Z Your answer is partially correct. Try again.t= 066 22>22 Q30 н27 V0.32 mF v(t)i(t)$46 NAfter the switch opens, this circuit can be represented by the differential equationd ?i(t)di(t)+ 48.882dt42.614 =+ 176.77 (t).dt2The natural response isin(t) = A,e 3.9326t-44.949t+ Aze'А.di(0)di(0)The coefficients A1 and A2 depend on the initial conditions, v(0) , (0) andis the derivative of t) evaluateddtdtdi(0)dtat time t = 0 ) Determine the values of V(0) , i(0) and:di(0)V(0)19.872i(0)0.108A and

Answered: t= 0 66 22 S22 N 30 н 27 V 0.32 mF v(t)…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Z Your answer is partially correct. Try again.
t= 0
66 22
>22 Q
30 н
27 V
0.32 mF v(t)
i(t)
$46 N
After the switch opens, this circuit can be represented by the differential equation
d ?i(t)
di(t)
+ 48.882
dt
42.614 =
+ 176.77 (t).
dt2
The natural response is
in(t) = A,e 3.9326t
-44.949t
+ Aze'
А.
di(0)
di(0)
The coefficients A1 and A2 depend on the initial conditions, v(0) , (0) and
is the derivative of t) evaluated
dt
dt
di(0)
dt
at time t = 0 ) Determine the values of V(0) , i(0) and
:
di(0)
V(0)
19.872
i(0)
0.108
A and

Expert Solution

Step 1

for t <0 circuit reached steady state, capacitor acts as open circuit and inductor acts as short circuit. now we have to find the steady state voltages and currents. as capacitor does not allow the sudden changes in voltages and inductor does not allow sudden change in the currents ,current in inductor is same as at t=0- and t=0+and same is true for capacitor voltage. Now we have to find out the new voltage across the inductor when switch is closed.

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