AP Physics C E&M- 11.3 Resistance, Resistivity, and Ohm's Law- Study Notes- New Syllabus
AP Physics C E&M- 11.3 Resistance, Resistivity, and Ohm’s Law – Study Notes
AP Physics C E&M- 11.3 Resistance, Resistivity, and Ohm’s Law – Study Notes – per latest Syllabus.
Key Concepts:
- Resistance of an Object
- Electrical Characteristics of Circuit Elements
Resistance of an Object
Resistance is a measure of how strongly an object opposes the flow of electric charge. It depends on both the material’s properties and the object’s geometry.
Uniform Resistor:
For a resistor with uniform cross-sectional area and constant resistivity:
\( \mathrm{R = \rho \dfrac{L}{A}} \)
- \(\mathrm{R}\) = resistance (\(\Omega\))
- \(\mathrm{\rho}\) = resistivity of the material (\(\Omega \cdot m\))
- \(\mathrm{L}\) = length of the conductor (m)
- \(\mathrm{A}\) = cross-sectional area (\(m^2\))
Resistivity:
A fundamental property of a material determined by its atomic and molecular structure.
- Quantifies how strongly the material opposes charge motion.
- For conductors, resistivity typically increases with temperature:
\( \mathrm{\rho(T) = \rho_0 [1 + \alpha (T – T_0)]} \)
where \(\mathrm{\alpha}\) is the temperature coefficient of resistivity.
Non-uniform Resistor:
If resistivity or cross-sectional area varies along the length, resistance must be calculated using an integral:
\( \mathrm{R = \displaystyle \int_0^L \dfrac{\rho(x)}{A(x)} \, dx} \)
- \(\mathrm{\rho(x)}\) = resistivity as a function of position
- \(\mathrm{A(x)}\) = cross-sectional area as a function of position
- This accounts for non-uniform materials or tapered geometries.
Key Idea: Resistance depends on material (resistivity), length, and cross-sectional area. Uniform objects follow the simple ratio, but non-uniform ones require integration of \(\rho(x)/A(x)\).
Example
A copper wire of length \(2.0 \, m\) and cross-sectional area \(1.0 \times 10^{-6} \, m^2\) has resistivity \( \rho = 1.7 \times 10^{-8} \, \Omega \cdot m\). Find its resistance.
▶️ Answer/Explanation
Step 1: Use formula: \( \mathrm{R = \rho \dfrac{L}{A}} \).
Step 2: Substitute: \( \mathrm{R = \dfrac{(1.7 \times 10^{-8})(2.0)}{1.0 \times 10^{-6}}} \).
Step 3: \( \mathrm{R = 3.4 \times 10^{-2} \, \Omega} \).
Final Answer: The wire’s resistance is \( \mathrm{0.034 \, \Omega} \).
Example
A resistor has resistivity that varies with position as \( \rho(x) = \rho_0 (1 + \dfrac{x}{L}) \), where \(L\) is the total length and cross-sectional area is constant. Find the total resistance of the resistor.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{R = \displaystyle \int_0^L \dfrac{\rho(x)}{A} dx} \).
Step 2: Substitute: \( \mathrm{R = \dfrac{1}{A} \int_0^L \rho_0 \left(1 + \dfrac{x}{L}\right) dx} \).
Step 3: Integrate: \( \mathrm{R = \dfrac{\rho_0}{A} \left[ x + \dfrac{x^2}{2L} \right]_0^L} \).
Step 4: \( \mathrm{R = \dfrac{\rho_0}{A} \left( L + \dfrac{L}{2} \right)} \).
Step 5: \( \mathrm{R = \dfrac{3}{2} \dfrac{\rho_0 L}{A}} \).
Final Answer: The total resistance is \( \mathrm{\dfrac{3}{2} \dfrac{\rho_0 L}{A}} \).
Electrical Characteristics of Circuit Elements
Ohm’s Law:
Ohm’s law relates the current through a circuit element to the potential difference across it and its resistance:
\( \mathrm{V = I R} \)
- \(\mathrm{V}\) = potential difference (volts)
- \(\mathrm{I}\) = current (amperes)
- \(\mathrm{R}\) = resistance (ohms)
Ohmic Materials:
A material that obeys Ohm’s law has a constant resistance for all values of current and voltage.
Graph of \( \mathrm{I} \) vs. \( \mathrm{V} \) is a straight line through the origin, with slope:
\( \mathrm{\dfrac{I}{V} = \dfrac{1}{R}} \)
Resistivity of an ideal ohmic material is constant (not dependent on temperature or current).
Non-Ohmic Materials:
- Some elements (e.g., diodes, filament lamps) do not follow Ohm’s law.
- Resistance may change with voltage, current, or temperature.
- Graph of \( \mathrm{I} \) vs. \( \mathrm{V} \) is nonlinear.
Energy Conversion in Resistors:
Resistors convert electrical energy into thermal energy due to collisions of moving charges with the lattice of the conductor.
Power dissipated in a resistor:
\( \mathrm{P = I V = I^2 R = \dfrac{V^2}{R}} \)
- This thermal energy may increase the resistor’s temperature, affecting both its resistance and the surrounding environment.
Experimental Determination of Resistance:
- Resistance can be determined from the slope of an \(\mathrm{I\text{–}V}\) graph.
- For ohmic elements: straight line, slope = \( 1/R \).
- For non-ohmic elements: curve, resistance varies with operating point.
Key Idea: Ohm’s law provides a simple relationship between current, potential difference, and resistance for ohmic materials, while deviations reveal non-ohmic behavior.
Example
A resistor has a constant resistance of \(10 \, \Omega\). If a potential difference of \(5.0 \, V\) is applied across it, find the current and the power dissipated.
▶️ Answer/Explanation
Step 1: Ohm’s law: \( \mathrm{I = \dfrac{V}{R} = \dfrac{5.0}{10}} = 0.50 \, A \).
Step 2: Power: \( \mathrm{P = I V = (0.50)(5.0) = 2.5 \, W} \).
Final Answer: The current is \(0.50 \, A\) and the resistor dissipates \(2.5 \, W\) of power as heat.
Example
A filament lamp draws \(0.2 \, A\) when \(2.0 \, V\) is applied, but \(0.5 \, A\) when \(6.0 \, V\) is applied. Show that the lamp is non-ohmic and calculate its resistance at each operating point.
▶️ Answer/Explanation
Step 1: Resistance at \(V = 2.0 \, V\): \( \mathrm{R = \dfrac{V}{I} = \dfrac{2.0}{0.2} = 10 \, \Omega} \).
Step 2: Resistance at \(V = 6.0 \, V\): \( \mathrm{R = \dfrac{6.0}{0.5} = 12 \, \Omega} \).
Step 3: Resistance increases with voltage → not constant.
Final Answer: The lamp is non-ohmic; resistance changes from \(10 \, \Omega\) to \(12 \, \Omega\) as current increases.