = \frac{\pi d^2}{4} ] [ R = \rho,\frac{L}{\pi d^2/4} = \frac{4\rho L}{\pi d^2} ]
2.4 Temperature Dependence
For metals (approximation near room temperature):
[
\rho_T = \rho_0,[1 + \alpha (T - T_0)]
]
- (\rho_0) – resistivity at reference temperature (T_0) (usually 20 °C).
- (\alpha) – temperature coefficient (per °C).
- For semiconductors, (\rho) decreases with temperature.
2.5 Conductivity
[ \sigma = \frac{1}{\rho}\quad(\text{S/m}) ]
2.6 Resistance in Series & Parallel
- Series: (R_{\text{tot}} = R_1 + R_2 + \dots)
- Parallel: (\displaystyle\frac{1}{R_{\text{tot}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots)
3. Worked Example
Problem
A copper wire of length (L = 2.0,\text{m}) and diameter (d = 0.5,\text{mm}) is placed at (20^\circ\text{C}).
- Resistivity of copper at (20^\circ\text{C}): (\rho_0 = 1.68\times10^{-8},\Omega\text{m}).
- Temperature coefficient: (\alpha = 3.9\times10^{-3},\text{°C}^{-1}).
(a) Find the resistance at (20^\circ\text{C}).
(b) Find the resistance at (100^\circ\text{C}).
Step‑by‑Step
-
Convert diameter to meters
[ d = 0.5,\text{mm}=5.0\times10^{-4},\text{m} ] -
Area of the wire
[ A = \frac{\pi d^2}{4} = \frac{\pi (5.0\times10^{-4})^2}{4} = 1.963\times10^{-7},\text{m}^2 ] -
Resistance at 20 °C
[ R_{20} = \rho_0,\frac{L}{A} = (1.68\times10^{-8}),\frac{2.0}{1.963\times10^{-7}} \approx 0.171,\Omega ] -
Resistivity at 100 °C
[ \rho_{100} = \rho_0,[1 + \alpha (100-20)] = 1.68\times10^{-8},[1 + 3.9\times10^{-3}\times80] = 1.68\times10^{-8},[1+0.312] = 2.