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March 28, 2017

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A capacitor in a single-loop RC circuit is charged to 64% of its final voltage in 1.5 s.
Find the time constant for the circuit.

Find the percentage of the circuit's final voltage after 3.9 s .
Vc/Vi= %

  • College physics - ,

    The voltage across the capacitor is
    Vi * [1 - e^-(t/RC)]
    in this type of circuit. Vi is the power supply voltage. RC is the time constant.

    The final voltage is Vi.
    When t = RC (the time constant), the voltage is
    Vc = Vi*[1 - e^-1] = 0.632 Vi

    When t = 1.5 s,
    0.64 Vi = Vs * [1 - e^(-t/RC)]
    e^(-t/RC)= 0.36
    t/RC = 1.02
    RC = 1.5/1.02 = 1.47 seconds
    That is the time constant.

    After 3.9 s

    Vc = Vi* [1 - e^(-3.9/1.47)]
    = 0.930 Vi

    Unless you show some work on future posts here, I will not be assisting you again. Perhaps someone else will.

    Vc = Vi* [1 - e^(-3.9/1.47)]
    = 0.930 Vi

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