Question

A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the capacitor takes to lose
(a) the first 1/8-th of its charge
(b)7/8-th of its charge?

Answer 1

Answer: a) 0.13*τ ; b) 2.08*τ

Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:

Q(t)=Qo* exp(-t/τ) for a lose of 1/8-th of its charge

in this case, Q(t)=7/8*Qo=7/8*exp(-t/τ)

ln(7/8)*τ=-t

then, t= -ln(7/8)*τ =0.13

For a lose of 7/8 th  of its charge , we have

Q(t)=1/7*Qo*exp(-t/τ)

t=-ln(1/8)*τ=2.08

Answer 2

In this exercise we will use the knowledge of electromagnetism to calculate the value of the charges of a resistor, so we have that:

a) 0.13*τ

b) 2.08*τ

Recalling the formula that must be used to discharge a capacitor through a resistor, we find that:

$Q(t)=Q_o* exp(-t/T)$

Where,

$Q(t)=7/8*Q_o=7/8*exp(-t/T)$

So for this case we will have that the calculations will be:

A) For 7/8 we have:

$ln(7/8)*T=-t\\t= -ln(7/8)*T =0.13$

B) For 1/8 we have:

$Q(t)=1/7*Qo*exp(-t/T)\\t=-ln(1/8)*T=2.08$

See more about electromagnetism at brainly.com/question/13803241