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?

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