Kirchhoff’s Voltage Law (KVL): In any single closed loop, is the algebraic sum of all voltages equal to zero? Interpret “sum around the loop” to include both rises and drops with sign.

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
Kirchhoff’s Voltage Law (KVL) is a fundamental law derived from energy conservation and electrostatic fields in lumped circuits. It states that the algebraic sum of voltages around any closed path is zero.


Given Data / Assumptions:

  • Lumped-circuit model applies; electromagnetic radiation and time-varying magnetic coupling are negligible for the loop in question.
  • Voltages are summed algebraically, with rises taken as positive and drops as negative (or vice versa consistently).
  • Applies to DC and AC steady-state analysis.


Concept / Approach:

Voltage is potential difference. Traversing a closed loop and summing all potential rises and drops should return to the starting potential, yielding a net sum of zero. If sources are present, the sum of drops equals the sum of rises; representing rises with opposite sign leads to a zero total.


Step-by-Step Solution:

Choose a loop direction and sign convention.Record each element’s voltage as +V (rise) or −V (drop) consistent with polarity.Sum all terms: ΣV_loop = 0.Interpretation: energy gained from sources equals energy lost in elements over one loop traversal.


Verification / Alternative check:

Example: A loop with a 10 V source and two drops of 6 V and 4 V. Algebraic sum is +10 − 6 − 4 = 0. Alternatively, “sum of drops equals source,” which is the same statement rearranged.


Why Other Options Are Wrong:

  • “False” ignores energy conservation in lumped circuits.
  • Limiting KVL to DC is incorrect; it applies in AC circuit theory as well.
  • Component equality or absence of sources is irrelevant; KVL still holds when properly signed.


Common Pitfalls:

Misreading “sum of drops” (all positive) as zero. The correct version is the algebraic sum (rises and drops with signs) equals zero. Maintain consistent polarity marking to avoid sign errors.


Final Answer:

True

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