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Calculating capacitance

Discussion in 'General Electronics Discussion' started by GuyMelvin, Jun 23, 2011.

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  1. GuyMelvin

    GuyMelvin

    1
    0
    Jun 23, 2011
    I am having to calculate some capacitances but i think i made an error with some of the voltages.

    Attatched is the question and below is my answers. But i have made a mistake on the voltage question i think? :confused:

    any help would be HUGELY appreciated!

    Many thanks

    Total capacitance

    Capacitors in parallel

    = C1 + C3
    = 3.3µF + 4.7µF
    = 3.3x10-6 + 4.7x10-6
    = 0.0000033 + 0.0000047
    = 8µF

    1
    1 + 1
    4µF 8µF

    = 2.66µF

    Charge stored by each capacitor

    C1 = 100 x 2.66µF
    C1 = 100 x 0.00000266
    C1 = 2.66x10-4
    C1 = 26.6mC

    C2 = 0.665 x 4 µF
    C2 = 0.665 x 0.000004
    C2 = 2.66x10-6
    C2 = 2.66µC

    C3 = 100 x 4.7µF
    C3 = 100 x 0.0000047
    C3 = 4.7x10-4
    C3 = 47mC

    Voltage across each capacitor

    V = Q / C

    C1 = 100V

    C2 = QT / C2
    C2 = 0.00000266 / 0.000004
    C2= 0.665
    C2 = 0.665V

    C3 = 100V

    Energy stored by each capacitor

    E1 = ½ x C1 x VT
    E1 = ½ x 3.3x10-6 x (100)2
    E1 = 1.65x10-2
    E1 = 16.5mJ

    E2 = ½ x C2 x V1
    E2 = ½ x 4µF x (0.665)2
    E2 = 8.8445x10-7
    E2 = 884.5nJ

    E3 = ½ x C3 x VT
    E3 = ½ x 4.7x10-6 x (100)2
    E3 = 2.35x10-2
    E3 = 23.5mJ

    Energy stored by whole circuit

    ET = E1 + E2 + E3
    ET = (1.65x10-2) + (8.8445x10-7) + (2.35x10-2)
    ET = 0.0165 + 0.00000088445 + 0.0235
    ET = 0.04000088445
    ET = 4.00x10-2
    ET = 40mJ
     

    Attached Files:

  2. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    When applying the basic capacitance relation C=Q/V just realize that for capacitors in parallel V is the same for all capacitors, and for capacitors in series Q is the same for all capacitors.
     
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