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The voltage(V) volts in a circuit

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data
    The voltage(V) volts in a circuit at time(t) is given by

    V(t)=18(1 - e -4t/9)

    Find the expression for dV/dt.

    Hence evaluate dV/dt at

    (a)t=0 and (b)t=5 seconds


    2. Relevant equations

    Just wondering if this is right and if not where did i go wrong

    3. The attempt at a solution

    dV/dt=18(-4/9)(1-e-4/9t)

    = -9(1-e-4/9t)

    (a) t=0 -8(1-e-4/9(0))

    =0mA

    (b)t=5 -8(1-e-4/9(5))

    =-7.13mA
     
  2. jcsd
  3. Aug 21, 2010 #2
    [tex]\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9}[/tex] what's the derivative?
     
  4. Aug 21, 2010 #3
    not to sure.Im sorry but im not that good at this hoping to get better!
     
  5. Aug 22, 2010 #4
    Well

    [tex]\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9} [/tex]

    Whats the derivative of 18 with respect to t?

    What about 18exp(-4t/9) with respect to t?

    I seperated the parts of the equation using the fact that [tex]\frac{d}{dt}[v(t) + w(t)] = \frac{d}{dt}[v(t)] + \frac{d}{dt}[w(t)][/tex] you can generally remember this rule as "the derivative of the sum is the sum of the derivatives".

    Does this make sense?
     
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