Kirchhoff's current law (KCL)

Surround the node by an imaginary boundary. During a small time interval, charge entering that region minus charge leaving it must equal the change of charge stored inside:

In the ideal lumped-node model used here, unresolved charge storage at the node itself is neglected. Intentional storage will later be represented by explicit circuit elements. Therefore

Equivalently, if every branch current is treated as a signed algebraic variable with its own arrow, the algebraic sum of currents at the node is zero.

Choose “entering positive” for this equation. The references in Figure 2.1 then give

You could instead choose “leaving positive” and obtain the negative of the entire equation. Both are identical physical statements. The error is mixing the two conventions inside one equation.

Using Figure 2.1, let I₁=3.0 mA enter and I₂=2.0 mA leave. Then

The answer is positive, so I₃ really flows in the direction drawn.

If the calculation had returned −1 mA, the physical branch current would be 1 mA upward toward the node.

Let the unknown node voltage be V relative to GND. Define the current through R1 from the 12 V node toward VOUT, and the currents through R2 and R3 from VOUT toward GND.

Ohm's law expresses each branch current in terms of V:

KCL at VOUT is I1=I2+I3:

The right side simplifies to V/2k:

Now recover the branch currents:

Independent closure:

Suppose a node equation is written with three currents defined as leaving:

If I₁=2 mA and I₂=1 mA, then

Nothing failed. Branch 3 actually carries 3 mA into the node, opposite to its leaving-reference arrow.

The conservation argument works for any chosen region in the lumped network. You may draw an imaginary boundary around several connected nodes and sum only the branch currents crossing that boundary. Internal branch currents cancel from the external balance.

This viewpoint later explains why “supernodes” and larger cut-set balances work: they are still charge conservation applied to a larger region.

The most general node-region statement is still dQ/dt=ΣIin−ΣIout. In ordinary lumped analysis, intentional storage is represented by explicit elements whose branch currents participate in KCL. The ideal connection node itself is treated as having no independent state variable.

This distinction will matter when capacitors are introduced: KCL does not stop working; a capacitor simply contributes a branch current related to its changing voltage.

1. 5 mA enters a node; 1.5 mA and 2.0 mA leave. Find the third leaving current.
2. All branch arrows are defined outward. I1=4 mA and I2=−7 mA. If only three branches exist, find I3.
3. At node V, a 10 V node connects through 2 kΩ, while V connects to ground through 4 kΩ. Write and solve the KCL equation for V.
4. Why is “current in equals current out” incomplete if the chosen region is allowed to accumulate charge?
5. Explain why currents inside an imaginary multi-node boundary need not appear in the KCL equation for that boundary.

KCL constrains currents at nodes. The next chapter develops the complementary closed-path constraint on voltage: KVL.