11.6 Kirchhoff's Loop Rule

Kirchhoff's loop rule says the sum of potential differences around a closed circuit loop is zero. It comes from conservation of energy: as charge moves around a complete loop, the energy gained from sources equals the energy lost across circuit elements.

$\sum \Delta V = 0$

Use it by choosing a loop direction, assigning signs to voltage rises and drops, and writing one equation that accounts for every element in the loop.

Why This Matters for the AP Physics C: E&M Exam

Loop-rule questions test conservation of energy in circuits. You may need to set up a symbolic loop equation, solve for current, compare potential differences, or interpret a graph of electric potential around a circuit loop.

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Energy Changes in Circuits

When charges move through a circuit, they experience changes in electric potential energy as they pass through different circuit elements. These energy changes are fundamental to understanding how circuits work.

The change in electric potential energy when a charge moves through a potential difference is given by:

$\Delta U_{\mathrm{E}} = q \Delta V$

Where:

• $q$ is the charge (in coulombs)
• $\Delta V$ is the electric potential difference (in volts)

This equation quantifies how much energy is transferred as charges move through circuit components. For example, when charges move through a resistor, electrical potential energy is converted to thermal energy, causing the resistor to heat up. Conversely, when charges move through a battery in the direction of increasing potential, the battery does work on the charges, increasing their potential energy.

Conservation of Energy

Kirchhoff's loop rule is a direct application of the principle of conservation of energy in electrical circuits.

The rule states that the sum of all potential differences around any closed loop in a circuit must equal zero:

$\sum \Delta V = 0$

This makes intuitive sense when we consider energy conservation. If a charge moves around a complete loop in a circuit and returns to its starting point, its potential energy must be the same as when it started. Otherwise, the loop equation would imply a net energy gain or loss without an external source.

In practical terms, this means:

• Voltage rises (from batteries or power sources) must be balanced by voltage drops (across resistors or other components)
• If you add up all voltage changes as you move around any closed loop, you'll always get zero

When applying Kirchhoff's loop rule, it's important to establish a consistent sign convention:

• Voltage rises (such as moving from negative to positive terminal of a battery) are typically counted as positive
• Voltage drops (such as across resistors) are typically counted as negative

Electric Potential Graphs

Electric potential graphs provide a visual representation of how potential changes as we move around a circuit loop.

These graphs plot electric potential (V) on the y-axis against position in the circuit on the x-axis. As you trace through the circuit:

• Upward slopes represent increasing potential (such as moving through a battery from negative to positive terminal)
• Downward slopes represent decreasing potential (such as moving through a resistor)
• The steepness of the slope indicates the rate of potential change

A key insight from these graphs is that when you complete a full loop and return to your starting point, the potential must return to its initial value. This graphical representation reinforces Kirchhoff's loop rule: the net change in potential around a closed loop is always zero.

These graphs are particularly useful for:

• Identifying the locations of voltage sources and loads in a circuit
• Visualizing the relative magnitudes of voltage changes across different components
• Confirming that energy is conserved throughout the circuit

Practice Problem 1: Applying Kirchhoff's Loop Rule

Solution

To solve this problem, we'll apply Kirchhoff's loop rule to find the current.

Step 1: Identify the voltage sources and resistors in the loop.

• Voltage source: 12V battery
• Resistors: 2Ω, 4Ω, and 6Ω in series

Step 2: Apply Kirchhoff's loop rule, where the sum of all potential differences equals zero. $\sum \Delta V = 0$

Step 3: Express the potential differences in terms of the current.

• Voltage rise across battery: +12V
• Voltage drop across 2Ω resistor: $-I \times 2\Omega$
• Voltage drop across 4Ω resistor: $-I \times 4\Omega$
• Voltage drop across 6Ω resistor: $-I \times 6\Omega$

Step 4: Set up the equation. $12V - I(2\Omega) - I(4\Omega) - I(6\Omega) = 0$

Step 5: Solve for the current. $12V - I(12\Omega) = 0$

$I = \frac{12V}{12\Omega} = 1A$

Therefore, the current in the circuit is 1 ampere.

Practice Problem 2: Electric Potential Graph Analysis

Solution

To solve this problem, we need to analyze how the electric potential changes as we move around the circuit.

Step 1: Calculate the current in the circuit using Ohm's law. Total resistance: $R_{total} = 3\Omega + 6\Omega = 9\Omega$ Current: $I = \frac{V}{R_{total}} = \frac{9V}{9\Omega} = 1A$

Step 2: Determine the potential differences across each component.

• Voltage rise across battery: +9V
• Voltage drop across 3Ω resistor: $V_{3\Omega} = I \times 3\Omega = 1A \times 3\Omega = 3V$
• Voltage drop across 6Ω resistor: $V_{6\Omega} = I \times 6\Omega = 1A \times 6\Omega = 6V$

Step 3: Sketch the electric potential graph.

• Starting at 0V (negative terminal of battery)
• Rise to +9V (positive terminal of battery)
• Drop by 3V across the 3Ω resistor (to +6V)
• Drop by 6V across the 6Ω resistor (back to 0V)

The potential difference across the 3Ω resistor is 3V. This can be verified using Kirchhoff's loop rule, as the sum of all potential differences around the loop equals zero: +9V - 3V - 6V = 0.