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Climaticsman
Lv 4
Climaticsman asked in Science & MathematicsMathematics · 10 years ago

Help with differentiation?

The formula for instantaneous voltage during a capacitor charge is v = 5(1 - e^ - (0.006/0.0047)) but I am getting the wrong answer. Not homework; genuine math problem. Any clues, please, including how you did it?

Update:

The variable t is 0.006s, or 6ms

Update 2:

I get the same answer as Joseph (thanks) but where's the differentiation in it -that's simple algebra

Update 3:

C is 100nF, R is 47kohm, thus T is 4.7ms

3 Answers

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  • Anonymous
    10 years ago
    Favorite Answer

    If you want the gradient at this point, you have to differentiate in terms of 't'.

    v = Vo(1 - e^ - (t/RC)), everything except t (assuming the set up is consistent) is exactly the same.

    dv/dt=RCVoe^ - (t/RC)

    At 6ms, with pd of 5V and time constant of 0.0047

    dv/dt=0.0047*5*e^(-0.006/0.0047)=6.56x10^-3 V/s

  • ?
    Lv 7
    10 years ago

    v = 5(1 - e^ - (0.006/0.0047))

    There is no variable t on the right hand side

  • Anonymous
    10 years ago

    RC=0.0047 if the units and value are correct, then the equation is set up right, are you sure its 47 microseconds?

    Vi = Vfinal * (1 -e^(-t/RC))

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