Assuming that the capacitor is initially uncharged, how long will it take for th

Question

Assuming that the capacitor is initially uncharged, how long will it take for the capacitor to charge to half the maximum charge? (Give your answer in the form of "ab.c s".)

or Figure 4, E = 35 kV , R = 3.5 k, and C = 6800.0 F. Determine the time constant for the circuit. (Give your answer in the form of "ab s".)

Assume that the capacitor has been fully charged.   It is then disconnected from the circuit and a 470 ohm resistor connected in parallel with it to discharge it. How long will it take to be 99% discharged?   Give your answer in the form "ab.c" s.

Explanation / Answer

Capacitor Is initiaily un charged,

so, there will be no charging voltage in it ,

Vc will be zero at initial time.

RC Time Constant, Tau FOr the circuit will be :

as per below:

The time constant, T is found using the formula T = R x C in seconds.

T = R x C=   3.5 k X 6800.0 F   = 23.8 s

RC Charging Table is as below,

now for half of full charge , means 50 %

it will Take 0.7 T     ( from table )

0.7 T = 0.7 x 23.8 ( from above result) =   16.66 s

Now , Assume that the capacitor has been fully charged.   It is then disconnected from the circuit and a 470 ohm resistor connected in parallel with it to discharge it. How long will it take to be 99% discharged?

Now RC Discharging Table is as below:

For 99% discharge time taken will be

5T ,   but we dont have T for new connected circuit,

T = RC = 6800 F  x 470 ohm = 6800F x 0.470 kohm =   3.196 sec

Time
Constant RC Value Percentage of Maximum Voltage Current 0.5 time constant 0.5T = 0.5RC 39.3% 60.7% 0.7 time constant 0.7T = 0.7RC 50.3% 49.7% 1.0 time constant 1T = 1RC 63.2% 36.8% 2.0 time constants 2T = 2RC 86.5% 13.5% 3.0 time constants 3T = 3RC 95.0% 5.0% 4.0 time constants 4T = 4RC 98.2% 1.8% 5.0 time constants 5T = 5RC 99.3% 0.7%

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