Available Fro Due Date Late submiss SaplingLarning Map You charge an initially u

Question

Available Fro Due Date Late submiss SaplingLarning Map You charge an initially uncharged 89.9-mF capacitor through a 42.3-0 resistor by means of a 9.00-v batev haeing nealicible internal resistance. Hind the time constant of the circuit. What is the charge of the capacitor 1.45 time constants after the circuit is closed? What is the charge after a long time? Points Possib Grade Catego Description: Time constant: Number You can chec You can iewW assignment You have thre There is no p Charge Number eTextbook O Help With T Charge after long time: O Web Help & Number O Technical Sup

Explanation / Answer

A.

Time constant is given by:

tau = R*C

tau = 42.3*89.9*10^-3

Time constant = 3.803 sec

B.

In RC circuit voltage after t time is given by:

Vc = V0*(1 - exp(-t/tau))

t = 1.45*tau

V0 = 9.00 V

Vc = 9*(1 - exp(-1.45)) = 6.89 V

At that time charge will be

Qc = C*Vc

Qc = 89.9*10^-3*6.89

Qc = 0.619 C

C.

After long time

Vc = V0 = 9.00 V

Qmax = C*Vmax

Qmax = 89.9*10^-3*9

Qmax = 0.809 C

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