Question

A 30.0 μF capacitor initially charged to 30.0 μC is discharged through a 2.50 kΩ resistor. How long does it take to reduce the capacitor's charge to 15.0 μC?

Answer 1

SOLUTION :

We uset the following formula :

Q = Q₀ e^ ( t /R C) 

where, 

Q = charge at time t secs. = 15 (µC)

Q0 = Initial charge at t₀ .= 30  (µC) 

t = time taken in secs for discharge from Q₀ to Q.

R = Resistance = 2.5*10^3 Ω

C = Capacitor capacitance = 30 µF.

Hence,

15 = 30 e^(- t/(2.5*10^3 * 30 * 10^(-6))

=> 15 / 30 = e^(- t/0.075) = e^(-13,3333 t)

Taking natural logarithm :

=> ln(0.5) = - 13.3333 t

=> t = ln(0.5) / (-13.3333) 

=> t = 0.051986 secs 

=> t = 51.986 m-sec (ANSWER).

Answer 2

0.052 s

e Q5X103x30x106 Capacitance of capacitar c= 30F Initial charge do = 3o4c Resistance R - 2.5x1632 - Q = 90 e - ERC - 154c = 3opec e 10.5= ė- 0.075 en o.s = -t t = 0.052 s or 51.986 ms (exact) t