Capacitance rule (repaired stem) — in a parallel combination that includes capacitor C4 along with other capacitors, the total capacitance is larger than the value of C4 alone. Is this statement correct?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Knowing how total capacitance compares to individual branch values helps validate designs quickly. In parallel, capacitances add directly, so the total must exceed each individual branch value.


Given Data / Assumptions:

  • Capacitors are connected in parallel, including C4 among them.
  • Capacitors behave linearly in the operating region.


Concept / Approach:
Parallel rule: CT = C1 + C2 + … + Cn. Since each Ci ≥ 0, CT ≥ any Ci, with strict inequality if at least one additional capacitor is present besides C4.


Step-by-Step Solution:

1) Write CT = C4 + Σ(other Ci).2) If any other Ci > 0, then CT > C4.3) Conclude the statement is correct for a genuine parallel group.4) Use this as a quick plausibility check after calculations.


Verification / Alternative check:
Example: C4 = 10 µF in parallel with 2.2 µF → CT = 12.2 µF > 10 µF.


Why Other Options Are Wrong:

Identical/AC/ESR caveats are unnecessary for the ideal addition rule; ESR affects losses, not the ideal capacitance sum.


Common Pitfalls:
Confusing series vs parallel behavior; in series, the total is less than the smallest element, which is the opposite trend.


Final Answer:
Correct

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