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Find an Expression For the Voltage

  1. Jan 29, 2010 #1
    Hey, so I have an assignment and this is one of the questions on it, I did it but I am not sure if what I did is right.

    What I did was find Ceq=(C1C2)/(C1+C2))

    So then Q=CV(t)=(C1C2)/(C1+C2))V(t)

    Then I found the voltage across C2=Q/C2=(C1C2)/(C1+C2))V(t)/C2=(C1)/(C1+C2))V(t)

    And then V0=V(t)-(C1)/(C1+C2))V(t)

    Can anyone tell me if what I did is right or if I am completely off?
  2. jcsd
  3. Jan 29, 2010 #2
    For starters, shouldn't it be Q = CVT = (C1C2VT)/(C1+C2)?
  4. Jan 29, 2010 #3
    You could have just found the voltage across [tex]c_1[/tex] like you did for [tex]c_2[/tex] and stopped there.: [tex]\frac{q}{c_1}=v_o=v(t) \frac{c_2}{c_1+c_2}[/tex]
    Your expression you found simplifies to that anyway:
    [tex]v_o=v(t)[1 - \frac{c_1}{c_1+c_2}][/tex]
    [tex]v(t)[\frac{c_1+c_2}{c_1+c_2} - \frac{c_1}{c_1+c_2}]= v(t) \frac{c_1+c_2 - c_1}{c_1+c_2}=v(t) \frac{c_2}{c_1+c_2}[/tex]
    I don't know if you know about complex impedance yet but you can also use that to arrive to the same answer:
    [tex]Z_T = Z_C_1 + Z_C_2[/tex]
    [tex]Z_T = \frac{1}{\omega c_1}+\frac{1}{\omega c_2}[/tex]
    These impedances work like resistance with v = IR. Therefore,voltage division works too.
    [tex]v_o = v(t) \frac{ Z_C_1}{Z_T}[/tex]
    [tex]v_o = v(t) \frac{ \frac{1}{\omega c_1}}{\frac{1}{\omega c_1}+\frac{1}{\omega c_2}}[/tex]
    Omegas cancel and you come to the same answer.
    Last edited: Jan 29, 2010
  5. Jan 31, 2010 #4


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    Just for the record: no, Q=CV for a capacitor, as the OP said.
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