AP Physics C E&M- 11.6 Kirchhoff's Loop Rule- Study Notes- New Syllabus
AP Physics C E&M- 11.6 Kirchhoff’s Loop Rule – Study Notes
AP Physics C E&M- 11.6 Kirchhoff’s Loop Rule – Study Notes – per latest Syllabus.
Key Concepts:
- Kirchhoff’s Loop Rule
Kirchhoff’s Loop Rule
Kirchhoff’s loop rule is based on the conservation of energy. It states that in any closed circuit loop, the total change in electric potential is zero, since a charge that completes a loop returns to the same potential from which it started.
Relevant Equation: 
\( \mathrm{\sum \Delta V = 0} \)
- \(\mathrm{\Delta V}\) = potential difference across each circuit element
- Sum taken over one complete closed loop
Interpretation:
- As charges move through elements like batteries, resistors, or capacitors, they gain or lose electric potential energy.
- A battery provides an increase in potential (\(+\mathcal{E}\)).
- Resistors represent a decrease in potential (\(-IR\)).
- The total around the loop must sum to zero because no net energy is gained or lost after a full loop traversal.
Graphical Representation:
- Potential at points in a circuit loop can be plotted as a function of position.
- The graph shows increases at sources (like batteries) and decreases across resistors or loads.
Key Idea: Kirchhoff’s loop rule ensures energy conservation in electrical circuits: energy supplied by sources is equal to the energy dissipated or stored in circuit elements.
Example
A \(12 \, V\) battery is connected in series with two resistors of \(4.0 \, \Omega\) and \(8.0 \, \Omega\). Use Kirchhoff’s loop rule to find the current in the circuit.
▶️ Answer/Explanation
Step 1: Loop equation: \( \mathrm{+\mathcal{E} – I R_1 – I R_2 = 0} \).
Step 2: Substitute: \( \mathrm{12 – I(4.0) – I(8.0) = 0} \).
Step 3: Simplify: \( \mathrm{12 = 12I} \quad \Rightarrow \quad I = 1.0 \, A} \).
Final Answer: The current in the loop is \( \mathrm{1.0 \, A} \).
Example
A \(9.0 \, V\) battery is connected across three series resistors of \(1.0 \, \Omega\), \(2.0 \, \Omega\), and \(3.0 \, \Omega\). Use Kirchhoff’s loop rule to find the voltage drop across each resistor.
▶️ Answer/Explanation
Step 1: Loop equation: \( \mathrm{+\mathcal{E} – IR_1 – IR_2 – IR_3 = 0} \).
Step 2: Total resistance: \( \mathrm{R_{total} = 1 + 2 + 3 = 6 \, \Omega} \).
Step 3: Current: \( \mathrm{I = \dfrac{V}{R_{total}} = \dfrac{9}{6} = 1.5 \, A} \).
Step 4: Voltage drops: \( \mathrm{V_1 = I R_1 = (1.5)(1.0) = 1.5 \, V} \). \( \mathrm{V_2 = (1.5)(2.0) = 3.0 \, V} \). \( \mathrm{V_3 = (1.5)(3.0) = 4.5 \, V} \).
Step 5: Check: \( 1.5 + 3.0 + 4.5 = 9.0 \, V \) = battery emf ✅
Final Answer: Voltage drops: \( \mathrm{1.5 \, V, \; 3.0 \, V, \; 4.5 \, V} \).