Kirchhoff’s Voltage Law

Part of Electrical Theory

The conservation of energy in circuit loops — how voltage rises and drops balance around every closed path.

Why This Matters

Kirchhoff’s Voltage Law (KVL) is the second of the two circuit analysis laws that form the foundation of all electrical engineering. While KCL tracks current at junctions, KVL tracks voltage around loops. Together they allow you to solve any circuit, no matter how many components or interconnections it contains.

KVL is directly derived from conservation of energy. If you walk around a complete loop in a circuit and return to your starting point, you must have gained exactly as much voltage as you lost. Otherwise, energy would be created or destroyed — which doesn’t happen.

For rebuilding applications: KVL tells you the voltage available at any point in a circuit, helps you size voltage dividers and regulators, explains why long wire runs cause voltage drop problems, and provides the tool to calculate any unknown voltage in a network.

The Statement of KVL

KVL: The algebraic sum of all voltage changes around any closed loop equals zero.

Or equivalently: The sum of voltage rises (sources) equals the sum of voltage drops (loads) around any loop.

Walk around a circuit loop and sum up every voltage change:

• Across a battery from − to +: voltage rises (add)
• Across a battery from + to −: voltage drops (subtract)
• Across a resistor in the direction of current: voltage drops (subtract, this is Ohm’s law drop)
• Across a resistor against the direction of current: voltage rises (add)

When you complete the loop and return to start: total = 0.

Simple Series Circuit: KVL in Action

Example: 12V battery, three series resistors R1=2Ω, R2=3Ω, R3=7Ω

Total resistance: 12Ω. Current I = 12V/12Ω = 1A.

KVL around the loop (starting at negative terminal, going clockwise):

• Battery: +12V (rise)
• R1: -I×R1 = -1×2 = -2V (drop)
• R2: -I×R2 = -1×3 = -3V (drop)
• R3: -I×R3 = -1×7 = -7V (drop)

Sum: +12 - 2 - 3 - 7 = 0 ✓

This confirms the circuit is correctly analyzed. More usefully, KVL lets us find unknown quantities.

Finding Unknown Voltages

Example: Same circuit, but we don’t know R3. We measure voltage across R3 = 7V. What is R3?

KVL: 12 - 2 - 3 - V_R3 = 0 V_R3 = 7V ✓

With I = 1A and V_R3 = 7V: R3 = V/I = 7Ω.

Finding voltage at any point: The voltage at any node can be found by tracing from a known reference (usually ground) and summing voltage changes along the path.

If ground (0V) is at the negative terminal:

• Point between R2 and R3: V = 12 - I×R1 - I×R2 = 12 - 2 - 3 = 7V
• Point between R1 and R2: V = 12 - I×R1 = 12 - 2 = 10V

These are called node voltages — the voltage at each node relative to ground.

Multiple Loop Circuits

When a circuit has multiple loops (mesh), you apply KVL to each independent loop.

Example: Ladder network

        R1          R2
+--[10Ω]--+--[20Ω]--+
|          |          |
12V      [30Ω]      [60Ω]
|          |          |
+-----------+---------+
           GND

Two loops. Assign mesh currents I1 (left loop, clockwise) and I2 (right loop, clockwise).

Loop 1 (KVL, clockwise): 12 - 10×I1 - 30×(I1 - I2) = 0 12 - 10I1 - 30I1 + 30I2 = 0 12 - 40I1 + 30I2 = 0 … equation (1)

Loop 2 (KVL, clockwise): -20×I2 - 60×I2 - 30×(I2 - I1) = 0 Note: no voltage source in loop 2 -20I2 - 60I2 - 30I2 + 30I1 = 0 30I1 - 110I2 = 0 … equation (2)

From equation (2): I1 = (110/30)×I2 = 3.67×I2

Substituting in equation (1): 12 - 40×(3.67×I2) + 30I2 = 0 12 - 146.7I2 + 30I2 = 0 12 = 116.7×I2 I2 = 0.103A

I1 = 3.67 × 0.103 = 0.378A

Now check: current through R1 = I1 = 0.378A. Voltage drop across R1 = 10×0.378 = 3.78V. Voltage at middle node = 12 - 3.78 = 8.22V. Verify with R3: current through R3 = I1 - I2 = 0.378 - 0.103 = 0.275A. Voltage across R3 = 30×0.275 = 8.25V ≈ 8.22V ✓ (small rounding error).

KVL and Voltage Drop in Wiring

One of the most practical applications of KVL in building wiring:

Problem: A 12V battery 50 meters from a 12V, 5A lamp. Wiring is 1mm copper wire. What voltage reaches the lamp?

1. Total wire resistance: 1mm copper wire = 0.0215Ω per meter; 50m each way = 100m total R_wire = 100 × 0.0215 = 2.15Ω

2. Total current: 5A

3. KVL around loop: 12V (battery) - 2.15×5 (wire drop) - V_lamp = 0 V_lamp = 12 - 10.75 = 1.25V

The lamp receives only 1.25V instead of 12V — it won’t light. The wire resistance causes a 10.75V drop.

Solution: Use thicker wire.

• 4mm² copper wire: 0.0054Ω per meter
• R_wire = 100 × 0.0054 = 0.54Ω
• V_lamp = 12 - 0.54×5 = 12 - 2.7 = 9.3V

Still some drop, but lamp will function. For critical loads, size wire to limit voltage drop to 3–5% of supply voltage.

General wire sizing rule: Maximum resistance = (acceptable voltage drop) / (load current)

KVL with Multiple Sources

When a circuit has multiple batteries or generators:

Example: Two batteries in series

• Battery 1: 6V
• Battery 2: 9V
• Load: 15Ω

KVL: +6 + 9 - 15×I = 0 I = 15/15 = 1A ✓ (sources add in series, same direction)

Example: Two batteries opposing each other

• Battery 1: 12V (left, positive up)
• Battery 2: 9V (right, positive up)
• Resistor: 1Ω between them

KVL: +12 - 9 - 1×I = 0 I = 3/1 = 3A (current flows from higher to lower voltage)

This is what happens when you accidentally connect batteries backward — the voltage difference drives current through the internal resistances, potentially causing overheating.

KVL as a Diagnostic Tool

Voltage budget analysis: For any circuit branch, the sum of voltage drops must equal the source voltage. If it doesn’t, you have a fault.

Hunting down voltage drops:

1. Measure voltage at source terminals (should equal rated voltage under load)
2. Measure voltage at each junction point
3. The difference between adjacent measurements is the voltage drop across that section
4. An unexpected large drop indicates high resistance — loose connection, corroded terminal, undersized wire

Example: You expect 12V at the lamp but measure 8V.

• Measure at battery: 12V ✓
• Measure at junction box: 11.8V (0.2V drop in main feed, OK)
• Measure at lamp terminal: 8V

The drop from junction box to lamp is 11.8 - 8 = 3.8V across 4A = 0.95Ω — far too high for a short wire run. Inspect that section for corrosion, loose terminal, or too-thin wire.

KVL makes voltage troubleshooting systematic: measure, calculate, locate the excess drop, fix it. No guessing required.