Kirchhoff’s Voltage Law (KVL) is the bedrock of circuit analysis, stating that the directed sum of the potential differences around any closed loop is zero. While the concept is simple, exam questions frequently introduce shared components and opposing sources to test your grasp of the passive sign convention. Below is a complete, step-by-step walkthrough of a classic two-loop DC exam problem, designed to show every algebraic step, highlight common traps, and verify the final answer using power balance.

The Exam Problem: Multi-Loop KVL Walkthrough

Problem Statement

Consider a circuit with two distinct loops sharing a central branch.

  • Left Branch: A 12V DC voltage source (positive terminal facing up) in series with a 4Ω resistor (R1).
  • Middle Branch: A shared 2Ω resistor (R3).
  • Right Branch: A 5V DC voltage source (positive terminal facing up) in series with a 6Ω resistor (R2).
Task: Calculate the mesh currents I1 (left loop) and I2 (right loop), determine the actual current flowing through the shared resistor R3, and verify the solution.

Which Method Applies and Why?

For this topology, Mesh Analysis (driven by KVL) is the optimal choice over Nodal Analysis (KCL). Mesh analysis is preferred when a circuit is dominated by voltage sources and the primary unknowns are branch currents. Nodal analysis would require converting the voltage sources to Norton equivalents or using supernodes, adding unnecessary algebraic overhead.

The Trap in This Problem

The most common failure point in these Kirchhoff's voltage law examples is the passive sign convention on the mutual resistor (R3). Students often forget that the net current through R3 is the superposition of both mesh currents. If you assign both I1 and I2 in a clockwise direction, they oppose each other in the shared branch. Failing to account for this opposition results in a sign error that cascades through the entire matrix.

Step-by-Step Solution and Algebra

We will use the standard mesh current method. For a deeper theoretical foundation on mesh matrices, refer to the Mesh Analysis tutorial on Electronics Tutorials.

  1. Assign Mesh Currents: Define I1 as the clockwise current in the left loop, and I2 as the clockwise current in the right loop.
  2. Write KVL for Loop 1 (Left): Start at the bottom-left node and traverse clockwise.
    • We experience a voltage rise of +12V across the source.
    • We experience a voltage drop across R1: 4 * I1.
    • We traverse R3 downwards. The current I1 flows down, but I2 flows up. The net downward current is (I1 - I2). The drop is 2 * (I1 - I2).
    • Equation: 12 - 4*I1 - 2*(I1 - I2) = 0
    • Simplify: 12 - 6*I1 + 2*I2 = 0 → 6*I1 - 2*I2 = 12 (Eq. 1)
  3. Write KVL for Loop 2 (Right): Start at the bottom-right node and traverse clockwise.
    • We traverse R3 upwards. Moving against I1 and with I2 yields a drop of 2 * (I2 - I1).
    • We traverse R2 downwards, yielding a drop of 6 * I2.
    • We traverse V2 downwards. Since the positive terminal is on top, moving down is a voltage drop of 5V (or a rise of -5V).
    • Equation: -2*(I1 - I2) - 6*I2 - 5 = 0
    • Simplify: -2*I1 + 2*I2 - 6*I2 - 5 = 0 → -2*I1 - 4*I2 = 5 Wait, let's re-verify the R3 drop. Standard matrix form: Self-resistance is 6+2=8. Mutual is -2. Equation: -2*I1 + 8*I2 = -5. Let's trace again: Up R3 (drop is 2*I2 - 2*I1), down R2 (drop 6*I2), down V2 (drop 5). Sum of drops = 0 → 2*I2 - 2*I1 + 6*I2 + 5 = 0 → -2*I1 + 8*I2 = -5. (Eq. 2)
  4. Solve the System of Equations:
    • From Eq. 1: 2*I2 = 6*I1 - 12 → I2 = 3*I1 - 6
    • Substitute into Eq. 2: -2*I1 + 8*(3*I1 - 6) = -5
    • Expand: -2*I1 + 24*I1 - 48 = -5
    • Combine: 22*I1 = 43 → I1 = 43/22 ≈ 1.955 A
    • Solve for I2: I2 = 3*(43/22) - 132/22 = -3/22 → I2 ≈ -0.136 A
  5. Calculate Branch Currents:
    • Current through R1 = I1 = 1.955 A (downwards)
    • Current through R2 = I2 = 0.136 A (upwards, due to the negative sign)
    • Current through R3 (downwards) = I1 - I2 = 43/22 - (-3/22) = 46/22 = 2.091 A

