Electronic Devices - Diode Applications - Discussion

Discussion Forum : Diode Applications - General Questions (Q.No. 1)
1.
Determine the total discharge time for the capacitor in a clamper having C = 0.01 F and R = 500 k.
5 ms
25 ms
2.5 ms
50 ms
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 2 of 3.

Vijay Shinde said:   1 decade ago
The discharge voltage for the capacitor is given by:
VC (t) = Vo exp(-t/RC) for t > or = 0

Here, t exp(-t/RC)
t=1*RC 0.37
t=2*RC 0.14
t=3*RC 0.05
t=4*RC 0.02
t=5*RC 0.01
So that practically we say that after 5 time constant capacitor is fully charged/Discharged.

Subhashis sharma said:   1 decade ago
What is the meaning of 98% in 5secs? how it can b derived?/.

Jumshed UET Lahore said:   1 decade ago
Actually capacitor discharges 98% in five time constant. You can verify using formula of capacitor discharge.

Vikram said:   1 decade ago
Time constant is the time required by Capacitor to get charge to or discharge to 63.2% of total voltage applied across it, so in one time constant it will be 0.63 of final value, in second it will be 0.63 of remaining aiming voltage. In this way capacitor will approximately take 5 time constants to get discharge.

Akila said:   1 decade ago
I want clear explanation why we said that 5 time constant.

Santosh aladakatti said:   1 decade ago
Discharging time t=5rc.

So ans=25ms.
(1)

Gangadhar reddy said:   1 decade ago
Yes, the answer is correct because we take 5 time constants in practical so the formula is 5*R*C.
(1)

Viswa said:   1 decade ago
Could any one explain How do we know capacitor charges and discharges in 5 time const ?

LOKESH said:   1 decade ago
Actually it takes infinite time since it is exponential process. But for practical purpose we take it as 5*time constant i.e, 5*RC

Annanya said:   1 decade ago
How do we know capacitor charges and discharges in 5 time const ?


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