Question

Part A

What value resistor will discharge a 2.60 μF capacitor to 20.0 % of its initial charge in 2.50 ms?

Answer 1

Answer 2

SOLUTION :

We use the following formula :

Q = Q0 e^(- t / R C) 

=> Q / Q0 = e^ ( - t / RC)

where, 

Q / Q0 = 0.20

t = time taken in secs for discharge to 0.20 Q0 = 2.50 ms = 2.50^10^(- 3) sec.

R = Resistance  in Ωs  to be determined .

C = Capacitor capacitance = 2.60 µF= 2.60*10^(-6) F .

Hence,

0.20 =  e^(- 2.50*10^(-3) / (R * 2.60*10^(-6)) 

=> 0.20 = e^(- 961.53846 / R)

Taking natural logarithm :

=> ln(0.2) = - 961.53846 / R

=> R = - 961.53846 / ln(0.2) 

=> R = 597.44 Ω   approx. 

=> Resistor of 597.44  Ω approx. (ANSWER).