Kirchhoff's Voltage Law
Imagine climbing a hill and returning to your starting point by a different route. No matter which path you take, if you end up exactly where you started, your net change in height is zero. You may have climbed higher in the middle, but the final tally of ups and downs always cancels out. Kirchhoff's Voltage Law (KVL) is the electrical equivalent of this observation, applied to voltage instead of height.
KVL is one of the two most powerful tools in circuit analysis — the other is Kirchhoff's Current Law, covered in the next lesson. Together they form the complete mathematical foundation for analysing any DC electrical circuit, no matter how complex. KVL is not an approximation or a rule that works most of the time — it is a direct consequence of the conservation of energy and is universally true in any circuit where the electric field is well-defined.
You have already used KVL without knowing it. Every time you verified that voltage drops across a series circuit summed to the supply voltage, you were applying KVL. This lesson makes that principle explicit, gives you a formal procedure to apply it, and shows you how it extends to circuits that cannot be solved by simple series and parallel rules.
- The Law and Why It Must Be True
- Sign Convention — The Critical Step
- The KVL Procedure Step by Step
- Worked Example 1: Single Loop, Unknown Voltage Drop
- Worked Example 2: Finding Unknown Current
- Worked Example 3: Two-Loop Circuit
- KVL and the Series Voltage Rule
- KVL Loop Calculator
- Experiment: Verify KVL in a Real Circuit
- KVL in Ham Radio Circuits
Traversing a closed loop and summing voltage rises and drops: the result must always be zero, by KVL.
View LargerThe Law and Why It Must Be True
Kirchhoff's Voltage Law states:
ΣV = V1 + V2 + V3 + ... + Vn = 0
The word "algebraic" is the key. Some voltages are positive (rises — the charge gains energy as it moves through the element, as through a battery from − to +) and some are negative (drops — the charge loses energy as heat in a resistor). KVL says that if you add all these signed voltages around a complete loop, the total is exactly zero.
Why must this be true? Because of the conservation of energy. Think of a unit positive charge moving around a closed loop. At a battery, it gains electrical potential energy (voltage rise). At a resistor, it loses potential energy as heat (voltage drop). When the charge completes the loop and returns to its starting point, it is at exactly the same potential as when it started — because potential is a property of position, and it has returned to the same position. Therefore, the total energy gained (from rises) must equal the total energy lost (from drops). Dividing by charge converts energy to voltage, and the statement becomes: total voltage rises = total voltage drops, or equivalently, their algebraic sum = 0.
There is no exception to KVL in any practical circuit. If you measure a loop and the voltages do not sum to zero, one of the following is true: you have made a measurement error, you have a meter loading issue (the meter itself changes the circuit), or you have a non-conservative field effect (significant inductance at high frequency). For all DC circuits and most low-frequency AC circuits, KVL holds exactly.
Sign Convention — The Critical Step
The sign convention is the step where most beginners struggle with KVL, so let us spend time on it carefully. The good news is that there is only one rule to memorise, and it applies consistently to every element in every loop.
First, pick a traversal direction for your loop — clockwise or counter-clockwise. Either works; the choice is completely arbitrary and does not affect the final answer. Once chosen, stick with it all the way around the loop.
Sign convention for KVL: entering a + terminal is a rise (+); entering a − terminal is a drop (−). Mark polarity on each element before writing the equation.
View LargerAs you walk around the loop in your chosen direction, you encounter each element and enter it from one side. The rule is:
Resistors: If you traverse the resistor in the same direction as the assumed current, write it as a drop: −IR. If you traverse against the assumed current, write it as a rise: +IR.
For resistors, you need to first assume a current direction. Choose any direction — it does not matter if it is wrong. If the solution gives a negative current value, it simply means the actual current flows the other way. Never change your assumed directions mid-calculation to try to fix a negative result; just interpret the negative sign as reversed direction when you state the final answer.
A practical shortcut for resistors in simple single-supply circuits: if you have already decided the conventional current flows from the supply's positive terminal around to the negative terminal (the most natural assumption), then every resistor in the path from + to − is a drop, and the battery is a rise. This makes the KVL equation for a series circuit trivially obvious: Vsupply − VR1 − VR2 − ... = 0.
The KVL Procedure Step by Step
Step 2. Assign current directions (any direction; guess if needed). Label unknown currents with symbols.
Step 3. Identify the closed loop(s) you want to analyse. For complex circuits, identify all independent loops.
Step 4. Choose a traversal direction (clockwise or counter-clockwise) for each loop.
