AP Physics 2- 11.6 Kirchhoff's Loop Rule- Study Notes- New Syllabus
AP Physics 2- 11.6 Kirchhoff’s Loop Rule – Study Notes
AP Physics 2- 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 states that the sum of potential differences (voltage gains and drops) around any closed loop in a circuit is zero.
\(\displaystyle \sum \Delta V = 0\)
Physical Basis:
This rule is a consequence of the conservation of energy. As charge carriers move around a closed loop, the total energy gained (from sources like batteries) is equal to the total energy lost (in resistors, capacitors, or other elements).
Applying the Rule:
- Assign a direction for traversing the loop (clockwise or counterclockwise).
- Assign a current direction in each branch (even if guessed, the math will correct it).
- Voltage rise: Moving across a battery from negative to positive terminal: \(+\text{V}\).
- Voltage drop: Moving across a resistor in the direction of current: \(-IR\).
- Sum all voltage changes around the loop and set equal to zero.
Energy Changes in Circuits:
Energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements:
\(\text{U} = q \, \Delta V\)
Example :
A 12 V battery is connected in series with two resistors, \(R_1 = 4 \, \Omega\) and \(R_2 = 6 \, \Omega\). Find the current in the circuit using Kirchhoff’s loop rule.
▶️ Answer/Explanation
Step 1: Assign loop direction (clockwise). Assume current \(I\) flows through the loop.
Step 2: Write loop equation: \(+12 – IR_1 – IR_2 = 0\).
Step 3: Substitute values: \(12 – I(4) – I(6) = 0\).
Step 4: Simplify: \(12 – 10I = 0 \implies I = 1.2 \, \text{A}\).
Answer: Current = \(1.2 \, \text{A}\).
Example :
In a loop, a 9 V battery is connected with two resistors: \(R_1 = 2 \, \Omega\) and \(R_2 = 1 \, \Omega\). Find the potential drop across each resistor using Kirchhoff’s loop rule.
▶️ Answer/Explanation
Step 1: Write loop equation: \(+9 – I(2) – I(1) = 0\).
Step 2: Combine: \(9 – 3I = 0 \implies I = 3 \, \text{A}\).
Step 3: Find voltage drops: Across \(R_1\): \(V_1 = IR_1 = 3(2) = 6 \, \text{V}\). Across \(R_2\): \(V_2 = IR_2 = 3(1) = 3 \, \text{V}\).
Step 4: Check: \(V_1 + V_2 = 6 + 3 = 9 \, \text{V}\), which matches the battery.
Answer: Voltage drops: \(V_1 = 6 \, \text{V}, V_2 = 3 \, \text{V}\).