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Introduction to Kirchhoff's Laws

Kirchhoff's Laws are the fundamental tools for complex DC circuit analysis. While Ohm's Law (V=IR) works for simple single-resistor circuits, Kirchhoff's Laws allow us to solve for currents and voltages in networks with multiple branches, nodes, and loops.

Kirchhoff's Current Law (KCL) - Junction Rule

KCL states that for any junction (node) in an electrical circuit, the sum of currents flowing into that junction is equal to the sum of currents flowing out of that junction.

ΣIin = ΣIout   (or ΣI = 0)

Physical Basis: Conservation of Charge. Charge cannot "pile up" at a junction; what goes in must come out.

Kirchhoff's Voltage Law (KVL) - Loop Rule

KVL states that the directed sum of the potential differences (voltages) around any closed loop is zero.

ΣV = 0   around any closed loop

Physical Basis: Conservation of Energy. If you travel in a loop and return to the starting point, the net change in potential energy must be zero.

Rule of Thumb: Crossing a battery from (-) to (+) is +E. Crossing a resistor in the direction of current is -IR.

Series & Parallel Resistors

These formulas are derived from Kirchhoff's Laws and simplify circuit analysis:

Series: Rtotal = R₁ + R₂ + R₃ + ...

In series, current is the same through all components, but voltage is divided.

Parallel: 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ + ...

In parallel, voltage is the same across all components, but current is divided.

💡 Solver Tip: Always define your current directions and loop paths before writing your equations. If a calculated current is negative, it simply means the actual current flows in the opposite direction to what you assumed.

Worked Examples

✅ Example 1: KCL at a Junction

Three wires meet at a node. Wire A carries 5A into the node, Wire B carries 2A away from the node. Find the current in Wire C.

Step 1: Identify In and Out. Let Wire C be IC.
Step 2: ΣIin = ΣIout → 5 = 2 + IC
Step 3: IC = 5 - 2 = 3A
Analysis: Since IC is positive, Wire C carries 3A away from the node.

✅ Example 2: Simple KVL Loop

A loop contains a 12V battery and two resistors, 4Ω and 8Ω, in series. Prove KVL sums to zero.

Step 1: Calculate total current I = V/R = 12 / (4+8) = 1A.
Step 2: Voltage drops: V1 = 1A×4Ω = 4V, V2 = 1A×8Ω = 8V.
Step 3: Sum around loop: +12V (battery) - 4V (R1) - 8V (R2) = 0V
KVL Verified.

✅ Example 3: Parallel Combination

Find the equivalent resistance of a 10Ω, 20Ω, and 30Ω resistor connected in parallel.

Formula: 1/Rp = 1/10 + 1/20 + 1/30
Common Denom (60): 1/Rp = 6/60 + 3/60 + 2/60 = 11/60
Invert: Rp = 60 / 11 ≈ 5.45Ω

✅ Example 4: Mesh Analysis Hint

In a two-loop circuit, Loop 1 has an 18V source and a shared 6Ω resistor. Loop 2 has 12Ω. If I1=2A and I2=0.5A, find V across the shared 6Ω resistor.

Context: Currents are flowing in opposite directions through shared R.
Net Current: Inet = I1 - I2 = 2.0 - 0.5 = 1.5A.
Ohm's Law: V = 1.5A × 6Ω = 9V

✅ Example 5: Conservation of Power

A 10V battery supplies 0.5A to a circuit. Calculate power supplied vs power dissipated by a 20Ω resistor.

Power Supplied: P = VI = 10V × 0.5A = 5W.
Power in R: P = I²R = (0.5)² × 20 = 0.25 × 20 = 5W.
Result: 5W = 5W. Power is conserved.