Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 35)
35.
How long would it take the capacitor in the given circuit to completely discharge if the switch was in position 2?
Discussion:
18 comments Page 1 of 2.
Ashok said:
1 decade ago
Can anyone explain this..?
Vid said:
1 decade ago
When switch is at 1,capacitor stores max value of voltage. When switches to 2
time constsnt=RC
=(6.2+2.2)*1000*10*10^-6
=84 ms
to completely discharge 5 time constants
=5RC
=84*5=420 ms
time constsnt=RC
=(6.2+2.2)*1000*10*10^-6
=84 ms
to completely discharge 5 time constants
=5RC
=84*5=420 ms
Nithya said:
1 decade ago
@Vid your explanation is correct.
Kazim said:
1 decade ago
Time constant =RC
=(6.2+2.2)*1000*10^-6
=84 ms.
To completely discharge 5 time constants
=5RC
=84*5=420 ms.
=(6.2+2.2)*1000*10^-6
=84 ms.
To completely discharge 5 time constants
=5RC
=84*5=420 ms.
Cornelius said:
1 decade ago
Why 5 time constants?
Divu said:
1 decade ago
To discharge completely capacitor requires 5RC time.
Sadhana said:
1 decade ago
Why it requires 5 RC time to discharge? Please explain.
AJAY REDDY P said:
1 decade ago
Can you explain about 5RC time discharge?
Abhiee said:
1 decade ago
Vc = vin*(1-e^(-t/rc)).
If we substitute t=1 rc then output of capacitor is 63.7%. Similarly if we substitute t=5*r*c in the equation. Then the output of the capacitor is 99.8%. i.e. it is fully charged.
If we substitute t=1 rc then output of capacitor is 63.7%. Similarly if we substitute t=5*r*c in the equation. Then the output of the capacitor is 99.8%. i.e. it is fully charged.
KIRAN V said:
9 years ago
Time constant = RC.
= (6.2 + 2.2) * 1000 * 10^-6.
= 8.4 ms.
To completely discharge 5 time constants.
= 5RC.
= 8.4 * 5 = 42 ms.
= (6.2 + 2.2) * 1000 * 10^-6.
= 8.4 ms.
To completely discharge 5 time constants.
= 5RC.
= 8.4 * 5 = 42 ms.
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