The flash unit in a camera uses a special circuit to “step up” the 3.0 V from the batteries to 290 V, which charges a capacitor. The capacit

Question

The flash unit in a camera uses a special circuit to “step up” the 3.0 V from the batteries to 290 V, which charges a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes 10 μs, and the average power dissipated in the flashlamp is 1.0×105 W. What is the capacitance of the capacitator?

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Mộc Miên 4 years 2021-08-26T20:37:34+00:00 2 Answers 29 views 0

Answers ( )

  1. Neala
    0
    2021-08-26T20:38:52+00:00
    Reply

    Answer:

    Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F

    Explanation:

    We apply the concepts related to Power and energy stored in a capacitor.

    By definition we know that power is represented as

    P = \frac{E}{t}

    Where,

    E= Energy

    t = time

    to find the Energy we have,

    E = P*t

    P = 1*10^5Wt = 10*10^{-6}s

    E= (1*10^5)(10*10^{-6})E = 1.0J

    With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is

    E=\frac{1}{2}CV^2

    C = \frac{2E}{V^2}\\\\\\\frac{2\times 1 }{290^2 - 3^2\\} \\= 2.38 \times 10^-^5F

    Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F

  2. Cherry
    0
    2021-08-26T20:39:13+00:00
    Reply

    Answer:

    24 μF.

    Explanation:

    Given:

    Time, t = 10 μs

    P = 1.0×105 W

    V1 = 3 V

    V2 = 290 V

    Power, P = Energy/time

    Energy = 1/2 × C × V^2

    1 × 10-5 × 1 × 10^-5 = 1/2 × C × (290^2 – 3^2)

    C = 2/84091

    = 2.38 × 10^-5 F

    = 24 μF

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