Question

What value resistor will discharge a 2.1 μF capacitor to 20 % of its initial charge in 2.4 ms ?

Answer 1

v = v₀e^(–t/RC)

0.2 = e^(–(2.4e-3)/R(2.1e-6))

solve for R

-1.609 = –(2.4e-3)/R(2.1e-6)
R = (2.4e-3)/(1.609)(2.1e-6)
R = 710 ohms

Answer 2

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 2.4 ms = 2.4^10^(- 3) sec.

R = Resistance  in Ωs  to be determined .

C = Capacitor capacitance = 2.10 µF= 2.10*10^(-6) F .

Hence,

0.20 =  e^(- 2.4*10^(-3) / (R * 2.10*10^(-6)) 

=> 0.20 = e^(- 1250 / R)

Taking natural logarithm :

=> ln(0.2) = - 1142.86 / R

=> R = - 1142.86 / ln(0.2) 

=> R = 710.10 Ω   approx. 

=> Resistor of 710.10  Ω approx. (ANSWER).