Two capacitors of 36 µF and 3.8 µF are connected parallel and charged with a 51 V power supply. Calculate the total energy stored in the two

Question

Two capacitors of 36 µF and 3.8 µF are connected parallel and charged with a 51 V power supply. Calculate the total energy stored in the two capacitors.

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Phúc Điền 5 months 2021-09-05T11:16:41+00:00 1 Answers 0 views 0

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    2021-09-05T11:17:50+00:00

    Answer:

     U=0.0517 J

    Explanation:

    Given that

    The values of capacitance

    C₁ = 36 µF

    C₂ =  3.8 µF

    We know that when capacitor are connected in the parallel then the equivalent value of the capacitance is given as

    C=C₁ +C₂

    Now by putting the values

    C= 36 + 3.8 µF

    C = 39.8 µF

    The voltage difference

    ΔV= 51 V

    The total stored energy in capacitor is given as

    U=\dfrac{1}{2}C\Delta V^2

    Now by putting the values in the above equation we get

    U=\dfrac{1}{2}\times 39.8\times 10^{-6}\times 51^2\ J

    U=0.0517 J

    Therefore the stored energy in the capacitors will be 0.0517 J.

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