Kirchhoff's Current Law
Think of a river flowing into a small lake, with several streams flowing out. In steady state — when the lake level is not rising or falling — the total flow rate in must equal the total flow rate out. Water cannot pile up indefinitely, and it cannot disappear. The lake level stays constant because inflow and outflow are in balance.
Kirchhoff's Current Law (KCL) is the electrical equivalent of this observation. In steady state, charge cannot accumulate at a node — a point where wires meet. Whatever current (charge per second) flows into a node must flow out again through other branches. This simple but powerful law, combined with Kirchhoff's Voltage Law from the previous lesson, gives you the complete mathematical toolkit to analyse any resistive circuit.
You have already used KCL every time you worked with parallel circuits and said "total current equals the sum of branch currents." That statement is KCL applied to the junction node of a parallel circuit. This lesson makes the principle explicit, shows you how to apply it systematically to any node, and introduces node analysis — a powerful method for solving complex multi-branch circuits.
- The Law and Why It Must Be True
- Sign Convention for Nodes
- The KCL Procedure Step by Step
- Worked Example 1: Four-Branch Node
- Worked Example 2: Finding Branch Current in a Parallel Circuit
- Introduction to Node Voltage Analysis
- Worked Example 3: Node Analysis with Two Unknowns
- KCL and the Parallel Current Rule
- KCL Node Calculator
- Experiment: Verify KCL at a Parallel Junction
- KCL in Ham Radio Circuits
KCL at a node: the sum of currents entering equals the sum of currents leaving. No charge accumulates at a junction in steady state.
View LargerThe Law and Why It Must Be True
Kirchhoff's Current Law states:
ΣI = 0 (with entering currents positive and leaving currents negative, or vice versa)
Equivalently: ΣIin = ΣIout
The physical basis is conservation of electric charge. Charge is a conserved quantity — it cannot be created or destroyed. In a metal conductor under steady conditions, electrons cannot pile up at a node or disappear from one. If electrons were accumulating at a junction, the junction's voltage would be continuously rising (more charge = more repulsion = higher potential). But "steady state" means voltages are constant — so charge is not accumulating anywhere, and therefore current in must equal current out at every node.
This is not a special rule that only applies in simple circuits. KCL holds at every node in every linear DC circuit, at every moment in time for AC circuits, and in any circuit where the charge carrier density at a node is not changing. The only exception is a capacitor plate, which temporarily stores charge during a transient — but even then, the current flowing into the capacitor's plate always equals the current flowing through the dielectric as displacement current. KCL is a universally true conservation law.
Sign Convention for Nodes
Unlike KVL which deals with loops, KCL deals with a single point — a node. The sign convention is simpler:
Convention A (most common): Currents entering the node are positive (+). Currents leaving the node are negative (−). Sum = 0.
Convention B (also valid): Currents leaving the node are positive (+). Currents entering are negative (−). Sum = 0.
Both conventions give the same final answer. The choice is a matter of personal preference. Most textbooks use convention A (entering = positive), and that is what this course uses.
When you do not know the actual direction of a current, assume a direction (in or out), assign a sign accordingly, and write the equation. If the solution gives a negative value, the current actually flows in the opposite direction from your assumption. Accept the negative result and reverse your direction interpretation — never change your assumed directions mid-calculation.
The KCL Procedure Step by Step
Step 2. List all branches connected to that node.
Step 3. For each branch, decide whether the current is entering or leaving the node. If unknown, assume a direction and stick with it.
Step 4. Write the KCL equation: Iin1 + Iin2 + ... − Iout1 − Iout2 − ... = 0.
Step 5. If currents are unknown, express them in terms of voltages and resistances using Ohm's Law: I = V / R.
Step 6. Solve for the unknown(s). A negative result means your assumed direction was reversed.
For a circuit with N nodes, you can write N−1 independent KCL equations. (The Nth equation is always the sum of all the others, so it is redundant — writing it gives no new information.) Combined with KVL loop equations, these independent equations are always sufficient to solve any linear DC circuit completely.
