Ask Question
18 April, 01:32

Suppose that the voltage V (t) of electricity at time t (in seconds) is draining from a capacitor at a rate that is proportional to its value. That is, V (t) satisfies the differential equation where k > 0 is the constant of proportionality. If k = 1/20, how long will it take the voltage to drop to 10 percent of its original value?

+3
Answers (1)
  1. 18 April, 01:46
    0
    it will take t = 12.64 seconds

    Step-by-step explanation:

    the voltage drop rate dV/dt will be

    dV/dt=-k*V

    then

    ∫dV/V=∫-kdt

    ln (V/V₀) = - k*t

    V=V₀*e^ (-k*t)

    in order for the voltage to drop 10 percent of its original value, then

    (V₀-V) / V₀ = 0.1 (10%)

    [V₀ - V₀*e^ (-k*t) ] / V₀ = 0.1

    1 - e^ (-k*t) = 0.1

    e^ (-k*t) = 0.9

    t = ( - ln 0.9) / k = 120 * ( - ln 0.9) = 12.64 seconds

    t = 12.64 seconds
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Suppose that the voltage V (t) of electricity at time t (in seconds) is draining from a capacitor at a rate that is proportional to its ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers