Question

What value resistor will discharge a 2.6?F capacitor to 20% of its initial charge in 2.9ms ?

PLEASE ONLY SERIOUS INQUIRIES ONLY THIS IS MY GRADE AND I WANT TO LEARN HOW TO DO IT RIGHT STEP BY STEP

Express your answer using two significant figures.

R =

?

Answer 1

while discharging in RC circuit

we have relation for charge as

Q = Q0(e^-t/RC)

e^-t/RC = 20/100 = 0.2

-t./RC = ln (0.2) = -1.609

RC = 0.0029/1.609

RC = 0.0018023

R = 0.0018023/2.6*10^-6

R   =693.19 ohms

Answer 2

I = I0 e

Answer 3

SOLUTION :

We use the following formula :

Q = Q0 e^(- t / R C) 

=> Q / Q0 = e^ ( - t / RC)

where, 

Q / Q0 = 0.20 (capacitor is discharged to 20% of initial value)

t = time taken for discharge to 0.20 Q0  is 2.9 ms = 9.22^10^(- 3) sec.

R = Resistance  in Ωs  to be determined .

C = Capacitor capacitance = 2.6 µF= 2.6 *10^(-6) F .

Hence,

0.20 =  e^(- 2.9*10^(-3) / (R * 2.6 *10^(-6)) 

=> 0.20 = e^(- 1115.3846/ R)

Taking natural logarithm :

=> ln(0.20) = - 1115.3846 / R

=> R = - 1115.3846 / ln(0.20) 

=> R = 693.03 = 693.03 Ω   approx. 

=> Resistor of 693.03  Ω approx. (ANSWER).