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**

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