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**Unformatted text preview: **EE100/42 – Spring 2011 – Prof. Poolla
HW5 Solution
P3.13
{{
Therefore:
{{ # " - { { ' 0 " {{ {{ {{ J{{
{{ {{ #
$ $ {{ The units are A, V, W and J for current, voltage, power and energy respectively. P3.22
{{ $""" . {{
J{ { 0 {. 0
{{{{ At t=0, J{ {
At t=0.5 ms, J{ { 0.
$""" . {0{ $"""
$""" $""" {
{ so it is neither absorbing nor delivering the power.
J{ { . 0 the capacitor is absorbing the power. #{ 0{ #{ % and because it is positive, P4.11
(a) After t=0 we have VC=VR and from KCL:
- 0 {{ $ 0 #"" % And we know that Vc(t=0)=10 V. Therefore A=10.
{{ #"" {{ $"" JJ {{ JJ 2 $"" {{ JJ (b)
#""
#"" {{ J{{ J{{ JJ 2 JJ (c)
∞ " ∞ J{{ " $"" . #
$"" #
$"" $"" ∞
É" H (d) The initial energy store in the capacitor is:
#
$ ${ #
$ { 0 0 $ H P.4.22
and therefore i=0) and At steady state the capacitance has no current (because because the switch is open, all the 10mA current has to pass through 1kΩ resistance. So
the steady-state value of vc is vc=10mA×1kΩ=10 V.
After opening of the switch at t=0:
- $ $ {{ % 0
0 0 0 % $ . #"" JJ 2 99% of the steady-state value of vc is 99%×10=9.9 V. Therefore:
.
. #""
#
#"" %%
J P4.27
The circuit has been connected for a long time. So it is in the steady-state condition and
we can replace the capacitance with open-circuit and the inductance with short-circuit.
Because the capacitance has zero current, all the 4mA current of the current source has
to pass through the resistance and inductance and since the inductance in steady-state is
short-circuit, iR=4 mA.
Writing KVL around the outer loop (resistance, inductance, voltage source and
capacitance):
0 -- . P4.28
Before t=0, the circuit is in steady-state and switch is closed. The capacitance current is
zero so all the 5mA current should divide between two 10kΩ resistances. Each resistance
will have 2.5mA and therefore
vC(before t=0)=2.5mA×10kΩ=25V.
A long time after t=0, the circuit again is in steady-state condition however this time, the
10kΩ resistance in series with the switch is not in the circuit anymore and its current is
zero. So
vC(long time after t=0)=5mA×10kΩ=50V .
The time-constant after the switch is open is:
RC=10 ×103×10×10-6=0.1s
- #"0#" 0 # # {{ % { -{ . #" JJ 2 IJ {∞{ ...

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- Fall '08
- CHUA
- Steady State, Volt, long time, Inductor, steady-state value