Step 5. Walk around each loop, writing down each voltage with its sign according to the convention above.
Step 6. Set the algebraic sum equal to zero: ΣV = 0.
Step 7. If you have multiple loops, you now have a system of simultaneous equations. Solve for the unknowns.
Step 8. Interpret negative results as reversed direction from your assumption.
Worked Example 1: Single Loop, Unknown Voltage Drop
A 9 V battery drives a series loop containing R1 = 100 Ω, R2 = 150 Ω, R3 = 200 Ω, and an LED. The measured loop current is 15 mA. Find the forward voltage of the LED.
KVL equation (clockwise traversal, starting at battery −):
+Vbattery − VR1 − VR2 − VR3 − VLED = 0
Calculate known voltage drops:
VR1 = I × R1 = 0.015 × 100 = 1.50 V
VR2 = I × R2 = 0.015 × 150 = 2.25 V
VR3 = I × R3 = 0.015 × 200 = 3.00 V
Substitute into KVL:
+9 − 1.50 − 2.25 − 3.00 − VLED = 0
VLED = 9 − 1.50 − 2.25 − 3.00 = 2.25 V
Interpretation: The LED's forward voltage is 2.25 V — consistent with a green or yellow LED (typical green: 2.0–2.5 V; red: 1.6–2.0 V). We found this without directly measuring the LED by applying KVL to the other elements in the loop.
Notice the power of this approach: KVL let us find the voltage across an element we could not easily calculate directly (the LED, which does not follow Ohm's Law), simply by knowing the other voltages in the loop. In troubleshooting, this is invaluable — measure everything you can easily measure, then use KVL to determine what you cannot directly measure.
Worked Example 2: Finding Unknown Current
A circuit has a 12 V supply in series with R1 = 220 Ω and R2 = 470 Ω. Find the current without knowing the answer in advance — use KVL with an assumed current and solve the equation.
KVL equation:
+12 − I × 220 − I × 470 = 0
12 = I × (220 + 470)
12 = I × 690
I = 12 / 690 = 17.39 mA
Verify — voltage drops:
VR1 = 0.01739 × 220 = 3.83 V
VR2 = 0.01739 × 470 = 8.17 V
Sum: 3.83 + 8.17 = 12.00 V ✓
This worked example shows that KVL naturally produces Ohm's Law for series circuits. The equation I × (R1 + R2) = Vsupply is just V = I × Rtotal rearranged. KVL is the deeper principle from which Ohm's Law calculations for series circuits follow.
Worked Example 3: Two-Loop Circuit
This example shows KVL solving a circuit that cannot be handled by simple series-parallel reduction. The circuit has two batteries: V1 = 12 V and V2 = 6 V, with three resistors: R1 = 100 Ω between the left battery and the centre node, R2 = 200 Ω between the centre node and the right battery, and R3 = 300 Ω from the centre node to ground. Find all three currents.
This is a two-mesh circuit. Define I1 as the clockwise mesh current in the left loop and I2 as the clockwise mesh current in the right loop. R3 is shared between both meshes — the current through R3 is I1 − I2 (if both flow downward through R3, they subtract).
+12 − I1 × 100 − (I1 − I2) × 300 = 0
12 − 100·I1 − 300·I1 + 300·I2 = 0
12 = 400·I1 − 300·I2 ... (equation 1)
KVL for right loop (clockwise):
−6 − I2 × 200 − (I2 − I1) × 300 = 0
(note: V2 = 6 V is entered from + to − going clockwise, so it is a drop: −6)
−6 − 200·I2 − 300·I2 + 300·I1 = 0
−6 = −300·I1 + 500·I2 ... (equation 2)
Solve the system:
From equation 1: 12 = 400·I1 − 300·I2
From equation 2: −6 = −300·I1 + 500·I2
Multiply equation 1 by 5 and equation 2 by 3:
60 = 2000·I1 − 1500·I2
−18 = −900·I1 + 1500·I2
Add: 42 = 1100·I1 → I1 = 42/1100 = 38.18 mA
Substitute back into equation 1:
12 = 400 × 0.03818 − 300·I2
12 = 15.27 − 300·I2
300·I2 = 3.27 → I2 = 10.91 mA
Individual branch currents:
IR1 = I1 = 38.18 mA
IR2 = I2 = 10.91 mA
IR3 = I1 − I2 = 38.18 − 10.91 = 27.27 mA
Check — KVL left loop:
12 − 38.18 × 0.1 − 27.27 × 0.3 = 12 − 3.82 − 8.18 = 0.00 ✓
Check — KVL right loop:
−6 − 10.91 × 0.2 − (−27.27 × 0.3) = −6 − 2.18 + 8.18 = 0.00 ✓
The two-loop example demonstrates mesh analysis — one of the standard techniques for solving circuits that cannot be reduced by series-parallel methods. The key insight: each loop produces one KVL equation, and each equation gives you one constraint on the unknowns. With as many independent equations as unknowns, the system is solvable.
