Electrical Engineering - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 4)
4.
In Question 6, the capacitor will reach full charge in a time equal to approximately
Discussion:
15 comments Page 2 of 2.
Yagnesh said:
1 decade ago
In @Priyanka's explanation she wrote current it must be replace by voltage.
Like 1 Ts 63.2% of max voltage.
Mathematical constant e, specifically 1-e^{-1}, more specifically as voltage to charge the capacitor versus time.
Charging V(t) = V0 (1-e^{-t/tau}).
Tau = RC.
Like 1 Ts 63.2% of max voltage.
Mathematical constant e, specifically 1-e^{-1}, more specifically as voltage to charge the capacitor versus time.
Charging V(t) = V0 (1-e^{-t/tau}).
Tau = RC.
Birendra Das said:
1 decade ago
V (Capacitor) = V(1-e^[-t/T]).
Put [t=T].
V(c) = V(1-e^[-1]).
As we know e^[-1] = 0.37 above answer will be 0.63.
Similarly put [t = 5T].
V(c) = V(1-e^[-5T/T]).
V(c) = V(1-e^[-5]).
V(c) = V(0. 9932) which is nearly equal to 100% it means at 5T we are getting the capacitor fully charged.
Put [t=T].
V(c) = V(1-e^[-1]).
As we know e^[-1] = 0.37 above answer will be 0.63.
Similarly put [t = 5T].
V(c) = V(1-e^[-5T/T]).
V(c) = V(1-e^[-5]).
V(c) = V(0. 9932) which is nearly equal to 100% it means at 5T we are getting the capacitor fully charged.
Mak said:
9 years ago
Hi, @Birendra Das.
How about it is in 6RC, 99.75 percent is much closer to 100 rather than 99.32 percent?
How about it is in 6RC, 99.75 percent is much closer to 100 rather than 99.32 percent?
Saurav said:
9 years ago
I don't understand the question clearly. Can anyone explain me?
(1)
Sindu said:
9 years ago
@Priyanka you have explained clearly.
(3)
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