Unit 11 · Topic 11.6

11.6 Kirchhoff’s Loop Rule

Kirchhoff's loop rule says the potential differences around any closed loop add to zero. It's conservation of energy written for circuits, and it lets you write equations for circuits that series and parallel rules can't handle and graph the potential around a loop.

Key terms

Kirchhoff’s loop rule
conservation of energy
potential difference
closed loop
potential vs. position graph

The rule and why it's true

Electric potential is a value at each point, like height on a hiking trail. If you walk around a closed loop and return to your starting point, you're back at the same potential, so all the ups and downs along the way must cancel: $\sum \Delta V = 0$ around any closed loop.

In energy terms, a charge that goes all the way around gains energy in the battery and gives the same amount away in the resistors. That's why the loop rule is a statement of conservation of energy. It works because the electrostatic field is conservative. (In Unit 13 you'll see that a changing magnetic field breaks this, which is why an inductor gets its own term.)

Sign rules

Pick a direction for the current in each branch (a guess is fine), then pick a direction to walk around the loop. Add up the change in potential across each element as you pass it:

Element you cross	Direction you walk	Change in potential
Battery	from − terminal to + terminal	$+\mathcal{E}$
Battery	from + terminal to − terminal	$-\mathcal{E}$
Resistor	same direction as the current	$-IR$
Resistor	against the current	$+IR$
Capacitor	from + plate to − plate	$-\frac{q}{C}$
Ideal wire	either	0

Using the rule

Write one equation per independent loop. Then solve for the unknowns. If a current comes out negative, it just flows opposite to your guess; keep the negative sign while you finish solving, then report the actual direction.

A real battery adds an internal resistance term, $-Ir$ , right next to its emf. Capacitors add $\pm\dfrac{q}{C}$ , which is how you'll set up RC circuits in 11.8.

The exam won't test circuits that put batteries of different emf directly in parallel, so you don't need to worry about that setup.

Potential versus position

You can graph the potential at each point as you travel around a loop. Choose one point as 0 V, often the negative terminal of the battery. Then step around the loop: the graph jumps up by the emf across a battery, drops by IR across each resistor, and stays flat along ideal wires.

When you get back to the start, the graph must return to the starting value. That's the loop rule drawn as a picture. A bigger drop means more potential difference across that element. For elements in series, which share one current, the bigger drop marks the larger resistance.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Two batteries in one loop
A single loop contains a 12 V battery, a 2.0 Ω resistor, a 3.0 V battery and a 1.0 Ω resistor, in that order. The batteries face opposite ways, so their positive terminals point in opposite directions around the loop. Find the current and its direction.
Show the solutionHide the solution
1. Step 1: The 12 V battery is stronger, so guess that the current flows the way it pushes: out of its positive terminal.
2. Step 2: Walk around the loop in that direction, starting at the 12 V battery's negative terminal. Cross it from − to +: +12 V. Cross the 2.0 Ω resistor with the current: −2.0I. The 3.0 V battery faces the other way, so you cross it from + to −: −3.0 V. Cross the 1.0 Ω resistor: −1.0I.
3. Step 3: Loop rule: $12 - 2.0I - 3.0 - 1.0I = 0$ , so $9.0 = 3.0I$ and $I = 3.0\text{ A}$ .
4. Step 4: The answer is positive, so the guess was right. The 3.0 V battery is being pushed backward and is absorbing energy, like a battery being recharged.
Answer: 3.0 A, in the direction the 12 V battery pushes
Example 2Calculator allowed
Walking the potential around a loop
Using the loop from the previous example (I = 3.0 A), set the 12 V battery's negative terminal at 0 V. Find the potential after each element as you walk around in the direction of the current.
Show the solutionHide the solution
1. Step 1: After the 12 V battery: 0 + 12 = 12 V.
2. Step 2: After the 2.0 Ω resistor: 12 − (3.0)(2.0) = 6.0 V.
3. Step 3: After the 3.0 V battery, crossed from + to −: 6.0 − 3.0 = 3.0 V.
4. Step 4: After the 1.0 Ω resistor: 3.0 − (3.0)(1.0) = 0 V. You're back at the start and back at 0 V, as the loop rule requires.
5. Step 5: On a graph of potential against position, that's a jump up of 12 V, a drop of 6 V, a drop of 3 V and a drop of 3 V, with flat segments for the wires.
Answer: 12 V, then 6.0 V, then 3.0 V, then back to 0 V

Common mistakes

Mixing up the sign for a resistor. Walking with the current is a drop (−IR); walking against it is a rise (+IR).
Giving a battery a sign based on the current's direction. A battery's sign depends only on which terminal you cross first.
Throwing away a negative answer or flipping it partway through. Keep the sign through the algebra, then interpret it as a direction.
Writing more loop equations than you need. Each new equation must include at least one element not already used, or it adds nothing.

On the exam

Free-response questions often say "write, but do not solve, equations that could be used to find the currents." Show a labeled current direction for every branch and one equation per loop, with consistent signs.
Potential-versus-position graphs and questions about the potential at a labeled point are common. Pick a zero point, then add the changes one element at a time.

Connected topics

Videos

Kirchhoff's Rules of Electrical Circuits
Flipping PhysicsWatch on YouTube (opens in a new tab)
Kirchoff's Loop Rule
Bozeman ScienceWatch on YouTube (opens in a new tab)
Kirchhoff's voltage law | Circuit analysis | Electrical engineering | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Kirchoff's Loop Rule | Physics with Professor Matt Anderson | M22-07
Physics with Professor Matt AndersonWatch on YouTube (opens in a new tab)
Kirchhoff's Loop and Junction Rules Theory | Doc Physics
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Check yourself

4 questions on 11.6 Kirchhoff’s Loop Rule. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A single loop contains a 12 V battery, a 3.0 V battery, a 2.0 Ω resistor and a 1.0 Ω resistor. The batteries are oriented so that they try to push current in opposite directions around the loop. What is the current in the loop?

Question 2 of 4Calculator allowed

The negative terminal of an ideal 12 V battery is grounded (V = 0). From the positive terminal, current passes through resistor R₁ = 2R and then resistor R₂ = R before returning to the negative terminal. What is the electric potential at the point between R₁ and R₂?

Question 3 of 4Calculator allowed

Kirchhoff’s loop rule says the potential differences around any closed loop of a circuit add to zero. Which principle is this rule a consequence of?

Question 4 of 4Calculator allowed

A student graphs electric potential versus position while tracing a single loop that contains a battery, two resistors and connecting wires. Some parts of the graph are horizontal. What do the horizontal parts represent?

0 of 4 answered