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Capacitor question

2023 · 13 Apr · Shift 2 · Q52
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  5. /2023 · 13 Apr · Shift 2 · Q52

Capacitor question

2023 · 13 Apr · Shift 2 · Q52

JEE MainPhysicsCapacitorMCQ+4 / −1
In the network shown below, the charge accumulated in the capacitor in steady state will be: JEE Main 2023 (Online) 13th April Evening Shift Physics - Capacitor Question 41 English
  1. A
    10.3 μ\muμ C
  2. B
    7.2 μ\muμ C
  3. C
    4.8 μ\muμ C
  4. D
    12 μ\muμ C
View written solutionFree

Correct answer: B

The circuit diagram is not actually visible in the prompt, so the exact network values/connections are missing. Without the figure, the capacitor charge in steady state cannot be derived uniquely from first principles.

However, for a steady-state capacitor problem in a DC resistive network, the standard method would be:

  1. Open-circuit the capacitor in steady state, since no current flows through it after a long time.
  2. Find the potential difference across the two terminals where the capacitor is connected using circuit analysis (series-parallel reduction, Kirchhoff laws, symmetry, or Thevenin equivalent).
  3. Use Q=CVQ = C VQ=CV where VVV is the steady-state voltage across the capacitor.

Because the actual circuit image is missing, I cannot compute VVV or verify which numerical option follows from the network.

Given the provided stored correct answer is B: 7.2 μC7.2\,\mu\text{C}7.2μC, the most reasonable response is to report that value while noting that it cannot be independently verified without the diagram.

So, based on the supplied answer key: Q=7.2 μCQ = 7.2\,\mu\text{C}Q=7.2μC

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