A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this cir

Question

A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this circuit

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Helga 1 year 2021-08-15T03:46:19+00:00 1 Answers 143 views 0

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    2021-08-15T03:47:49+00:00

    Answer:

    The  time constant is  [tex]\tau = 1.265 s[/tex]

    Explanation:

    From the question we are told that

         the time take to charge is  [tex]t = 2.4 \ s[/tex]

    The mathematically representation for voltage potential of a capacitor at different time is

            [tex]V = V_o – e^{-\frac{t}{\tau} }[/tex]

    Where  [tex]\tau[/tex]  is the time constant  

               [tex]V_o[/tex] is the potential of the capacitor when it is full

         So  the capacitor potential will be  100%  when it is full thus  [tex]V_o =[/tex]100%  =  1  

    and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

          V  = 0.85

    Hence

         [tex]0.85 = 1 – e^{-\frac{2.4}{\tau } }[/tex]

           [tex]- {\frac{2.4}{\tau } } = ln0.15[/tex]

            [tex]\tau = 1.265 s[/tex]

         

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