Topic 11.1011 min read
Kirchhoff's Loop Rule
The algebraic sum of potential changes around a closed loop is zero.
Choose Your Level
Imagine carrying one coulomb of charge around a complete circuit and returning to the starting point. Its final electric potential must equal its starting potential, so all voltage rises and drops must balance.
A cell normally provides a voltage rise, while a resistor produces a voltage drop in the direction of current.
Recommended for Class 11
Kirchhoff's loop rule states that the algebraic sum of potential changes around every closed loop is zero. It extends energy conservation to circuits containing several sources and resistors.
Why a closed-loop sum is zero
Potential difference is energy transferred per unit charge. After a charge travels around a closed path and returns to its starting point, its net change in electric potential energy must be zero.
Equivalently, the energy per coulomb supplied by sources equals the energy per coulomb transferred in loads and internal resistances.
Choosing the loop direction
You may travel clockwise or anticlockwise. The physical answer does not depend on this choice because reversing the traversal reverses every term in the equation and leaves the same solution.
Potential change across a resistor
Conventional current passes through a resistor from higher to lower potential. Moving through the resistor with the current therefore gives a drop; moving against current gives a rise.
Potential change across a source
Moving through an ideal source from its negative terminal to its positive terminal raises potential by ε. Moving from positive to negative lowers potential by ε.
| Element crossing | Term in loop equation |
|---|---|
| Resistor, with current | −IR |
| Resistor, against current | +IR |
| Source, negative to positive | +ε |
| Source, positive to negative | −ε |
Single-source loop
A 12 V ideal source drives current through 2 Ω and 4 Ω in series. Traverse from the negative to positive source terminal and then through both resistors with the current.
The resistor drops are 4 V and 8 V. Together they equal the 12 V rise supplied by the source.
Opposing sources
Suppose 12 V and 6 V sources oppose each other in one loop with total resistance 3 Ω. Choose the direction favored by the larger source.
The smaller source is crossed from positive to negative in this traversal, so its emf appears with a minus sign.
Including internal resistance
Internal resistance behaves like an ordinary series resistor for sign purposes. A cell of emf ε and internal resistance r connected to external resistance R obeys:
Several loops and unknown currents
In a network, first assign a current to each branch. Use junction equations to relate branch currents, then write loop equations for enough independent closed paths to solve the remaining unknowns.
A component shared by two loops carries the algebraic combination of the chosen loop currents. Its voltage term must use the actual branch-current expression.
Equation-solving checklist
- Draw and label every emf, resistance and current arrow.
- Choose a loop traversal direction.
- Walk around the loop once and write each voltage change in order.
- Use −IR with current and +IR against current.
- Use +ε from negative to positive and −ε from positive to negative.
- Set the algebraic sum equal to zero.
- Combine with junction equations when the circuit has branches.
- Check that the final voltage rises equal the final voltage drops.
Common conceptual checks
A closed-loop equation must have voltage units in every term. If all resistor terms have the same sign in a one-current loop, the emf terms must supply the balancing rise. Reversing the complete loop direction should produce the same equation multiplied by −1.
Exam-ready response
- State the rule and its energy-conservation basis.
- Show the four traversal sign cases.
- Write the loop equation before substituting or simplifying.
- Preserve units in the final answer.
- Interpret negative current direction rather than discarding the result.
Exam Focus
Exam Focus
Important points
- Statement
- Energy-conservation basis
- Loop direction
- Resistor signs
- Source signs
- Internal resistance
- Negative current
- Multi-loop method
Important formulas
- ∑Δ V=0
- Δ VR=± IR
- \Delta V_\varepsilon=\pm\varepsilon
- I=εR+r
Common exam traps
- Separate loop direction from assumed current direction.
- Reverse all signs consistently if reversing traversal.
Long question structure
Energy basis → rule → sign convention → labelled loop → equation → solution → direction and voltage check
Common Mistakes
Common Mistakes
- Mistake
Mixing signs after choosing a traversal direction.
- Mistake
Assigning emf signs from current rather than source polarity.
- Mistake
Using −IR when traversing a resistor against current.
- Mistake
Discarding a negative current instead of reversing its direction.
- Mistake
Writing dependent loop equations that add no new information.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Kirchhoff's loop rule follows from conservation of energy.
The algebraic sum of potential changes around a closed loop is zero.
Across a resistor: −IR with current and +IR against current.
Across a source: +ε from negative to positive and −ε from positive to negative.
The algebraic sum of potential changes around a closed loop is zero.
∑Δ V=0
- Definition
- The algebraic sum of potential changes around a closed loop is zero.
- Formula
- ∑Δ V=0
- Example
- For a 12 V source with 2 Ω and 4 Ω, 12−2I−4I=0 gives I=2 A.
- Critical warning
- Component signs depend on traversal direction and marked polarity/current.
Did you understand this topic?
Your answer is saved with your learning record. It does not mark the topic as mastered.
Ready to test yourself?
You've completed the learning section.
Exam Practice
Statement
Q1. State Kirchhoff's loop rule and name its physical basis.
Signs
Q2. State the sign convention for resistors and sources in a loop equation.
Numerical
Q3. A 12 V source drives 2 Ω and 4 Ω in series. Use the loop rule to find current.
Reasoning
Q4. Why may either clockwise or anticlockwise traversal be chosen?
Application
Q5. How are Kirchhoff's rules used together in a multi-loop circuit?