- #1
opticaltempest
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Hello,
I am working on the following problem:
http://img155.imageshack.us/img155/2478/thevpo9.jpg
I am stuck on this seemingly simple circuit problem. I've worked many Thevenin circuit problems in my introductory circuit analysis. This problem is from a new class and seems different. I realize that I must first find the open-circuit voltage at the terminals. After that I need to the Thevenin equivalent resistance by using
[tex]\LARGE R_{Th}=V_{Th}/I_{sc}[/tex].
First let me find the open-circuit voltage [tex]v_{oc}[/tex]. Is this correct?
We know the circuit current is [tex]g_mv[/tex]. Thus the open circuit voltage which is the voltage across the voltage-controlled current source is
Using KVL:
[tex]\LARGE -v_s+v-v_{oc}=0 \implies v_{oc}=v-v_s[/tex]
and
[tex]\LARGE v=(g_mv)R_1[/tex].
So we have
[tex]
\LARGE v_{oc} = (g_mv)R_1-v_s \implies
\LARGE v_{oc} = (0.002v)(50k\Omega)-v_s \implies
\LARGE v_{oc} = 100v-v_s
[/tex]
Does this look correct for the open circuit voltage?
I am working on the following problem:
http://img155.imageshack.us/img155/2478/thevpo9.jpg
I am stuck on this seemingly simple circuit problem. I've worked many Thevenin circuit problems in my introductory circuit analysis. This problem is from a new class and seems different. I realize that I must first find the open-circuit voltage at the terminals. After that I need to the Thevenin equivalent resistance by using
[tex]\LARGE R_{Th}=V_{Th}/I_{sc}[/tex].
First let me find the open-circuit voltage [tex]v_{oc}[/tex]. Is this correct?
We know the circuit current is [tex]g_mv[/tex]. Thus the open circuit voltage which is the voltage across the voltage-controlled current source is
Using KVL:
[tex]\LARGE -v_s+v-v_{oc}=0 \implies v_{oc}=v-v_s[/tex]
and
[tex]\LARGE v=(g_mv)R_1[/tex].
So we have
[tex]
\LARGE v_{oc} = (g_mv)R_1-v_s \implies
\LARGE v_{oc} = (0.002v)(50k\Omega)-v_s \implies
\LARGE v_{oc} = 100v-v_s
[/tex]
Does this look correct for the open circuit voltage?
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