Question

A 2.1-mF capacitor is discharged through a 4.0-kΩ resistor. How long will it take for the capacitor to lose half its initial stored energy?
(A) 5.82 s
(B) 2.91 s
(C) 0.660 s
(D) 4.20 s
(E) 8.40 s

Answer 1

Answer:

the answer is a)5.82s

Explanation:

Hello!

The first thing we must understand to solve this exercise is that a capacitor has the following characteristic function

$V=(Vo)e^{-t/RC}$

where

Vo=initial voltage

V=voltage in an instant of time

t=time

R=resistance=4000Ω

C=capacitance=2.1-mF=0.0021F

we must consider that the final voltage is half of the initial voltage so we deduce the following equation

V=Vo/2

Now we replace in the initial equation, and start an algebraic process to find t

$\frac{Vo}{2} =(Vo)e^{-t/RC}\\0.5=e^{-t/RC}\\ln(0.5)=ln(e^{-t/RC})\\ln(0.5)=\frac{-t}{RC} \\t=-(RC)ln0.5$

finally with the equation ready, we use the resistance and capacitance values ​​to find the time value

$t=-(4000)(0.0021)ln(0.5)=5.82s$

the answer is a)5.82s