Question

A 7.7-?F capacitor is connected in series with 3.4 M? resistor, and this combination is connected across an ideal 16-V DC battery. What is the current in the circuit when the capacitor has reached 29% of its maximum charge?

Answer 1

given
C = 7.7 micro F = 7.7*10^-6 F
R = 3.4 Mohm = 3.4*10^6 ohm

Time constant of the circuit, T = R*C

= 3.4*10^6*7.7*10^-6

= 26.18 s

let at time t, Q = 29% of Qmax

we know,

Q = Qmax*e^(-t/T)

0.29*Qmax = Qmax*(1 - e^(-t/T))

0.29 = 1 - e^(-t/T)

e^(-t/T) = 1 - 0.29

-t/T = ln(1 - 0.29)

t = - T*ln(1 - 0.29)

= -26.18*ln(1 - 0.29)

= 8.97 s

we know, Imax = V/R = 16/(3.4*10^6) = 4.71*10^-6 A

at time t = 8.97 s

current through the ckt,

I = Imax*e^(-t/T)

= 4.71*10^-6*e^(-8.97/26.18)

= 3.34*10^-6 A or 3.34 micro A <<<<<<<------------Answer