
#1
Oct1807, 10:42 PM

P: 48

1. The problem statement, all variables and given/known data
See attachment 2. Relevant equations We are to assume the voltage drop across diodes is .7v (don't use the diode equation!). 3. The attempt at a solution I first was assuming that both diodes were "on". Then my work is as follows: V1 is the voltage directly above diode D1 (the left most diode)... [tex]I+ \frac{V_{1}.7(5)}{5k\Omega}=\frac{5V_{1}}{10k\Omega}[/tex] [tex]I+ \frac{.7.7+5}{5k\Omega}=\frac{5.7}{10k\Omega}[/tex] [tex]I=.57mA[/tex] Now since I is negative, one of my assumptions is wrong? So now I assume the rightmost one is "off" (D2). So.. then Solving for I [tex]I = \frac{5.7}{10k\Omega}[/tex] [tex]I = .43 mA[/tex] Solving for V Well.. since theres no current, v=5? Am I approaching this correctly? 


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