KVL and the Series Voltage Rule
It is worth making the connection explicit: the series circuit voltage rule you learnt two lessons ago — that voltage drops across series components sum to the supply voltage — is simply KVL applied to a single-loop series circuit. KVL is the more general statement; the series rule is a special case.
Similarly, the checking step in series-parallel circuit analysis (verifying that voltage drops along any path from supply to ground sum to the supply voltage) is KVL. Every time you used that check, you were doing KVL. Now you have the formal framework and the sign convention that allows you to apply it to any loop in any circuit — even circuits with multiple supplies, branches, and cross-connections.
KVL also explains why you cannot have a voltage source directly in parallel with another voltage source of different voltage. Suppose you connected a 12 V battery and a 9 V battery in parallel. The KVL equation around the loop formed by the two batteries would be: +12 − 9 = 3 V ≠ 0. KVL is violated — which in practice means an enormous current flows to try to equalize the voltages, limited only by internal resistance, until one or both batteries is destroyed. Never connect batteries of different voltages directly in parallel without current-limiting resistance.
KVL Loop Calculator
KVL Loop Calculator
Enter up to five known loop voltages (positive = voltage rise, negative = voltage drop). Leave exactly one field blank to solve for that unknown voltage. Fill all fields to verify that the loop balances (sum should be zero).
⚖ Experiment: Verify KVL in a Real Circuit
Measure the voltages around a real loop and confirm that the algebraic sum is zero. This experiment directly confirms the conservation of energy principle underlying KVL.
- 9 V battery with clip connector
- Three resistors: 470 Ω, 1 kΩ, 2.2 kΩ in series
- One red LED (optional — for an extra element with non-ohmic voltage)
- Breadboard and jumper wires
- Digital multimeter
- Predict first. Calculate the expected current: I = 9 / (470 + 1000 + 2200) = 9 / 3670 = 2.45 mA. Calculate expected voltage drops: V470 = 1.15 V, V1k = 2.45 V, V2.2k = 5.40 V. Total: 1.15 + 2.45 + 5.40 = 9.00 V.
- Build the circuit: battery + → 470 Ω → 1 kΩ → 2.2 kΩ → battery −. All in series on the breadboard.
- Measure battery terminal voltage. Touch probes directly to the battery terminals (with circuit connected). This is your actual supply voltage — it may be slightly less than 9 V if the battery is not fresh. Record Vbattery.
- Measure each resistor voltage. Place the red probe on the end of each resistor that is closer to the battery positive terminal, black probe on the other end. Record VR1 (470 Ω), VR2 (1 kΩ), VR3 (2.2 kΩ). These should all be positive readings.
- Enter into the KVL calculator. V1 = +Vbattery (positive: it is a rise as you traverse from − to +). V2 = −VR1, V3 = −VR2, V4 = −VR3 (negative: drops). Confirm the sum is approximately zero.
- Add the LED (optional). Insert the LED in series (observe polarity — longer leg to positive side). Now four elements are in the loop. Measure each voltage and confirm KVL still holds, even though the LED does not follow Ohm's Law.
- Test a sub-loop. Measure voltage across R2 and R3 together (one probe at the R1/R2 junction, other probe at battery −). Confirm this equals VR2 + VR3. This is KVL applied to the sub-loop of just those two elements.
The sum Vbattery − VR1 − VR2 − VR3 equals zero within a few millivolts (meter accuracy and resistor tolerance). Vbattery exactly equals the sum of the three resistor voltages. The sub-loop check confirms that KVL applies to any closed path in the circuit — not just the full outer loop. The LED, despite not following Ohm's Law, still obeys KVL: the voltage across it is simply whatever remains after the resistor drops are subtracted from the supply.
KVL in Ham Radio Circuits
Voltage drop budgeting in DC wiring
When a transceiver is installed in a vehicle or at a remote location, the DC wiring itself has resistance. A typical mobile installation: 13.8 V supply → 0.5 Ω wiring resistance → 0.3 V fuse holder drop → 13 V at the transceiver input. KVL for this loop: 13.8 − 0.5×I − 0.3 − 13 = 0, giving I = (13.8 − 0.3 − 13) / 0.5 = 1 A. If the transceiver draws more than 1 A, the voltage at its terminals will drop below 13 V. KVL gives you the complete voltage budget immediately — you can see exactly where each fraction of a volt is being consumed.
