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Capacitor question

2021 · 27 Aug · Shift 1 · Q48
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  5. /2021 · 27 Aug · Shift 1 · Q48

Capacitor question

2021 · 27 Aug · Shift 1 · Q48

JEE MainPhysicsCapacitorMCQ+4 / −1
Calculate the amount of charge on capacitor of 4 μ\muμ F. The internal resistance of battery is 1 Ω\OmegaΩ : JEE Main 2021 (Online) 27th August Morning Shift Physics - Capacitor Question 79 English
  1. A
    8 μ\muμ C
  2. B
    zero
  3. C
    16 μ\muμ C
  4. D
    4 μ\muμ C
View written solutionFree

Correct answer: A

  1. Key idea

    For a capacitor connected to a battery in a DC circuit, after a long time the current becomes zero. Hence, there is no voltage drop across the internal resistance of the battery.

  2. Final potential difference across capacitor

    In steady state, I=0I=0I=0 so the drop across internal resistance r=1 Ωr=1\,\Omegar=1Ω is Ir=0.Ir=0.Ir=0.

    Therefore, the capacitor gets charged to the full emf of the battery.

  3. Use capacitor formula

    Charge on capacitor is Q=CVQ=CVQ=CV

    Here, C=4 μFC=4\,\mu FC=4μF and from the figure/context, the battery emf is V=2 V.V=2\,V.V=2V.

    So, Q=(4 μF)(2 V)=8 μC.Q=(4\,\mu F)(2\,V)=8\,\mu C.Q=(4μF)(2V)=8μC.

  4. Check options

    • A: 8 μC8\,\mu C8μC ✅
    • B: zero ❌
    • C: 16 μC16\,\mu C16μC ❌
    • D: 4 μC4\,\mu C4μC ❌

Thus, the correct answer is A.

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