Kirchhoff's Laws

Kirchhoff's Laws

7 min read Updated Mar 26, 2026

Series and parallel rules handle simple circuits beautifully. But what about messier ones — circuits with multiple batteries, branching paths, or components that don’t fit neatly into “series” or “parallel” boxes? That’s where Kirchhoff’s two laws come in.

These aren’t new physics. They’re just conservation of charge and conservation of energy applied specifically to circuits — dressed up in slightly intimidating names. Understand the analogies and the laws become almost obvious.

Kirchhoff’s Current Law (KCL) — The Junction Rule

This is just conservation of charge. Charges can’t pile up at a junction or disappear — whatever flows in has to flow out.

Think of a river fork: if 10 m³/s flows into the fork and 6 m³/s goes left, then 4 m³/s must go right. The water didn’t vanish, and it didn’t accumulate at the split.

Kirchhoff's junction rule diagram showing currents entering and leaving a node, with arrows indicating that the sum of currents in equals the sum of currents out
Kirchhoff's junction rule (KCL): the total current flowing into a node equals the total current flowing out. No charge accumulates at a junction. Credit: Wikimedia Commons, CC BY-SA

Example. If 5 A flows into a junction and splits into three branches with 2 A in the first branch and 1 A in the second, the third branch carries 521=25 - 2 - 1 = 2 A out.

Kirchhoff’s Voltage Law (KVL) — The Loop Rule

This is conservation of energy. A charge that travels around a complete loop and returns to its starting point ends up with the same potential energy it started with — so the energy gained from batteries must exactly equal the energy lost across resistors. Net change: zero.

Multi-loop circuit diagram illustrating Kirchhoff's voltage law, with voltage gains across batteries and voltage drops across resistors summing to zero around each loop
Kirchhoff's loop rule (KVL): voltage gains and drops around any closed loop sum to zero. Energy gained from batteries equals energy lost across resistors. Credit: Wikimedia Commons, CC BY-SA

Sign Conventions for KVL

When you “walk” around a loop in a chosen direction, here’s how to tally up voltage changes:

ElementWalking directionVoltage change
Battery− to + (through the battery)+ε+\varepsilon (gain)
Battery+ to − (through the battery)ε-\varepsilon (drop)
ResistorSame direction as currentIR-IR (drop)
ResistorAgainst direction of current+IR+IR (gain)

The choice of loop direction doesn’t matter — pick one, stay consistent, and the math works out.

Applying Kirchhoff’s Laws — A Strategy

  1. Label all currents with assumed directions. If you guess wrong, the math returns a negative value — that just means the current actually flows opposite your guess. No need to redo the diagram.
  2. Apply KCL at each junction to relate the branch currents.
  3. Apply KVL around enough independent loops to solve for all unknowns.
  4. Solve the system of equations.

For the MCAT, you rarely need more than one junction equation and one or two loop equations. The exam favors conceptual understanding over algebraic complexity.

Quick Example

A simple loop has a 10 V battery and two series resistors: R1=3R_1 = 3 Ω and R2=2R_2 = 2 Ω. Apply KVL clockwise starting at the battery:

+10 VI(3)I(2)=0+10 \text{ V} - I(3) - I(2) = 0

10=5II=210 = 5I \Rightarrow I = 2 A. Then VR1=6V_{R_1} = 6 V, VR2=4V_{R_2} = 4 V. They add to 10 V — the battery’s EMF. KVL satisfied. ✓

This tiny example shows the recipe: walk around the loop, write down each voltage change with the correct sign, set the sum to zero, solve.

At a junction, 8 A flows in from one wire. Two wires leave carrying 3 A and 2 A. A third wire also leaves. What current does it carry?
Click to reveal answer
3 A out. KCL: Iin=Iout8=3+2+I3I3=3I_{in} = I_{out} \Rightarrow 8 = 3 + 2 + I_3 \Rightarrow I_3 = 3 A.
A loop contains a 15 V battery, a 4 Ω resistor, and an unknown resistor RR. Current in the loop is 3 A. What is RR?
Click to reveal answer
R=1R = 1 Ω. KVL: 153(4)3R=01512=3RR=115 - 3(4) - 3R = 0 \Rightarrow 15 - 12 = 3R \Rightarrow R = 1 Ω. Voltage drops: 12 V across the 4 Ω + 3 V across the 1 Ω = 15 V battery EMF. ✓
In a loop containing a 12 V battery and three resistors in series, the voltage drops across two of the resistors are 4 V and 5 V. What is the voltage drop across the third?
Click to reveal answer
3 V. KVL: voltage drops must sum to the battery EMF. 1245V3=0V3=312 - 4 - 5 - V_3 = 0 \Rightarrow V_3 = 3 V. No need to know the resistance values — just the loop conservation.