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 5 months 2021-08-15T03:46:19+00:00 1 Answers 25 views 0

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

    Answer:

    The  time constant is  \tau  = 1.265 s

    Explanation:

    From the question we are told that

         the time take to charge is  t = 2.4 \  s

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

            V  =  V_o  - e^{-\frac{t}{\tau} }

    Where  \tau  is the time constant  

               V_o is the potential of the capacitor when it is full

         So  the capacitor potential will be  100%  when it is full thus  V_o  =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

         0.85 =  1 -  e^{-\frac{2.4}{\tau } }

           - {\frac{2.4}{\tau } }  =  ln0.15

            \tau  = 1.265 s

         

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