KVL in series networks — the sum of the individual voltage drops across series resistors must equal the total applied source voltage. Is this fundamental statement accurate?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
This statement encapsulates Kirchhoff’s Voltage Law (KVL) applied to a series resistive circuit. KVL is one of the two cornerstone conservation laws for lumped circuits, ensuring energy consistency around closed loops. The item tests whether you can connect KVL to observable behavior in series resistor chains.


Given Data / Assumptions:

  • Ideal source and ideal resistors in series.
  • Closed loop such that KVL applies.
  • Steady-state conditions (but the principle also holds instant-by-instant in AC).


Concept / Approach:
KVL states the algebraic sum of voltages around any closed loop is zero. Traversing a loop with a source and series drops yields V_source − V1 − V2 − … − Vn = 0, or equivalently V_source = V1 + V2 + … + Vn. Each drop V_k follows Ohm’s law V_k = I*R_k with common current I in series. Thus the sum of the drops equals the applied source voltage regardless of individual resistor values.


Step-by-Step Solution:

Assume series chain R1..Rn with loop current I.Compute each drop: V_k = I*R_k.Apply KVL: V_source = Σ V_k.Therefore, the statement holds for all series networks.


Verification / Alternative check:
Bench test with a DMM across each resistor and across the source shows that measured drops sum to the source voltage within meter tolerance, validating KVL experimentally for DC and snapshot values in AC.


Why Other Options Are Wrong:

  • Incorrect: Violates energy conservation around the loop.
  • Only true for DC: KVL holds for AC as an instantaneous law in lumped circuits.
  • Only when resistors are equal: Equality is irrelevant; drops scale with R.


Common Pitfalls:
Sign convention mistakes when summing rises and drops; forgetting that meter polarity affects the sign, not the magnitude relation that the algebraic sum around the loop is zero.


Final Answer:
Correct — by KVL, individual drops sum to the applied source voltage.

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