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Circuit-check and help

  1. Apr 24, 2008 #1

    ~christina~

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    Gold Member

    1. The problem statement, all variables and given/known data
    A photographer's electronic flash unit consists of a 150.0-kΩ resistor in series with a 22-µF capacitor and a 12.0 V source of emf. The flashbulb is placed in parallel with the capacitor so that when a sufficient charge is stored on the capacitor switch S2 can be closed and the capacitor quickly discharges through the small resistance of the bulb, causing the flash. Suppose switch S2 remains open and at t = 0 s switch Sl is closed.

    (a) At what time after the capacitor begins to charge has the current decreased to one-half its initial value?

    (b) What is the charge on the capacitor at this time?

    (c) What is the voltage across the capacitor at this time?

    (d) Sketch graphs of the current through this circuit and the charge on the capacitor as functions of time (two graphs

    [​IMG]

    2. Relevant equations
    [tex]\tau= RC [/tex]

    [tex]I_i= \frac{\epsilon} {R} [/tex]
    [tex]I(t)= \frac{\epsilon} {R} e^{\frac {-t} {RC}} [/tex]

    3. The attempt at a solution

    a) At what time after the capacitor begins to charge has the current decreased to one-half its initial value?

    [tex]\tau= RC= (22x10^{-6} F)(1.50x10^5 \omega )= 3.3s [/tex]

    max current: [tex]I_i= \frac{\epsilon} {R} = 12.0V/ 1.50x10^5 \omega= 8x10^{-5} A[/tex]

    current as a function of time: [tex]I(t)= \frac{\epsilon} {R} e^{\frac {-t} {RC}} [/tex]

    half of the total current: [tex]4x10^{-5}A [/tex]

    I think I would find the time this way but I'm not sure if it is right.
    time it takes for the current to half
    [tex]I(t)= (8x10^-5A)\frac e^{\frac {-t} {3.3s}} [/tex]
    [tex]0.5A= \frac e^{\frac {-t} {3.3s}} [/tex]
    [tex]ln 0.5A= \frac{-t} {3.3s}} [/tex]
    [tex]t= 2.287s[/tex]

    (b) What is the charge on the capacitor at this time?

    [tex]q(t)= C \epsilon (1-e^{-\frac{t} {RC}}) [/tex]
    [tex]C \epsilon = (22x10^{-6}F)(12.0V)= 2.64x10^{-4}C[/tex]
    [tex]q(t)= (2.64x10^{-4}C)(1-e^{-\frac{2.287s} {3.3s}}) [/tex]
    [tex]q(t)= (2.64 x10^ {-4}C)(0.499941)=1.319844x10^{-4}C [/tex]

    (c) What is the voltage across the capacitor at this time?

    Not sure how to find this..


    (d) Sketch graphs of the current through this circuit and the charge on the capacitor as functions of time (two graphs

    wouldn't this look the typical graph of

    charge vs time (exponential curve going up)=> |/|
    current vs time (exponential curve going down)=> |\|

    I'd appreciate it if someone could help me out with part c and also checking whether I did the other parts correctly.

    Thanks alot :smile:
     
  2. jcsd
  3. Apr 24, 2008 #2
    part a. the method and the result is Ok. 0.5 shouldn't have a u unit of amperes as this is a dimensionless number.

    part b. Ok. A shorter method would be to notice that since the current is now half the initial current, half of the voltage of the voltage source is the drop across the resitor.

    part c. you already gave q(t) in part b. Q = CV

    part d. you already computed I(t) and Q(t)
     
  4. Apr 24, 2008 #3

    ~christina~

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    Gold Member

    ok

    yep, didn't notice that untill you mentioned it.

    oh...so I'd just find the voltage by using the charge I found for the specific time.

    okay.


    Thanks for your help kamerling :smile:
     
  5. Apr 1, 2010 #4
    ya thanks
     
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