Worked Example 1: Four-Branch Node
A node has four branches. The currents entering are I1 = 50 mA and I2 = 30 mA. One current leaving is I3 = 20 mA. The fourth branch current I4 is unknown. Find I4 and its direction.
KCL equation (entering positive, leaving negative):
+I1 + I2 − I3 − I4 = 0
+50 + 30 − 20 − I4 = 0
I4 = 50 + 30 − 20 = 60 mA
Result: I4 = 60 mA, leaving the node (positive result confirms our assumed direction was correct).
Check: In = Out
Entering: 50 + 30 = 80 mA
Leaving: 20 + 60 = 80 mA ✓
This is the direct statement of KCL: current in = current out. The node is in balance — no charge is accumulating.
Worked Example 2: Finding Branch Current in a Parallel Circuit
Three resistors — R1 = 1 kΩ, R2 = 2.2 kΩ, R3 = 4.7 kΩ — are in parallel across a 9 V supply. The total supply current is measured as 16.2 mA. The currents through R1 and R3 are calculated as 9 mA and 1.91 mA respectively. Use KCL to find the current through R2.
KCL: Isupply = IR1 + IR2 + IR3
16.2 = 9.0 + IR2 + 1.91
IR2 = 16.2 − 9.0 − 1.91 = 5.29 mA
Check using Ohm's Law:
IR2 = V / R2 = 9 / 2200 = 4.09 mA
The small discrepancy (5.29 vs 4.09 mA) arises from a rounding error in the stated total current. If we calculate the exact total: IR1 = 9/1000 = 9.00 mA, IR2 = 9/2200 = 4.09 mA, IR3 = 9/4700 = 1.91 mA, total = 15.00 mA. Entering 15.00 into KCL: IR2 = 15.00 − 9.00 − 1.91 = 4.09 mA ✓
This example shows how KCL can both confirm calculated currents and reveal measurement or rounding errors.
Introduction to Node Voltage Analysis
Node voltage analysis (often simply called "nodal analysis") is a systematic method for solving any resistive network by writing KCL equations at each node and expressing the branch currents in terms of node voltages. It is the foundation of SPICE circuit simulators, which analyse circuits with thousands of nodes by solving the node voltage equations numerically.
The procedure builds on a simple insight: if you know the voltage at every node (relative to a chosen reference point), you know everything about the circuit. The current through any resistor between two nodes is simply the voltage difference divided by the resistance: I = (VA − VB) / R.
1. Choose a reference node (usually the node with the most connections, or ground). Assign it V = 0 V.
2. Label all other nodes with unknown voltages: VA, VB, VC, etc.
3. For each non-reference node, write a KCL equation. Express every branch current in terms of node voltages using Ohm's Law: current from node A to node B through R = (VA − VB) / R.
4. Solve the resulting system of simultaneous equations for the unknown node voltages.
5. Calculate individual branch currents from the solved node voltages using Ohm's Law.
Node analysis: label node voltages, express each branch current as (V_high − V_low) / R, then write KCL at each non-reference node. Solve the resulting equations for the node voltages.
View LargerWorked Example 3: Node Analysis with Two Unknowns
Circuit: 12 V supply at node A (left), reference (ground) at the right. R1 = 1 kΩ between node A and node B. R2 = 2.2 kΩ between node B and ground. R3 = 3.3 kΩ between node A and ground. Find the voltage at node B and all branch currents.
In this circuit, node A is set by the voltage source: VA = 12 V. Node B is unknown. Reference (ground) = 0 V.