Sanity Checks and Independent Verification

Never hand in an exam paper without running a sanity check. In circuit theory, the ultimate verification tool is Tellegen’s Theorem, which dictates that the total power delivered by sources must exactly equal the total power absorbed by the resistors.

Order of Magnitude Check

If R3 and the right loop were entirely removed, I1 would be 12V / 4Ω = 3A. Because R3 adds series resistance and the right loop draws some current, I1 must be strictly less than 3A. Our answer of 1.955 A passes this logical bound check perfectly.

Power Balance Verification

Power Delivered by Sources:

  • V1: Current I1 flows up through V1 (from negative to positive terminal). It is delivering power.
    P_V1 = 12V * (43/22)A = 516/22 = 23.454 W
  • V2: Because I2 is negative, the actual physical current flows UP through the right branch. It enters the negative terminal of V2 and leaves the positive terminal. Therefore, V2 is also delivering power.
    P_V2 = 5V * (3/22)A = 15/22 = 0.682 W
  • Total Delivered: 23.454 + 0.682 = 24.136 W

Power Absorbed by Resistors:

  • R1: (43/22)² * 4 = 15.281 W
  • R2: (-3/22)² * 6 = 0.112 W
  • R3: (46/22)² * 2 = 8.744 W
  • Total Absorbed: 15.281 + 0.112 + 8.744 = 24.137 W (Allowing for minor rounding in the decimal display; exact fractions yield 5841/242 for both sides).

The power balance matches to the exact fraction. The solution is verified independently.

Frequently Asked Questions on Kirchhoff's Voltage Law Examples

How do you apply Kirchhoff's Voltage Law to parallel circuits?

KVL does not apply to the parallel branches as a single collective entity; it applies to the loops formed by those branches. In a simple parallel circuit with a single voltage source and three parallel resistors, you actually have three distinct loops. KVL dictates that the voltage drop across each parallel resistor must equal the source voltage. For complex networks, you must identify independent closed paths (meshes) and write a KVL equation for each, just as we did in the walkthrough above.

What is the most common mistake in Kirchhoff's voltage law examples?

The most frequent error is misapplying the passive sign convention when traversing a shared component. When writing the KVL equation for Loop 1, students often forget to subtract the opposing current from Loop 2 when calculating the voltage drop across the mutual resistor. A secondary, equally fatal mistake is mixing up voltage rises and drops across sources. Always define your traversal direction (e.g., clockwise) and strictly adhere to: entering a source's positive terminal is a rise (or negative drop), and entering the negative terminal is a drop.

How does Kirchhoff's Voltage Law apply to AC circuits with inductors?

KVL holds true for AC circuits, but you must transition from scalar algebra to complex phasor math. Resistors remain real numbers, while inductors and capacitors introduce imaginary impedance (jωL and -j/ωC). The KVL equation becomes a vector sum: the phasor sum of all voltage drops around the loop equals zero. For a deep dive into AC circuit fundamentals and phasor domain analysis, the MIT OpenCourseWare Circuits and Electronics curriculum provides excellent foundational lectures.

Can Kirchhoff's Voltage Law be violated in real life?

Yes, under specific conditions. KVL is derived from Maxwell’s equations under the assumption of a conservative electric field (the lumped element model). If a time-varying magnetic field passes through the area enclosed by your circuit loop, it induces an electromotive force (EMF) according to Faraday’s Law of Induction. In this scenario, the electric field is non-conservative, and the sum of the voltages around the loop will not equal zero; it will equal the negative rate of change of the magnetic flux. This is the fundamental principle behind transformers and inductors, where KVL must be modified to include the induced EMF as a source term.