Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 69)
69.
If a capacitor is placed across a 20 V source, what will be the amount of charge after 5 tc?
Discussion:
7 comments Page 1 of 1.
KIRAN V said:
5 years ago
Given: Time constant (T / Æ®) = (T / RC) = 5 TC.
(Æ®) = RC.
VC = [VIN (1- e^-(T/(R*C))].
VC = [20 * (1- e^-(T/ Æ®)].
VC = [20 * (1- e^-(5))].
VC = [20 * (1- e^-(5))].
[Where: e^1 = 2.7182; e^-5 = (1/e5) = [1 / (2.7182)^5] = [1 / 148.41]= 0.0067].
VC = [20 * (1- 0.0067)].
VC = [20 * (0.9933)] = 19.866 V.
VC = 19.866 V => 20 V.
(Æ®) = RC.
VC = [VIN (1- e^-(T/(R*C))].
VC = [20 * (1- e^-(T/ Æ®)].
VC = [20 * (1- e^-(5))].
VC = [20 * (1- e^-(5))].
[Where: e^1 = 2.7182; e^-5 = (1/e5) = [1 / (2.7182)^5] = [1 / 148.41]= 0.0067].
VC = [20 * (1- 0.0067)].
VC = [20 * (0.9933)] = 19.866 V.
VC = 19.866 V => 20 V.
Pallavi said:
8 years ago
Here, rc is time constant, τ time constant is 5t. So the (t/rc) is (t/5t)
e^(-(1/5) then solve you get the answer.
e^(-(1/5) then solve you get the answer.
Mahi said:
9 years ago
(1-e^(-t/rc)) what are the values of this equation?
Prasad said:
1 decade ago
Why this rc values?
G B said:
1 decade ago
Voltage = 20 v.
So, the amount of charge after 5tc= Vin * (1-e^(-t/rc))
= 20*0.993
=19.86
And its app. = 20 v.
So, the amount of charge after 5tc= Vin * (1-e^(-t/rc))
= 20*0.993
=19.86
And its app. = 20 v.
(1)
Hanish said:
1 decade ago
Sry now I undrstood. Only after 5tc capacitor is fully charged so it attains 20v.
Rama said:
1 decade ago
Please send me the solution of the problem.
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