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<h1>Kirchhoff's Voltage Law (KVL)</h1>
<p>KVL states that the sum of all voltages around any closed loop in a circuit equals zero. This is conservation of energy: if you travel around a loop and return to your starting point, the energy gained and lost must balance exactly. In practice, it means the voltage drops across components in a loop must sum to the supply voltage. Together with KCL, it forms the complete foundation for analysing any DC circuit.</p>

<h2>The Law</h2>
<pre>
∑V = 0   around any closed loop
</pre>

<table>
  <thead><tr><th>Convention</th><th>Voltage rise</th><th>Voltage drop</th></tr></thead>
  <tbody>
    <tr><td>Travelling through a source from − to +</td><td>+V (gain)</td><td>—</td></tr>
    <tr><td>Travelling through a source from + to −</td><td>—</td><td>−V (loss)</td></tr>
    <tr><td>Travelling through a resistor in current direction</td><td>—</td><td>−IR (loss)</td></tr>
    <tr><td>Travelling through a resistor against current direction</td><td>+IR (gain)</td><td>—</td></tr>
  </tbody>
</table>

<p>Pick a starting point, pick a direction (clockwise or counter-clockwise), and be consistent. The algebra takes care of itself.</p>

<h2>Simple Series Loop</h2>
<pre>
    +──[R1=4Ω]──[R2=8Ω]──+
    |                     |
   [12V]                  |
    |                     |
    +─────────────────────+

Clockwise loop starting at battery positive terminal:
  +12 − V_R1 − V_R2 = 0

Total resistance: 4 + 8 = 12Ω
Current: I = 12/12 = 1A

V_R1 = 1A × 4Ω = 4V
V_R2 = 1A × 8Ω = 8V

Verify KVL: 12 − 4 − 8 = 0 ✓
</pre>

<h2>Three-Resistor Series Circuit</h2>
<pre>
    +──[R1=100Ω]──[R2=220Ω]──[R3=330Ω]──+
    |                                    |
   [9V]                                  |
    |                                    |
    +────────────────────────────────────+

R_total = 100 + 220 + 330 = 650Ω
I = 9V / 650Ω = 13.85 mA

V_R1 = 0.01385 × 100 = 1.38V
V_R2 = 0.01385 × 220 = 3.05V
V_R3 = 0.01385 × 330 = 4.57V

KVL: 9 − 1.38 − 3.05 − 4.57 = 0.00 ✓ (rounding)
</pre>

<h2>Mesh Analysis — Multiple Loops</h2>
<p>Mesh analysis applies KVL to every independent loop, with each loop assigned an assumed mesh current. The loop currents may overlap through shared components:</p>
<pre>
      R1=2Ω         R3=3Ω
+──────[  ]────┬────[  ]──+
|              |          |
[10V]         [R2=4Ω]   [5V]
|              |          |
+──────────────┴──────────+

Two meshes: I1 (left loop, clockwise), I2 (right loop, clockwise)

KVL left loop:
  10 − 2I1 − 4(I1 − I2) = 0
  10 − 2I1 − 4I1 + 4I2 = 0
  10 − 6I1 + 4I2 = 0   ... (1)

KVL right loop (note: 5V battery opposes clockwise direction):
  −4(I2 − I1) − 3I2 − 5 = 0
  4I1 − 4I2 − 3I2 = 5
  4I1 − 7I2 = 5          ... (2)

From (1): I1 = (10 + 4I2) / 6
Substitute into (2): 4 × (10 + 4I2)/6 − 7I2 = 5
  (40 + 16I2)/6 − 7I2 = 5
  40 + 16I2 − 42I2 = 30
  −26I2 = −10
  I2 = 0.385A

  I1 = (10 + 4×0.385)/6 = (10 + 1.54)/6 = 1.923A

Currents:
  R1: 1.923A
  R2: I1 − I2 = 1.923 − 0.385 = 1.538A
  R3: 0.385A
</pre>

<h2>KVL and the Voltage Divider</h2>
<p>The voltage divider formula is a direct consequence of KVL and Ohm's Law:</p>
<pre>
+──[R1]──●──[R2]──+
|        |        |
[Vs]   Vout      |
|                |
+────────────────+

KVL: Vs − V_R1 − V_R2 = 0     → Vs = V_R1 + V_R2
Ohm: Same current I through both → V_R1 = I×R1, V_R2 = I×R2

Solve for Vout = V_R2:
  I = Vs / (R1 + R2)
  Vout = I × R2 = Vs × R2 / (R1 + R2)
</pre>

<h2>Supermesh</h2>
<p>If a current source appears in a branch shared between two meshes, you can't write a KVL equation that includes it (current sources don't have a defined voltage drop). The trick is a <em>supermesh</em>: exclude the current source branch, write KVL around the outer perimeter of the combined loops, then add the current source as an additional constraint:</p>
<pre>
If a 2A current source shares a branch between mesh I1 and I2:

Constraint: I1 − I2 = 2A   (or I2 − I1 = 2A, depending on direction)
KVL: written around outer perimeter ignoring the shared branch
</pre>

<h2>Summary — KVL vs KCL</h2>

<table>
  <thead><tr><th></th><th>KVL</th><th>KCL</th></tr></thead>
  <tbody>
    <tr><td>Governing principle</td><td>Conservation of energy</td><td>Conservation of charge</td></tr>
    <tr><td>Applied to</td><td>Closed loops</td><td>Nodes (junctions)</td></tr>
    <tr><td>Solves for</td><td>Voltages and currents (mesh analysis)</td><td>Node voltages (nodal analysis)</td></tr>
    <tr><td>Equation type</td><td>∑V = 0 around loop</td><td>∑I = 0 at node</td></tr>
    <tr><td>Number of equations</td><td>One per independent loop</td><td>One per non-reference node</td></tr>
  </tbody>
</table>

<h2>References</h2>
<ul>
  <li>Hayt, Kemmerly &amp; Durbin — <em>Engineering Circuit Analysis</em>, 8th ed.</li>
  <li>Nilsson &amp; Riedel — <em>Electric Circuits</em>, 10th ed.</li>
  <li><a href="https://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-voltage-law-kvl/" target="_blank" rel="noopener">All About Circuits — KVL ↗</a></li>
</ul>