Current into B from A through R1: IR1 = (VA − VB) / R1 = (12 − VB) / 1000
Current out of B to ground through R2: IR2 = VB / R2 = VB / 2200
KCL at node B: IR1 − IR2 = 0 (only two branches at this node)
(12 − VB) / 1000 = VB / 2200
Cross multiply:
2200 × (12 − VB) = 1000 × VB
26400 − 2200·VB = 1000·VB
26400 = 3200·VB
VB = 26400 / 3200 = 8.25 V
Branch currents:
IR1 = (12 − 8.25) / 1000 = 3.75 / 1000 = 3.75 mA
IR2 = 8.25 / 2200 = 3.75 mA
IR3 = 12 / 3300 = 3.636 mA (this is the current through R3 from node A to ground directly)
Total supply current (KCL at node A):
Isupply = IR1 + IR3 = 3.75 + 3.636 = 7.386 mA
Check — KCL at node B:
Entering (from A through R1): 3.75 mA
Leaving (to ground through R2): 3.75 mA
Sum = 0 ✓
Check — KVL around the outer loop (A → R3 → ground → supply → A):
12 − 3.636 × 3.3 − 0 = 12 − 12.0 = 0 ✓
KCL and the Parallel Current Rule
As with KVL and the series voltage rule, the connection between KCL and the parallel circuit rule you learned earlier is worth making explicit. The rule "total current equals the sum of branch currents" in a parallel circuit is simply KCL applied to the top junction node of the parallel circuit.
At the top junction: Isupply enters; IR1, IR2, IR3 leave. KCL says: Isupply = IR1 + IR2 + IR3. That is exactly the parallel current rule. KCL is the more general statement; the parallel rule is a special case.
Similarly, the check you performed in series-parallel circuit analysis — verifying that currents into each node equal currents out — was applying KCL. Now you have the formal framework that lets you use it methodically in any circuit topology.
KCL Node Calculator
KCL Node Calculator
Enter currents at a node in milliamps. Use positive values for currents entering the node and negative values for currents leaving the node. Leave exactly one field blank to solve for the unknown current. Fill all fields to verify the node balances.
⚖ Experiment: Verify KCL at a Parallel Junction
Use a three-branch parallel circuit to confirm that total supply current equals the sum of all branch currents — a direct test of KCL at the junction node where all branches connect.
- 9 V battery with clip connector
- Three resistors: 1 kΩ, 2.2 kΩ, 4.7 kΩ
- Breadboard and jumper wires
- Digital multimeter
- Predict first. Calculate expected branch currents: I1k = 9/1000 = 9.00 mA, I2.2k = 9/2200 = 4.09 mA, I4.7k = 9/4700 = 1.91 mA. Total = 15.00 mA. Write these down.
- Build the parallel circuit. Connect all three resistors in parallel between the top and bottom breadboard rails. Connect the battery across the rails.
- Measure total supply current. Break the wire from the battery positive terminal to the top rail. Insert the multimeter (set to DC milliamps) in series. Record Itotal.
- Measure each branch current. With all three resistors still connected, break one end of R1 (1 kΩ) from the rail, insert the meter in series with that resistor only. Record IR1. Reconnect R1, then repeat for R2 (2.2 kΩ) and R3 (4.7 kΩ).
- Sum the branch currents. Add your three measured values: IR1 + IR2 + IR3. Compare this sum to your measured Itotal. The difference should be less than 5% (resistor and meter tolerance).
- Enter into the KCL calculator. Enter Itotal as a negative number (leaving the supply's positive terminal node), and each branch current as a positive number (entering the load). Confirm the sum is approximately zero.
- Test an intermediate node. Connect R1 and R2 together at one end, leaving R3 separate. Measure the current at the node where R1 and R2 join — it should equal IR1 + IR2. This confirms KCL at an intermediate junction.
The sum of measured branch currents equals the total supply current within about 5–10% (resistor tolerance and meter accuracy). Predicted values: I1k ≈ 9 mA, I2.2k ≈ 4.1 mA, I4.7k ≈ 1.9 mA, total ≈ 15 mA. KCL is confirmed at every node you test. The 1 kΩ branch carries the most current (lowest resistance) and the 4.7 kΩ branch carries the least — exactly as predicted by Ohm's Law applied to each independent branch.
KCL in Ham Radio Circuits
Station power supply current budgeting
Every station power supply must deliver the total current demanded by all parallel loads. KCL at the supply output node: Isupply = IRX + IPA + Idisplay + Iaccessories. A typical 100 W HF station on transmit: receiver chain 0.8 A + PA 18 A + accessories 1.2 A = 20 A total supply current needed. KCL defines the minimum current rating your power supply must have. If the supply draws more current than expected, KCL tells you an additional current path exists — perhaps a relay coil stuck closed, a leakage path through a failed component, or an unintended feedback path through shared wiring.
