Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you

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Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you

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Minh Khuê 6 months 2021-08-13T16:54:32+00:00 1 Answers 10 views 0

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    2021-08-13T16:55:34+00:00

    Answer:

     I = E/R   e^{-t/RC}

    Explanation:

    In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,

                      E -q / c-IR = 0

    we replace the current by its expression and divide by the resistance

                       I = dq / dt

                     

                       dq / dt = E / R  -q / RC

                       dq / dt = (CE -q) / RC

    we solve the equation

                       dq / (Ce-q) = -dt / RC

    we integrate and evaluate for the charge between 0 and q and for the time 0 and t

                       ln (q-CE / -CE) = -1 /RC   (t -0)

    eliminate the logarithm

                  q – CE = CE e^{-t/RC}

                   q = CE (1 + 1/RC  e^{-t/RC} )

    In general the teams measure the current therefore we take the derivative to find the current

                   i = CE (e^{-t/RC} / RC)

                   I = E/R   e^{-t/RC}

    This expression is the one that describes the charge of a condensate in a DC circuit

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