AP Physics 2- 11.3 Resistance, Resistivity, and Ohm's Law- Study Notes- New Syllabus
AP Physics 2- 11.3 Resistance, Resistivity, and Ohm’s Law – Study Notes
AP Physics 2- 11.3 Resistance, Resistivity, and Ohm’s Law – Study Notes – per latest Syllabus.
Key Concepts:
- Resistance and Resistivity
- Ohm’s Law and Electrical Characteristics of Circuit Elements
Resistance and Resistivity
Resistance (\(\text{R}\)):
Resistance is the property of a conductor that opposes the flow of electric current. It depends on the material, length, and cross-sectional area of the conductor.
Relevant equation:
\(\text{R} = \rho \dfrac{\text{L}}{\text{A}}\)
- \(\text{R}\) = resistance (ohms, \(\Omega\))
- \(\rho\) = resistivity of the material (\(\Omega \, \text{m}\))
- \(\text{L}\) = length of conductor (m)
- \(\text{A}\) = cross-sectional area (\(\text{m}^2\))
Resistivity (\(\rho\)):
Resistivity is a fundamental property of materials that quantifies how strongly a material opposes current flow.
- Low resistivity → good conductor (e.g., copper, silver).
- High resistivity → poor conductor or insulator (e.g., rubber, glass).
- Resistivity generally increases with temperature for metals.
Example :
A copper wire has length \(\text{L} = 2.0 \, \text{m}\), cross-sectional area \(\text{A} = 1.0 \times 10^{-6} \, \text{m}^2\), and resistivity \(\rho = 1.7 \times 10^{-8} \, \Omega \, \text{m}\). Find its resistance.
▶️ Answer/Explanation
Step 1: Use the resistance formula: \(\text{R} = \rho \dfrac{\text{L}}{\text{A}}\)
Step 2: Substitute values: \(\text{R} = \dfrac{(1.7 \times 10^{-8})(2.0)}{1.0 \times 10^{-6}}\)
Step 3: Calculate: \(\text{R} = 3.4 \times 10^{-2} \, \Omega\)
Answer: Resistance ≈ \(0.034 \, \Omega\).
Example :
Two wires of the same material and length have different cross-sectional areas: \(\text{A}_1 = 2 \times 10^{-6} \, \text{m}^2\) and \(\text{A}_2 = 4 \times 10^{-6} \, \text{m}^2\). Compare their resistances.
▶️ Answer/Explanation
Step 1: Resistance formula: \(\text{R} = \rho \dfrac{\text{L}}{\text{A}}\).
Step 2: Since both wires have the same \(\rho\) and \(\text{L}\), resistance is inversely proportional to area: \(\text{R} \propto \dfrac{1}{\text{A}}\).
Step 3: Ratio of resistances: \(\dfrac{\text{R}_1}{\text{R}_2} = \dfrac{\text{A}_2}{\text{A}_1} = \dfrac{4}{2} = 2\).
Answer: The thinner wire (\(\text{A}_1\)) has twice the resistance of the thicker wire (\(\text{A}_2\)).
Ohm’s Law and Electrical Characteristics of Circuit Elements
Ohm’s Law:
Ohm’s law describes the relationship between the potential difference (\(\text{V}\)), current (\(\text{I}\)), and resistance (\(\text{R}\)) of an electrical element.
\(\text{V} = \text{IR}\)
- \(\text{V}\) = potential difference across the element (volts, V)
- \(\text{I}\) = current through the element (amperes, A)
- \(\text{R}\) = resistance of the element (ohms, \(\Omega\))
Ohmic vs. Non-Ohmic Elements:
- Ohmic elements: Obey Ohm’s law; their \(\text{V-I}\) graph is a straight line (e.g., resistors).
- Non-ohmic elements: Do not obey Ohm’s law; their \(\text{V-I}\) graph is nonlinear (e.g., diodes, filament bulbs).
Electrical Characteristics of Circuit Elements:
- Resistors: Convert electrical energy into heat; obey Ohm’s law.
- Batteries/EMF sources: Provide potential difference; chemical energy is converted to electrical energy.
- Capacitors: Store energy in the electric field between plates; current depends on charging/discharging process.
- Lightbulbs: Convert electrical energy into light and heat; resistance increases with temperature (non-ohmic).
- Diodes: Allow current flow primarily in one direction; nonlinear \(\text{V-I}\) relationship.
- Switches: Control the open or closed state of a circuit.
- Ammeters: Measure current (placed in series, ideally with negligible resistance).
- Voltmeters: Measure potential difference (placed in parallel, ideally with very high resistance).
Example :
A resistor of \(\text{R} = 10 \, \Omega\) is connected to a battery of \(\text{V} = 12 \, \text{V}\). Find the current through the resistor.
▶️ Answer/Explanation
Step 1: Apply Ohm’s law: \(\text{V} = \text{IR}\).
Step 2: Rearrange: \(\text{I} = \dfrac{\text{V}}{\text{R}} = \dfrac{12}{10}\).
Step 3: Calculate: \(\text{I} = 1.2 \, \text{A}\).
Answer: Current = \(1.2 \, \text{A}\).
Example :
A filament bulb has a resistance of \(20 \, \Omega\) when cold, but its resistance rises to \(80 \, \Omega\) at operating temperature. If a voltage of \(120 \, \text{V}\) is applied across the bulb at operating temperature, find the current.
▶️ Answer/Explanation
Step 1: Use Ohm’s law: \(\text{I} = \dfrac{\text{V}}{\text{R}}\).
Step 2: Substitute values: \(\text{I} = \dfrac{120}{80} = 1.5 \, \text{A}\).
Step 3: Note: Since the bulb is non-ohmic, the resistance changes with temperature, but once stabilized, Ohm’s law still applies for the operating resistance.
Answer: Current = \(1.5 \, \text{A}\).