Question

The capacitor in an RC circuit is discharged with a time constant of 10.0ms. a) At what time after the discharge begins is the charge on the capacitor reduced to half its initial value? b) At what time after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

Answer 1

(a) we know that for discharging , the equation is

v(t) = v*exp(-t/a) , a= time constant.

given v(t)= 0.5 *v

=> 0.5= exp(-t/10)

=>t=6.9315 ms

(b) E= 0.5*c*v^2

=> E1/E2= (V1/V2)^2

given , E2= 0.5*E1

=> 2= (V1/V2)^2

=>sqrt(2)= V/V(t)

=>V(t)= V/sqrt(2)

but V(t)= V*exp(-t/10)

=> V*exp(-t/10) = V/sqrt(2)

=>t= 3.4657 ms

Answer 2

The time after the discharge when charge is reduced to half is given by, -| | = ln (0.5) t-(-0.693) τ = (0.693)(10 ms - 6.93 ms Therefore, the time after the discharge when charge is reduced to half is 6.93 ms The time after the discharge when energy is reduced to half is given by, 2 C 2 2C 2t t =-(-0.693) ( ,) 10 ms = (0.693) | = 3.46 ms Therefore, the time after the discharge when energy is reduced to half is 3.46 ms

Answer 3

At what time after the discharge begins is the charge on the capacitor reduced to half its inial value?

Q= Qo *e^-(t/(RC))

substitute 1/2 of initial value of Qo for Q
and the Qo on both sides of equation cancel

1/2 Qo= Qo*e^-(t/10)
1/2 = e^-(t/10)
ln 1/2 = -t/10
-10 * ln 1/2 = t

t = 6.93 ms

At what time after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

sqrt1/2 =e^-(t/10)
.7071 = e^-(t/10)
-10 * ln .7071 = t
t=3.47 ms