Given that the voltage across the capacitor as a function of time is V(t)=q0Ce−t/(ROC)V(t)=q0Ce−t/(RC), what is the current I(t) flowing thr

Question

Given that the voltage across the capacitor as a function of time is V(t)=q0Ce−t/(ROC)V(t)=q0Ce−t/(RC), what is the current I(t) flowing through the resistor as a function of time (for t>0t>0)? It might be helpful to look again at Part A of this problem. Express your answer in terms of ttt and any quantities given in the problem introduction.

in progress 0
Mộc Miên 3 years 2021-09-05T01:23:50+00:00 1 Answers 294 views 0

Answers ( )

  1. lethu
    0
    2021-09-05T01:24:55+00:00
    Reply

    Answer:

    i(t) = -\frac{q_0C}{R}e^{-t/RC}

    Explanation:

    If the voltage across a capacitor is given by

    V(t) = q_0Ce^{-t/RC}

    then by definition of the capacitance (C = Q/V), the charge on the capacitor is

    q(t) = CV(t) = q_0C^2e^{-t/RC}

    Now we can relate the charge on the capacitor to the current through the capacitor by

    i(t) = \frac{dq(t)}{dt} = -\frac{q_0C^2}{RC}e^{-t/RC}

    Since this is a closed current with capacitor and resistor are connected in series, the current through the capacitor is equal to that of resistor.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )