Question

A 11.5-µF capacitor is charged to a potential of 55.0 V and then discharged through a 155-Ω resistor.

(a) How long does it take the capacitor to lose half of its charge?
  ms

(b) How long does it take the capacitor to lose half of its stored energy?
  ms

Answer 1

given

C = 11.5 micro F
V = 55.0 V
R = 155 ohms

a) Time constant of the ckt, T = R*C

= 155*11.5*10^-6

= 1.78*10^-3 s or 1.78 ms

we know, potential diffrencer across capacitor,

VC = Vmax*e^(-t/T)

at time t=t, VC = Vmax/2

Vmax/2 = Vmax*e^(-t/T)

1/2 = e^(-t/T)

ln(1/2) = -t/T

== > t = -T*ln(1/2)

= -1.78*ln(1/2)

= 1.23 ms

b) when Vc becomes Vmax/sqrt(2) , energy stored becomes half.

so, VC = Vmax*e^(-t/T)

at time t=t, VC = Vmax/sqrt(2)

Vmax/sqrt(2) = Vmax*e^(-t/T)

1/sqrt(2) = e^(-t/T)

ln(1/sqrt(2)) = -t/T

== > t = -T*ln(1/sqrt(2))

= -1.78*ln(1/sqrt(2))

= 0.617 ms