- #1
stefan10
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Homework Statement
Problem A
http://imagizer.imageshack.us/v2/800x600q90/855/6v0c.png
Problem B
http://imagizer.imageshack.us/v2/800x600q90/853/1m2c.png
http://imagizer.imageshack.us/v2/800x600q90/34/wfe4.png
Homework Equations
Kirchhoff's Rules/Ohm's Law
Voltage Divider Formula
[tex] V_{ab} = (V_{0}*R_{2})/(R_{1} + R_{2}) [/tex]
The Attempt at a Solution
Problem A:
I'm unsure how much information is sufficient for what the question is asking. I derived these equations, and relations for different currents:
[tex]I_{1} = I_{AB}=I_{EF}[/tex]
[tex]I_{3} = I_{BC}=I_{CD} = I_{DE} [/tex]
[tex]I_{2}=I_{BE}[/tex]
Loop 1 (Clockwise)
[tex] 0 = V_{0} - I_{1}R - I_{2}R - I_{1}R = V_{0} - 2I_{1}R - I_{2}R [/tex]
Loop 2 (Clockwise)
[tex] 0=V_{0} - 1/2 V_{0} - I_{3}R \ \mbox{(used Voltage Divider Eq for the voltage of GH)} [/tex]
[tex] \Rightarrow 0 = V_{0} - 2I_{3}R [/tex]
Node Equation
[tex] 0 = I_{1} - I_{3} - I_{2} [/tex]
Solving these equations I get
[tex] 3 I_{1} = I_{3} [/tex]
[tex] I_{2} = -2 I_{1} [/tex]
[tex] I_{3} = - (3/2) I_{2} [/tex]
So for the questions asked:
a.) V_{GH} = V_{0}/2
b.) I_{GH} = V_{0}/2R
c.) I'm not sure how to find this.
Problem B:
V Thevenin is equal to V_{GH} right? That means V_{th} = V_{GH} = V_{0}/2
R Thevenin is R parallel to R, which can be found with 1/R_{th} = 1/R + 1/R which give us R_{th} = 2/R.
I'm not as confident of Problem B as I am of Problem A.
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