Electronics - Time Response of Reactive Circuits - Discussion

Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 12)
12.

What is the voltage across the inductor in the given circuit on the falling edge of the first input pulse?

–0.2 V
0.2 V
–9.8 V
9.8 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Claude said:   4 years ago
No, it should be VL = Vin(e^-Rt/L) since it is an inductor.

Esketit said:   4 years ago
Since it is an inductor, at t=0, VL = Vmax = 10 V. The voltage across the inductor decreases as the current increases opposite of that of the capacitor therefore inductor discharges in the opposite direction thus VL = -Vin^-t/(L/R) = -0.183 = -0.2 V.
(1)

Marie said:   6 years ago
VR = Vin(1-(e^(-Rt/L)).
VR = 10(1-(e^(-((24x10^(-3))(50))/(300x10^-3)),
VR = 9.8V,

VL = VR - VIN.
VL = 9.8-10 = -0.2V.

Retsam said:   7 years ago
Why in negative?

Pooja said:   9 years ago
@Meenu 0.8-10.0 = -0.2?

Meenu said:   10 years ago
Toff = L/R = 6 ms.

Ton = 24 ms.

T = Toff+Ton = 24 ms+6 ms = 30 ms.

VL = Ton/T = 24 ms/30 ms = 0.8.

Vf = Vl-Vin = 0.8-10.0 = -0.2 ms.

Ramesh said:   1 decade ago
Voltage across inductor VL = vin(1-e^-Rt/L).

R = 50 ohms, L= 300 mH, t = 24 ms.

Vr = VL-Vin.

Muddassir said:   1 decade ago
Toff= RL=RC.

Ton is given.

T=Toff+Ton.

Ton/T=V (voltage across the inductor).

Put the Value & get the result.

Vasu said:   1 decade ago
Can anyone explain this please ?

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