BJT transistor current relationships
For a bipolar junction transistor (BJT), three currents meet at the device: IC (collector), IB (base), and IE (emitter). Applying KCL at the transistor gives: IE = IC + IB. Since the current gain β = IC / IB, the relationships are: IC = β × IB and IE = IC + IB = β × IB + IB = (β + 1) × IB. For a typical β of 100: IE ≈ 101 × IB ≈ IC. KCL at the transistor node is the starting point for every bias calculation, gain calculation, and fault diagnosis involving BJTs.
Node analysis in multi-stage amplifiers
A multi-stage amplifier has several internal nodes where the output of one stage connects to the input of the next. Each coupling network — typically a capacitor and bias resistors — creates a node. KCL at each node determines the bias voltage and quiescent current of that stage. When a stage is biased incorrectly (say, the collector voltage is not at the expected value), KCL immediately identifies which branch current is wrong — the bias resistors, the collector resistor, or the transistor itself — even before you know what has failed.
Wiring fault detection
KCL is a powerful debugging tool for wiring problems. If the current entering a connector does not equal the current leaving it, current is flowing somewhere unexpected — through a cable shield, through a chassis ground loop, or through a shorted insulation point. A station where the current drawn from the power supply does not match the sum of individually measured load currents has a hidden current path. KCL tells you this immediately; your job is then to find the hidden path and eliminate it.
Frequently Asked Questions
What is the difference between KCL and KVL?
KVL is a conservation of energy law applied to closed loops: the algebraic sum of all voltages around any loop equals zero. KCL is a conservation of charge law applied to nodes: the algebraic sum of all currents at any node equals zero. KVL gives you one equation per independent loop; KCL gives you one equation per non-reference node. Together, these two laws provide exactly enough independent equations to solve any linear DC circuit completely — regardless of complexity.
Does KCL apply to AC and RF circuits?
Yes — KCL applies at any frequency for lumped-circuit analysis. At AC and RF, currents are phasors (having both magnitude and phase angle), and the KCL equation becomes a phasor sum equal to zero. The real parts of all branch currents sum to zero, and the imaginary parts also sum to zero independently. At RF frequencies where the circuit dimensions approach a significant fraction of the wavelength, the lumped assumption breaks down — you must treat the circuit as a distributed system using transmission line theory. For practical ham radio circuit analysis up to low VHF, lumped KCL is accurate.
How many KCL equations do I need to solve a circuit with N nodes?
You need N−1 independent KCL equations, where N is the total number of nodes including the reference (ground) node. The KCL equation at the reference node is always the negative sum of all other node equations — redundant and giving no additional information. Combined with the KVL equations, you always have exactly enough constraints to solve for all unknown currents and voltages. For the example in this lesson with three nodes, you needed N−1 = 2 KCL equations (though one node had a known voltage from the supply, reducing the unknowns to one).
Does KCL hold if a capacitor is connected at a node?
Yes, always. In DC steady state, a capacitor blocks direct current, so the current through the capacitor branch is zero — KCL holds trivially because there is no current in or out of that branch. During a transient (while the capacitor is charging or discharging), a real current flows into or out of the capacitor plate: I = C × dV/dt. This current is part of the KCL equation just like any other branch current. During AC operation, the capacitor passes a sinusoidal current (displacement current) determined by its reactance. KCL is satisfied at all times and under all conditions.
Can I use KCL to find a current that I cannot directly measure?
Yes, and this is one of the most practical applications of KCL. If you can measure all other currents at a node, the unknown current is simply the negative sum of all the measured ones. For example, if three currents are known at a four-branch node, the fourth follows immediately from KCL. In a transistor circuit, you can find the base current by measuring the emitter and collector currents and applying KCL: IB = IE − IC. This is often easier than trying to measure the tiny base current directly, which could itself disturb the circuit.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.