A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected f

Question

A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 0.9 µF capacitor then drops to 2 V. What is the capacitance of the second capacitor?

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Hồng Cúc 4 years 2021-09-03T14:58:40+00:00 1 Answers 14 views 0

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  1. Dau
    0
    2021-09-03T15:00:36+00:00
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    Answer:

    3.6μF

    Explanation:

    The charge on the capacitor is defined by the formula

    q = CV

    because the charge will be conserved

    q₁ = C₁V₂

    q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

    q = q₁ + q₂ = C₁V₂ + C₂V₂

    CV = CV₂ + C₂V₂

    CV – CV₂ = C₂V₂

    C ( V – V₂) = C₂V₂

    C ( V/ V₂ – V₂ /V₂) = C₂

    C₂ = 0.9 ( 10 /2) – 1) = 0.9( 5 – 1) = 3.6μF

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