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<h1>Kirchhoff's Current Law (KCL)</h1>
<p>KCL states that the total current entering any node in a circuit equals the total current leaving it. This follows directly from conservation of charge — current can't accumulate at a junction. It sounds obvious stated plainly, but it's the formal rule that lets you write and solve equations for circuits with multiple branches and unknown currents. Gustav Kirchhoff published both his current and voltage laws in 1845.</p>

<h2>The Law</h2>
<pre>
∑I_in = ∑I_out

Or equivalently, with a sign convention (entering = positive):
∑I = 0  (at any node)
</pre>

<table>
  <thead><tr><th>Convention</th><th>Entering current</th><th>Leaving current</th></tr></thead>
  <tbody>
    <tr><td>Entering positive</td><td>+I</td><td>−I</td></tr>
    <tr><td>Leaving positive</td><td>−I</td><td>+I</td></tr>
  </tbody>
</table>

<p>Either convention works — pick one and stick with it throughout the problem. The result is the same.</p>

<h2>Single-Node Example</h2>
<pre>
         I1 = 5A
         ↓
    ─────●─────→ I3 = ?
         ↑
         I2 = 3A

KCL (currents in = currents out):
  I1 + I2 = I3
  5A + 3A = 8A

I3 = 8A
</pre>

<h2>Multi-Branch Node</h2>
<pre>
         I1 = 10A
         ↓
    ─────●─────→ I3 = 4A
    ↑         ↓
    I4        I2 = 3A

KCL: I1 = I2 + I3 + I4
  10 = 3 + 4 + I4
  I4 = 3A (leaving upward)
</pre>

<h2>Using KCL in Nodal Analysis</h2>
<p>Nodal analysis is the systematic application of KCL to find node voltages in a circuit. The procedure:</p>

<table>
  <thead><tr><th>Step</th><th>Action</th></tr></thead>
  <tbody>
    <tr><td>1</td><td>Choose a reference node (ground). Assign it 0 V.</td></tr>
    <tr><td>2</td><td>Assign a voltage variable (V1, V2 …) to every other node.</td></tr>
    <tr><td>3</td><td>At each non-reference node, write KCL. Express each branch current using Ohm's Law: I = (V_node − V_adjacent) / R.</td></tr>
    <tr><td>4</td><td>Solve the resulting system of linear equations.</td></tr>
  </tbody>
</table>

<h3>Worked Nodal Analysis Example</h3>
<pre>
        12V
         |
        [R1=4Ω]
         |
    ─────●─────[R3=6Ω]──── 0V (GND)
         |
        [R2=3Ω]
         |
         0V (GND)

One unknown node: V1

KCL at V1 (sum of currents leaving the node = 0):
  (V1 − 12) / 4   +   V1 / 3   +   V1 / 6   = 0

Multiply through by 12 (LCM of 4, 3, 6):
  3(V1 − 12) + 4(V1) + 2(V1) = 0
  3V1 − 36 + 4V1 + 2V1 = 0
  9V1 = 36
  V1 = 4V

Back-calculate currents:
  I through R1 = (12 − 4) / 4 = 2A (into node)
  I through R2 = 4 / 3 = 1.33A (out)
  I through R3 = 4 / 6 = 0.67A (out)

Check: 2A in = 1.33 + 0.67 = 2A out ✓
</pre>

<h2>Two-Node Example</h2>
<pre>
        10V             5V
         |               |
        [R1=5Ω]         [R3=5Ω]
         |               |
    ─────●───[R2=10Ω]───●─────
    V1                       V2

KCL at V1 (currents leaving):
  (V1 − 10) / 5   +   (V1 − V2) / 10   = 0

KCL at V2 (currents leaving):
  (V2 − V1) / 10   +   (V2 − 5) / 5   = 0

Multiply first equation by 10:
  2(V1 − 10) + (V1 − V2) = 0
  3V1 − V2 = 20   ... (1)

Multiply second equation by 10:
  (V2 − V1) + 2(V2 − 5) = 0
  −V1 + 3V2 = 10  ... (2)

From (1): V2 = 3V1 − 20
Sub into (2): −V1 + 3(3V1 − 20) = 10
  8V1 = 70 → V1 = 8.75V
  V2 = 3(8.75) − 20 = 6.25V
</pre>

<h2>KCL and Parallel Resistors</h2>
<p>KCL directly explains why parallel resistors add currents: each branch draws current from the same node independently, so total supply current is the sum of all branch currents. With V=12V across three parallel resistors of 6Ω, 12Ω, and 4Ω:</p>
<pre>
  I_R1 = 12/6  = 2A
  I_R2 = 12/12 = 1A
  I_R3 = 12/4  = 3A
  I_total = 6A (KCL: all currents leaving the top node must come from the source)
</pre>

<h2>Supernode</h2>
<p>When a voltage source connects two non-reference nodes, you can't write a branch current directly through it. The trick: treat the two nodes as a <em>supernode</em> — write KCL for the combined region, then add the voltage source as an additional constraint equation:</p>
<pre>
Voltage source V_s between nodes V1 and V2:

Additional constraint: V1 − V2 = V_s

Write KCL for the supernode (treat V1 and V2 together,
  sum all currents flowing in/out of the combined region)
</pre>

<h2>References</h2>
<ul>
  <li>Hayt, Kemmerly &amp; Durbin — <em>Engineering Circuit Analysis</em></li>
  <li>Nilsson &amp; Riedel — <em>Electric Circuits</em></li>
  <li><a href="https://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-current-law-kcl/" target="_blank" rel="noopener">All About Circuits — KCL ↗</a></li>
</ul>
