Consider the fallowing circuit After the switch is dosed how

Consider the fallowing circuit: After the switch is dosed, how long does it take lev the capacitor to reach ft of its final charge? After the switch has been dosed for a long time, the switch is then opened. How long does it take for the capacitor to discharge to V* of its fully charged value?

Solution

a)

equivalent resistance

1/R_eq =1/122 +1/88

R_eq = 51.12 ohms

Time Constant

T=R_eqC=51.12*550*10^-6 =0.028 s

For a RC circuit ,charge as a function of time for charging is given by

Q=Q_max[1-e^-t/T]

(1/2)Q_max =Q_max[1-e^-t/0.028]

-t/0.028 =ln(1/2) =-0.693

t=0.0194 s or 19.4 ms

b)

TIme Constant

T=(R₁+R₂)C=(122+88)*550*10^-6

T=0.11 s

For a RC circuit ,charge as a function of time for discharging is given by

Q=Q_max[e^-t/T]

(1/2)Q_max =Q_max[e^-t/0.028]

-t/0.11=ln(1/2) =-0.693

t=0.0762 s or 76.2 ms