Class JEE Physics ALL Q #1144
KNOWLEDGE BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
Two capacitors $C$ and $2C$ charged to $V$ and $2V$ respectively are connected in parallel with opposite polarity. The common potential is:
(A) $V$
(B) $\frac{V}{2}$
(C) $2V$
(D) $3V$
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AI Tutor Explanation

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Step-by-Step Solution

  1. Calculate the initial charge on each capacitor:

    • Capacitor C: $Q_1 = CV$
    • Capacitor 2C: $Q_2 = (2C)(2V) = 4CV$

  2. Since the capacitors are connected with opposite polarity, the net charge will be the difference between the two charges:

    $Q_{net} = |Q_2 - Q_1| = |4CV - CV| = 3CV$

  3. The total capacitance of the parallel combination is:

    $C_{total} = C + 2C = 3C$

  4. The common potential $V_{common}$ is given by:

    $V_{common} = \frac{Q_{net}}{C_{total}} = \frac{3CV}{3C} = V$

Correct Answer: V

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AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of capacitance, charge, and potential difference to solve a problem involving capacitors connected in parallel with opposite polarity.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to calculate the common potential, involving calculating initial charges, considering opposite polarities, and applying the principle of charge conservation.
Syllabus Audit: In the context of JEE, this is classified as KNOWLEDGE. The question directly tests the understanding and application of textbook concepts related to capacitors and circuits.

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