Electronics Simplified
📚 Course contents

Electronics Concepts, lesson 16 of 19

Kirchhoff's Laws Explained Simply: KCL and KVL

intermediate8 min read

Short answer: Kirchhoff's two laws are about bookkeeping.

  • KCL (current law): at any junction, the current flowing in equals the current flowing out. Charge doesn't pile up or disappear.
  • KVL (voltage law): go all the way round any closed loop and the voltage rises and drops add up to zero. You end where you started.

With these two rules plus Ohm's law you can solve almost any DC circuit in your syllabus, from two resistors to a full mesh-analysis problem.

KCL: what goes in must come out

Current is the flow of charge, and charge behaves just like that water. At any junction (a node):

Sum of currents in = sum of currents out

or, counting "in" as + and "out" as −: ΣI = 0.

Example. A 9 V battery feeds two resistors in parallel, 1 kΩ and 3 kΩ.

  1. Both resistors have the full 9 V across them (that's what parallel means).
  2. Current through 1 kΩ: 9 ÷ 1,000 = 9 mA.
  3. Current through 3 kΩ: 9 ÷ 3,000 = 3 mA.
  4. KCL at the junction: the battery must supply 9 + 3 = 12 mA.

Check with the combined resistance: 1 kΩ ∥ 3 kΩ = (1 × 3) ÷ (1 + 3) = 0.75 kΩ, and 9 ÷ 750 = 12 mA. Same answer.

Series: two bumpy roads one after the other. Their resistance adds up: 1000 + 1000 = 2000 Ω. Less electricity flows, so the LED is dim.

KVL: a walk round the mountain

Around any closed loop:

Sum of voltage rises = sum of voltage drops, or ΣV = 0.

Example. A 9 V battery with 1 kΩ and 2 kΩ in series.

  1. Total resistance = 1 + 2 = 3 kΩ, so the current is 9 ÷ 3,000 = 3 mA (the same everywhere in a series loop).
  2. Drop across 1 kΩ: 0.003 × 1,000 = 3 V.
  3. Drop across 2 kΩ: 0.003 × 2,000 = 6 V.
  4. KVL round the loop: +9 − 3 − 6 = 0 ✔

Notice the bigger resistor took the bigger share. That's exactly how a voltage divider works.

Using both together: a two-loop circuit

Exam questions usually mix them. Here's a classic:

   R1 = 2 kΩ        R3 = 1 kΩ
 +──/\/\/──┬──────/\/\/──+
 |         |             |
10 V    R2 = 2 kΩ        5 V
 |         |             |
 +─────────┴─────────────+

Two batteries (10 V on the left, 5 V on the right, both with + at the top), sharing the middle resistor R2. Find the current in each resistor.

Step 1: name the currents. Let I₁ flow through R1 (left to right), I₃ through R3 (right to left, from the 5 V battery), and both flow down through R2. KCL at the top middle node: I₂ = I₁ + I₃.

Step 2: KVL round the left loop (10 V, R1, R2): 10 − 2I₁ − 2I₂ = 0 → 2I₁ + 2I₂ = 10 (currents in mA, resistors in kΩ).

Step 3: KVL round the right loop (5 V, R3, R2): 5 − 1I₃ − 2I₂ = 0 → I₃ + 2I₂ = 5.

Step 4: solve. Put I₂ = I₁ + I₃ into both:

  • 2I₁ + 2(I₁ + I₃) = 10 → 4I₁ + 2I₃ = 10
  • I₃ + 2(I₁ + I₃) = 5 → 2I₁ + 3I₃ = 5

From the second: I₁ = (5 − 3I₃) ÷ 2. Into the first: 2(5 − 3I₃) + 2I₃ = 10 → 10 − 4I₃ = 10 → I₃ = 0 mA. Then I₁ = 2.5 mA and I₂ = 2.5 mA.

Step 5: check. Voltage across R2 = 2.5 mA × 2 kΩ = 5 V, exactly the right battery's voltage, so no current flows through R3. KVL holds round both loops. A good exam habit: always check, it catches sign mistakes.

Why the laws are true

  • KCL is conservation of charge. Charge can't be stored in a plain wire junction, so whatever arrives must leave.
  • KVL is conservation of energy. Voltage is energy per unit charge; a charge that goes round a loop and comes back to the same point has the same energy it started with.

They hold for any circuit with lumped parts (resistors, batteries, LEDs, transistors), not only for resistors. Ohm's law is the one that only works for resistors.

Common mistakes

  • Mixing up signs. Pick a direction to walk round the loop and stick to it. Going through a battery from − to + is a rise (+); going through a resistor in the direction of the current is a drop (−).
  • Guessing a current's direction wrong. That's fine! If you get a negative answer, the current just flows the other way. Don't redo the problem.
  • Counting a node twice. A long wire with no parts in it is all one node, however many corners it has.
  • Mixing mA and A. Either use mA with kΩ throughout (then volts come out right), or A with Ω. Don't mix them.

Check what you learnt

  1. At a junction, 15 mA flows in through one wire and 6 mA flows out through another. What happens in the third wire?

  2. A 12 V battery drives three resistors in series. Two of them drop 4 V and 5 V. What does the third drop?

  3. You solve a circuit and get I = −2 mA for a current you guessed flowed left. What does that mean?

Conclusion

Kirchhoff's laws are two pieces of common sense: water that enters a junction has to leave it, and a walk that ends at your door has as much downhill as uphill. Write one KCL equation per junction and one KVL equation per loop, use Ohm's law for each resistor, and solve.

What next?