When you encounter a multi-loop DC circuit on an exam or at the workbench, guessing current directions and eyeballing voltage drops will not yield reliable results. You need a systematic framework. This Kirchhoff example walks through a classic two-mesh (three-branch) DC circuit, demonstrating exactly how to apply Kirchhoff’s Voltage Law (KVL) without falling into the sign-error traps that cost students and junior technicians valuable time.

The Problem Statement and Component Specifications

Problem Statement:
Given a two-mesh DC circuit with a shared central branch, calculate the exact branch currents I1 (left loop), I2 (right loop), and I3 (shared middle branch). Assume both mesh currents flow in a clockwise direction. Verify your answer using Tellegen’s Theorem (power balance).

Before writing a single equation, a competent engineer maps the physical components to their theoretical models. In real-world prototyping, the resistor you pull from a bin has a specific tolerance and power rating that dictates whether it will survive the calculated current. Below is the spec sheet for the components in this Kirchhoff example.

Component Designation Nominal Value Tolerance Max Power Circuit Role
DC Source 1 V1 15.0 V N/A N/A Mesh 1 primary driver
DC Source 2 V2 9.0 V N/A N/A Mesh 2 opposing load
Carbon Film R1 100 Ω ±5% 0.25 W Mesh 1 series limiting
Carbon Film R2 150 Ω ±5% 0.25 W Mesh 2 series limiting
Metal Film R3 220 Ω ±1% 0.50 W Shared middle branch

Method Selection and The Common "Trap"

For this topology, Mesh Analysis (KVL) is vastly superior to Nodal Analysis (KCL). Why? Because the circuit is driven by voltage sources. Nodal analysis would require you to define a reference ground, solve for node voltages, and then back-calculate the branch currents using Ohm's Law. Mesh analysis yields the loop currents directly in fewer steps. For a deeper dive into when to choose KCL over KVL, refer to the foundational guides on Kirchhoff's Voltage Law at All About Circuits.

⚠ The Trap: The most frequent point of failure in this specific Kirchhoff example is the shared resistor, R3. When tracing Mesh 2 clockwise, you travel up through R3. Because we defined both I1 and I2 as clockwise, I2 flows up through R3, while I1 flows down. The net current through R3 in the direction of Mesh 2 traversal is (I2 - I1). Furthermore, V2 is oriented with its positive terminal at the top; traversing it clockwise means going from + to -, which is a voltage drop, not a rise.

Step-by-Step Algebraic Execution

We will now write the KVL equations for both meshes, summing the voltage drops to equal the supplied voltage. We assume standard passive sign convention.

  1. Write the raw KVL equation for Mesh 1 (Left Loop):
    Starting at the negative terminal of V1 and moving clockwise:
    +15 - 100(I1) - 220(I1 - I2) = 0
  2. Expand and group the Mesh 1 terms:
    15 - 100I1 - 220I1 + 220I2 = 0
    -320I1 + 220I2 = -15
    Multiply by -1 for standard positive-leading form:
    Equation A: 320I1 - 220I2 = 15
  3. Write the raw KVL equation for Mesh 2 (Right Loop):
    Starting at the bottom of R3 and moving clockwise:
    -220(I2 - I1) - 150(I2) - 9 = 0
  4. Expand and group the Mesh 2 terms:
    -220I2 + 220I1 - 150I2 - 9 = 0
    220I1 - 370I2 = 9
    Equation B: 220I1 - 370I2 = 9
  5. Solve the system using elimination:
    Multiply Equation A by 22: 7040I1 - 4840I2 = 330
    Multiply Equation B by 32: 7040I1 - 11840I2 = 288
    Subtract the modified B from the modified A:
    (7040 - 7040)I1 + (-4840 - (-11840))I2 = 330 - 288
    7000I2 = 42
    I2 = 42 / 7000 = 0.006 A (or 6 mA)
  6. Back-substitute I2 into Equation A to find I1:
    320I1 - 220(0.006) = 15
    320I1 - 1.32 = 15
    320I1 = 16.32
    I1 = 16.32 / 320 = 0.051 A (or 51 mA)
  7. Calculate the shared branch current (I3):
    Assuming downward flow through R3 is positive:
    I3 = I1 - I2 = 51 mA - 6 mA = 45 mA

Sanity Checks and Independent Verification

An answer is only as good as its verification. In professional DC circuit analysis, we never trust a raw calculation without running a power balance check (Tellegen's Theorem). The total power supplied by the sources must exactly equal the total power dissipated by the resistors.

1. Order of Magnitude and Unit Check

Our currents are in the milliamp range (6 mA to 51 mA). Given a 15V source and a minimum resistance path of roughly 320 Ω (100 + 220), Ohm's law estimates I ≈ 15 / 320 = 46 mA. Our calculated 51 mA is perfectly in the expected order of magnitude, accounting for the parallel draw of the second mesh. Units are strictly Amperes.

2. Power Balance (Tellegen's Theorem)

Power Supplied / Absorbed by Sources:

  • V1 (Supplying): P = V × I1 = 15 V × 0.051 A = 0.765 W
  • V2 (Absorbing): Because I2 (6 mA) flows clockwise, it enters the positive terminal of V2. Therefore, V2 is acting as a load (like a charging battery). P = 9 V × 0.006 A = 0.054 W absorbed.
  • Net Power Supplied to the resistive network = 0.765 W - 0.054 W = 0.711 W

Power Dissipated by Resistors (P = I²R):

  • R1: (0.051)² × 100 = 0.002601 × 100 = 0.2601 W
  • R2: (0.006)² × 150 = 0.000036 × 150 = 0.0054 W
  • R3: (0.045)² × 220 = 0.002025 × 220 = 0.4455 W
  • Total Power Dissipated = 0.2601 + 0.0054 + 0.4455 = 0.711 W

The net supplied power (0.711 W) perfectly matches the total dissipated power (0.711 W). The algebra is verified.

Bench Tip: Notice that R3 is dissipating 0.4455 W. If you had built this circuit using standard 1/4 W (0.25 W) carbon film resistors for all positions, R3 would overheat, drift in value, and potentially fail open. This is exactly why the spec table designated a 0.50 W metal film resistor for the shared branch. Theory must always inform physical component selection.

Frequently Asked Questions

What if I assumed the wrong direction for I2?
If you define I2 counter-clockwise and the math yields a negative number, it simply means the physical current flows opposite to your assumption. The magnitude remains correct. Always stick to your initial convention throughout the entire algebraic process; do not flip signs mid-calculation.

Does this Kirchhoff example apply to AC circuits?
Yes, but you must replace scalar resistance (R) with complex impedance (Z), and scalar voltages/currents with phasors. The KVL and KCL topology rules remain identical, but the algebra requires complex number arithmetic.