Since the switch is closed for a longtime initially the circuit was in steady state
Whenall sources are DC capacitors act like open circuit in steady state Using this
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information let’s find the voltage across the capacitor before the switch was opened
1002 2001A
Apply nodalanalysis
g 1 Lajos WEIL
multiply bothft 0 sidesby200 VA S t kVA 2VA 20 0
Thus the voltage across the capacitor right before the switch was opered is
Vc10 I Vc Ot 2 V
Next let’s analyze the circuit after the switch was opened and find its Theverin
equivalent seen by the capacitor
ECE 2020 AU22
Hw6 Solutions
a WTH b 12TH
Find opencircuit voltage seenby Sincethere are no dependent sources kill
the load capacitor all thesources in thecircuit find the equivalent
resistance seen by the load capacitor
From voltage division Req 1005211502
V to 1030 1
Thus the Theverin equivalent of this circuit fort 0 seen by the capacitor is
Time constant 2 RC 10 140 10 bMET
Then velt is givenby
vclH rel I e veto vel e th for t o
where veCoo is the final condition i e thevoltage across the capacitor after
the system reaches steady state It can be determined directly from theTherein circuitas
Vel 1 103V
Remember that Vc101 was calculated previously hence inserting veld Vc oo 2 into
this equation gives the following result
HI log 2 je e Et
Xt Derivation of VcLH Expression
Apply KVL in Theverin circuit
log ich KH DMET t
Remember that i v relation of a capacitor is
100 140 106 It Volt D
H volt log
volt consists of two parts VenLH i.e the natural response and UcfLH i.e
the forced response Natural response is the response of the circuit if there were
no excitation WTH D It is the solution of the following differential equation
ducal vcnlt.no
and it is in the exponential form
where Z RC for this problem and A is a constant that depends on initial
condition Velo
The other part of the solution is the forced response Vef It It is thevoltage
that is induced by the sources in the circuit Therefore it has the same form as the
inputs Since Kit is a DC voltage source in this problem the forced response is
some constant
Ther the total response is
volt VanLH vet t A e
constants A B must be determined from the initial final conditions
Note Natural response is generally notalways transient i e it eventually dies out
Let’s plug in into the differential equation
Ae Kit B t Ae Kit B
3 B I B101414 3
which can also be obtained from the finalcondition as
Vc oo him Velt lim B B
If we substitute B into VaLH expression and apply the initial condition
clot A B At log
Hence velt becomes
vcLH Ae Ttt B e It log
which is the same result obtained from equation L B
When S is open for a long time all of 5mA flows through the 3052 resistor
and no current can flow through the rest of the circuit Hence
il 10 1 0 A
Next find the Theverin Equivalent of the circuit seen by the inductor for t 0
division rule
is x 5 GImA30 20 30
TH 30 iz 45 22 mV
Since all sources are independent kill them find theequivalent resistance seer by
the load inductor
Iforj son FEI
Req 505L11 301
Io leg E r4
Then the Theuerin Equivalent is
5 r Time constant Z Lz 4ms
To find final condition assume the system reaches steady state find the current
across the inductor Since inductors act like short circuit in steady state
4 From Ohm’s law
224.52mV I 225 103
il oo 4 3mA
steady state
Then ICHI is givenby
iLlH idol i10 ii lool e th for t 0
Substituting i40 ie lo I 2 into this equation yields
ii t 3 O 3 e
3 3e mAfort oltinm
Differential Equation Solution
Apply KVL in the therein circuit
fifty 2425 741 ii HI viLH Oc
Remember that i v relation of an inductor is given by
L IH L dit
Substituting this into the KVL equation and arranging terms yields
75 fifth 7 ii IH 22
fifth ii IH 34
in H A e Natural response
if IH B Forced response
Ther ilLH Ae 44 B
il 10 At B D
icelool B 3 mA A 3 ii It 3e tht c 3 mA for t 0 It in ms
In DC steady state capacitors act like open circuit inductors act like
short circuit Based on this if we redraw the circuit
Notice that since C is opencircuit
14 0 A end there is no voltage drop
Li Sr across the 101 resistor
v I c f f4r
61 Since Sr Gr Lcr resistors are in
its series from Ohm’s law
i Vc 30 Siu 20V
i v Relations
Resistors Vr Rip
inIv ieIv tic
Capacitors ie C de
Inductor ve L
Let ie ie il be the currents and Ve be the voltages of the resistor
capacitor and inductor respectively Since they are connected inparallel
ir IH ieLH tilt Is Itt
Vinet C 41 ii HI Is Hl Express ir ie in terms of
11 C 41 ii HI Is It Substitute Va instead of
f dit Cdt Ltd KAI Ish Substitute vi Ldt
t d I Rearrange terms
LxLH a 4 b 1 c R
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