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    Physics: Potential Difference of a Capacitor

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please solve and explain problem.

    Part 1:

    A 0.410 mF (C1) and a 0.630 mF (C2) capacitor are connected in series to a 12.00 V battery. Calculate the potential difference across C1.

    Part 2:

    Calculate the charge on C2.

    Part 3:

    Now assume that the two capacitors are connected in parallel. Calculate the voltage difference across C1.

    Part 4:

    Calculate the charge on C2.

    © BrainMass Inc. brainmass.com December 24, 2021, 8:38 pm ad1c9bdddf
    https://brainmass.com/physics/capacitors/physics-potential-difference-capacitor-296220

    SOLUTION This solution is FREE courtesy of BrainMass!

    1. The total capacitance in series is:
    C = (0.41mF)(0.63mF)/(0.41mF+0.63mF)
    C = 0.248mF
    The total stored charge is:
    q = CVbatt
    q = (0.248mF)(12V)
    q = 2.976mC
    The voltage at C1 is:
    V1 = q/C1
    V1 = 2.976mC / 0.410mF
    V1 = 7.259 V

    2. Since the charge in series is constant, and we have solved it in the previous problem, the charge at C2 is:
    q2 = q = 2.976mC

    3. When the capacitors are connected in parallel, the voltage between 2 capacitors are constant. So we have:
    V1 = Vbatt = 12V

    4. The charge across C2 is:
    q2 = C2 Vbatt
    q2 = (0.630mF)(12V)
    q2 = 7.56mC

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:38 pm ad1c9bdddf>
    https://brainmass.com/physics/capacitors/physics-potential-difference-capacitor-296220

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