Electronics - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - Filling the Blanks (Q.No. 1)
1.
The output voltage in the given circuit is _____ at the end of the second pulse.
Discussion:
3 comments Page 1 of 1.
Divyanshu Shekhar said:
9 years ago
RC = 50 * 10^3 * 56 * 10^-9 = 2800 * 10^-6 s =2.8 ms.
For first pulse, Vc = Vo[1-e^(-τ/RC)] = 5(1-e^-2.8 ms / 2.8 ms) = 5(1-0.367) = 5 * 0.933 = 4.665 V.
Between first and second pulse, Vc = [Vi * e^(-τ/RC)] = 4.665 * (e^-5.6 ms / 2.8 ms) = 4.665 * 0.135 = 0.629 V = 0.63 V(approx).
For second pulse, Vc = [Vf + (Vi - Vf) * e^(-τ/RC)] = 5 + (0.63-5)e^-2.8ms /2.8ms = 5 +(-4.37) * 0.367 = 3.39 V.
For first pulse, Vc = Vo[1-e^(-τ/RC)] = 5(1-e^-2.8 ms / 2.8 ms) = 5(1-0.367) = 5 * 0.933 = 4.665 V.
Between first and second pulse, Vc = [Vi * e^(-τ/RC)] = 4.665 * (e^-5.6 ms / 2.8 ms) = 4.665 * 0.135 = 0.629 V = 0.63 V(approx).
For second pulse, Vc = [Vf + (Vi - Vf) * e^(-τ/RC)] = 5 + (0.63-5)e^-2.8ms /2.8ms = 5 +(-4.37) * 0.367 = 3.39 V.
Kannan said:
9 years ago
T = RC = 2.8 ms.
VC = 5 * 63.3 / 100 = 3.165 v.
Offtime, 2nd time constant (2*2.8 = 5.6).
So, capacitor will discharge 86.5% of 3.165 v = 2.74 v.
Next cycle it will again charge,
Total = (3.165 - 2.74) + 5 * 63.3/100 = 0.425 + 3.165 = 3.59 v.
VC = 5 * 63.3 / 100 = 3.165 v.
Offtime, 2nd time constant (2*2.8 = 5.6).
So, capacitor will discharge 86.5% of 3.165 v = 2.74 v.
Next cycle it will again charge,
Total = (3.165 - 2.74) + 5 * 63.3/100 = 0.425 + 3.165 = 3.59 v.
Sree said:
1 decade ago
Explain me please.
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