LED indicator current calculation
Every transceiver panel has LED indicators. To calculate the resistor needed to set the LED current at (say) 10 mA from a 5 V rail with a red LED (Vf ≈ 1.8 V): KVL for the loop: 5 − I × R − VLED = 0 → 5 − 0.010 × R − 1.8 = 0 → R = 3.2 / 0.010 = 320 Ω. Use 330 Ω (nearest standard value). This exact calculation is KVL applied to a simple two-element series loop.
Transistor bias point verification
KVL is the essential tool for verifying a BJT bias point. For an NPN common-emitter stage with VCC = 12 V, RC = 4.7 kΩ, RE = 470 Ω, and a measured collector current IC = 1.5 mA: KVL for the collector loop: 12 − IC × RC − VCE − IC × RE = 0. Solving: VCE = 12 − (0.0015 × 4700) − (0.0015 × 470) = 12 − 7.05 − 0.705 = 4.25 V. If the measured VCE is not approximately 4.25 V, the bias point is wrong and the stage will not amplify correctly.
Battery charger and load analysis
A 13.8 V regulated supply, connected through 0.15 Ω of wiring to a 12 V battery being maintained and a transceiver load drawing 5 A. KVL for the supply-to-load loop: 13.8 − 5 × 0.15 − Vbattery = 0 → Vbattery = 13.8 − 0.75 = 13.05 V. The battery terminal voltage under load is 13.05 V, not 13.8 V — the difference is dropped across the wiring resistance. If you need at least 13.5 V at the transceiver input, you must reduce the wiring resistance to below (13.8 − 13.5) / 5 = 0.06 Ω — which requires heavier gauge cable.
Frequently Asked Questions
Does it matter whether I traverse the loop clockwise or counter-clockwise?
No. If you reverse the traversal direction, every voltage in the loop changes sign — what was a rise becomes a drop and vice versa. But the equation −(sum of original terms) = 0 is mathematically identical to (sum of original terms) = 0. Both give the same solution for any unknowns. Choose whichever direction makes the signs most intuitive — usually the direction that matches your assumed current flow, so that resistors naturally produce drops without needing to think about "going against" the current.
Does KVL apply to AC circuits and RF circuits?
Yes — KVL applies at any frequency in lumped-circuit analysis. At AC and RF frequencies, inductors have inductive reactance (XL = 2πfL) and capacitors have capacitive reactance (XC = 1/2πfC). The voltages across these reactive elements are 90° out of phase with the current, so they must be treated as complex (phasor) quantities. The KVL equation becomes a phasor sum equal to zero — the real parts sum to zero and the imaginary parts sum to zero separately. At very high frequencies where the wavelength is comparable to circuit dimensions, lumped KVL becomes approximate and must be replaced by distributed transmission line analysis.
What if I get a negative voltage result from a KVL equation?
A negative result simply means the actual polarity of that voltage is opposite to what you assumed when you set up the equation. The magnitude is correct. For example, if you assumed current flowed left-to-right through a resistor and got VR = −3.5 V, the resistor actually has 3.5 V across it with the left end negative (current flows right-to-left). Never go back and change your assumed directions to avoid negative results — accept negative results and interpret them correctly. This is one of the most common traps beginners fall into.
How many independent KVL equations can I write for a circuit?
The number of independent KVL equations equals the number of independent loops (meshes) in the circuit, given by L = B − N + 1, where B is the number of branches and N is the number of nodes. For a simple series circuit (1 loop): L = 2 − 2 + 1 = 1. For a circuit with three branches between two nodes (like a parallel circuit plus its supply): L = 3 − 2 + 1 = 2. Writing more equations than independent loops simply produces equations that are sums or multiples of the others — redundant, not adding new information.
Why can you not connect two different-voltage batteries directly in parallel?
KVL tells you why. The loop formed by two batteries of different voltages in parallel contains only the two voltage sources — no resistors to absorb voltage. The KVL equation for that loop is V1 − V2 = 0, which requires V1 = V2. If V1 ≠ V2, KVL is violated in the ideal circuit model. In practice, what happens is that an enormous current flows limited only by the batteries' internal resistance, rapidly discharging the higher-voltage battery into the lower-voltage one — generating heat and potentially causing damage or fire. Always connect batteries of the same chemistry and the same state of charge when paralleling.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.