Question

What value resistor will discharge a 2.1 μF capacitor to 20 % of its initial charge in 2.8 ms ?

Answer 1

To Clo Ginen (-21MF = 2.14100F Let pretial chave on capacitor is to as shown in figure, and R be the resistance which discharge capacitor to 20% of pro initial nelve, in time 2.8m6 find Chaule on capacitor q = 20 20 = 20 R fine faken for this process t= 2.8 ms we know for discharging cisant chaye on capautor at any time of is given by q=2 e ERC. Accrocling to question. To - hoe RX218106 or 2 x 24 = 5 on this process t= 2.8ms =2.881035 2.8810

faking loge both sides, au, le 112.8x10311 7 - w R = 2184103 Our in leRx243) 1115) 2.8x103 -- ins = 828.45 ano 24 .: R=828.452

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 2.8 ms = 2.8^10^(- 3) sec.

R = Resistance  in Ωs  to be determined .

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

Hence,

0.20 =  e^(- 2.8*10^(-3) / (R * 2.10*10^(-6)) 

=> 0.20 = e^(- 1333.3333 / R)

Taking natural logarithm :

=> ln(0.2) = - 1333.3333 / R

=> R = - 1333.3333 / ln(0.2) 

=> R = 828.45 Ω   approx. 

=> Resistor of 828.45  Ω approx. (ANSWER).