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Voltage and capactiors

  1. Jul 4, 2006 #1
    I don't understand what this problems means....do I need some more equations?

    The voltage in a circuit is given in SI units by V = 80sin200t. What is the voltage at t = (1/60) s?
    A. 0V
    B. 15V
    C. 80V
    D. -15V

    Can you even have a negative voltage? I guess I just don't understand the complexities of this problem.

    Also guys....Do you guys know what the equations needed to help find charge and voltage through capacitors in series and parallel??
  2. jcsd
  3. Jul 4, 2006 #2


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    Just plug in the value of t in the equation (I am guessing that the argument of the sine is in radians here)
    Sure it can be negative (when one talks about voltage in a circuit, one always means a *difference* of voltage between two points, V_a- V_b. depending on whether the voltage is higher at point a than at point b or vice versa, the difference of voltage will be positive or negative.
    [tex] Q = C (\Delta V) [/tex]
    where deta V is the difference of potential between the two plates and Q is the charge on the positively charged plate (so the two plates have +q and -Q).

    For two capacitors in series
    [tex] {1 \over C_{eq}} ={1 \over C_1} + { 1\over C_2} [/tex]
    For two capacitors in parallel,
    [tex] C_{eq} = C_1 + C_2 [/tex]

    And the rule for voltage is the same as for resistors. The voltage across two capacitors in parallel is the same and the volate across two capacitors in series is the sum of the voltages across each capacitor.

    Hope this helps

    Last edited: Jul 4, 2006
  4. Jul 4, 2006 #3
    ha.....sorry Patrick didn't mean to waste you time with the first one....for some reason I didn't come up with any of those answers(calculator might have been set wrong).....so yeah that was as easy as I thought. Thanks for those equations though!!!!
  5. Jul 4, 2006 #4


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    No problem!

    It *is* important to make sure that the calculator is in the correct mode (radian vs degrees) though! I have seen a lot of students making mistakes because of this. (and if one of the multiple choices correspond to the answer with th ecalculator in degrees, it's easy to fall in the trap!)

    best luck

  6. Jul 4, 2006 #5
    uhm, you got the capcitors in series wrong... its [tex]\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}[/tex]
  7. Jul 4, 2006 #6


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    Absolutely. Sorry for the mistake:cry: .
    I edited my post. Thanks for the correction.
    I typed the series one and then cut and pasted and put the C_eq as a denominator but forgot to change the C_1 and C_2. Sorry!

    Thanks again

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