A 6.93 µF capacitor charged to 94 V and a 1.76 µF capacitor charged to 81 V are connected to each other, with the two positive plates connec

Question

A 6.93 µF capacitor charged to 94 V and a 1.76 µF capacitor charged to 81 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the final potential difference across the 1.76 µF capacitor? Answer in units of V.

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1 year 2021-09-04T06:29:50+00:00 1 Answers 2 views 0

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    2021-09-04T06:31:28+00:00

    Answer:

    225.56V

    Explanation:

    Q=cv

    6.93 µF capacitor was charged to 94 V, therefore Q₁ =6.93 *10^-6 x 94 =651.42µC

    1.76 µF capacitor charged to 81 V, therefore Q₂ =1.76 *10^-6 x 81 =142.56µC

    When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5(142.56µC + 651.42µC )= 396.99µC

    The final potential difference across the 1.76 µF capacitor =  Q/c = 396.99µC/1.76 µF = 225.56V

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