graphnewbox
graphtempnewdimen
=
=by 1by 2
by .5exby 1.083in
to 0pt-
=.5exby 0.833in
to 0ptVs
=by -1by 2
by .5exby 0.583in
to 0pt+
=by 1by 2
by .5exby 0.167in
to 0ptR1
=by -1by 2
by .5exby 0.083in
to 0ptI1
=.5exby 0.833in
to 0ptR2
=.5exby 0.400in
to 0ptI2
=.5exby 0.833in
to 0ptR3
=by -1by 2
by .5exby 0.583in
to 0pt+
=.5exby 0.833in
to 0ptV3
=by 1by 2
by .5exby 1.083in
to 0pt-
=.5exby 0.400in
to 0ptI3
depth1.603in width0pt height 0pt
1.
Determine a general expression for the current I1 and the voltage V3 as a function of the
input voltage and the resistor values.
2.
Give numeric values for I1 and V3 if Vs=5V,
,
.
3.
Now consider the following AC circuit:
graphnewbox
graphtempnewdimen
=
=by 1by 2
by .5exby 1.083in
to 0pt-
=.5exby 0.833in
to 0ptvs(t)
=by -1by 2
by .5exby 0.583in
to 0pt+
=by 1by 2
by .5exby 0.167in
to 0ptR1
=by -1by 2
by .5exby 0.083in
to 0pti1(t)
=.5exby 0.833in
to 0ptR2
=.5exby 0.400in
to 0pti2(t)
=.5exby 0.833in
to 0ptC
=by -1by 2
by .5exby 0.783in
to 0pt+
=.5exby 0.833in
to 0ptvc(t)
=by 1by 2
by .5exby 0.883in
to 0pt-
=.5exby 0.533in
to 0pti3(t)
depth1.603in width0pt height 0pt
Assume that the input signal is a complex exponential,
.
Then, the voltage vc(t) can be expressed in the form
.
Determine the paramters A, ,
and .