The circuit has been in operation for a long time. At t = = 0, thevoltage source drops from 100 V to 25 V and the currentsource reverses direction. (Figure 1)Figure100 V5 ΚΩ2 ΚΩ20 kΩ Σ 10 mA ( 1 ) 12 kΩ { 0.05 με1 of 1VoPart ASelect the correct expression for v。 (t) for t ≥ 0.vo(t) = 30 +105e-4000t Vvo(t) = 15+105e Vvo(t) = −15+105e-4000t Vvo(t) =30-105e-4000t Vvo(t): = -15- 105evo(t) =Submit=Provide Feedback-4000t==4000tV-30+105e-4000t Vvo(t) = −30 – 105e-4000t VO vo(t) = 15 — 105e-4000t VRequest Answer

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Select the correct expression for vo (t) for t > 0.

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter18: Resistive-inductive Parallel Circuits

Section: Chapter Questions

Problem 1PA: Incandescent lighting of 500 W is connected in parallel with an inductive load. A clamp-on ammeter...

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Question

100%

The circuit has been in operation for a long time. At t = = 0, the
voltage source drops from 100 V to 25 V and the current
source reverses direction. (Figure 1)
Figure
100 V
5 ΚΩ
2 ΚΩ
20 kΩ Σ 10 mA ( 1 ) 12 kΩ { 0.05 με
1 of 1
Vo
Part A
Select the correct expression for v。 (t) for t ≥ 0.
vo(t) = 30 +105e-4000t V
vo(t) = 15+105e V
vo(t) = −15+105e-4000t V
vo(t) =
30-105e-4000t V
vo(t): = -15- 105e
vo(t) =
Submit
=
Provide Feedback
-4000t
==
4000t
V
-30+105e-4000t V
vo(t) = −30 – 105e-4000t V
O vo(t) = 15 — 105e-4000t V
Request Answer

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