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 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?

Answer 1

Answer:

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