charging capacitor's capacitance

Discussion in 'Homework Help' started by lemon, May 24, 2010.

  1. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    Hi:
    I'm revising for an exam coming this Wednesday and have this question in my revision notes I cannot answer.

    A circuit consists of an uncharged capacitor connected in series with a 15kΩ resistor and has a time constant of 30 ms. Determine
    a) the capacitance of the capacitor and
    b) the voltage drop across the capacitor
    20 ms after connecting the circuit to a 100 V d.c. supply.

    I can find a voltage of 51.3V after 20 ms. And can find the initial current of 6.67mA. But am not sure how to find the capacitance after this time.
    I know how to find the capacitance of capacitor in series, but there is only one capacitor here.

    Should I use C=Q/V? Where the voltage is 51.3V? But then what would Q be.

    any help appreciated
    thank you
     
    Last edited: May 24, 2010
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    time constant = R * C = 30ms = 15K * C
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well you could uses this equation to.
    C = Q/V = I*t / V = (6.67mA * 30ms)/ 100V = 2uF
     
  4. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    ahh! I think I should have gotten that one.
    Thank you Jony130
     
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