Question

If you want to discharge a 8.00 μF capacitor to 55.0% of its initial charge in 9.22 ms, what value resistor will you need?

Question 2 If you want to discharge a 8.00 μ F capacitor to 55.0% of its initial charge in 9.22 ms, what value resistor will you need? Answer:

Answer 1

c = 8 times 10^-6 F t = 9.2 times 10^- 3 q (t) = Q _0 e^- t /RC rightarrow 0.55 = e^- t /RC rightarrow ln (0.55) = -t /RC rightarrow R = - t/c ln (0.55) = -9.22 times 10^- 3 /8 times 10^- 6 ln (0.55) = 1.92778 times 1000 = 1927.78 ohm R = 1927 .8 ohm

Answer 2

SOLUTION :

We use the following formula :

Q = Q0 e^(- t / R C) 

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

where, 

Q / Q0 = 0.55 (capacitor is discharged to 55% of initial value)

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

R = Resistance  in Ωs  to be determined .

C = Capacitor capacitance = 8.0 µF= 3.0 *10^(-6) F .

Hence,

0.55 =  e^(- 9.22*10^(-3) / (R * 8.0 *10^(-6)) 

=> 0.55 = e^(- 1152.5 / R)

Taking natural logarithm :

=> ln(0.55) = - 1152.5 / R

=> R = - 1152.5 / ln(0.55) 

=> R = 1927.78   approx. 

=> Resistor of 1927.78  Ω approx. (ANSWER).