Physics Electrostatics PMDC 2021
PMDC Verified Question 93 of 131
Q.120 For n numbers of Capacitors, each of the capacitance 'C' what will be the ratio between maximum and minimum capacitor?
A
n
B
\( n^2 \)
C
\( n^3 \)
D
\( n^4 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( n^2 \)
1. Concept:

Identical capacitors yield their absolute maximum total capacitance when wired in parallel, and absolute minimum when wired in series.

2. Formula:

$$ C_{max} = nC \quad \text{and} \quad C_{min} = \frac{C}{n} $$

3. Solution:

  • We must calculate the ratio: \( \frac{C_{max}}{C_{min}} \).


  • Substitute the expressions: \( \frac{nC}{C/n} \).


  • The 'C' variables algebraically cancel out.


  • The 'n' in the denominator's fraction flips up to multiply the numerator: \( n \times n = n^2 \).


4. Why other options are incorrect:

Option A is the answer if asked only for the maximum multiplier. The true ratio compares the extremes, forcing the \( n \) term to square.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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