Electric circuit with negative reference voltages and diodes

[canadiansmith]

va would equal 5-VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the -10 draw and the +5 push. therefore the total current in the left side would be 5+10-.7 =14.3V/R1+R2 = 1.43mA
VR1 = 1.43mA*R1 = 7.15 V
and Va = 5-7.15V = -2.15V is this right?

 

[gneill]

va would equal 5-VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the -10 draw and the +5 push. therefore the total current in the left side would be 5+10-.7 =14.3V/R1+R2 = 1.43mA
VR1 = 1.43mA*R1 = 7.15 V
and Va = 5-7.15V = -2.15V is this right?

It is. So in conclusion...?

 

[canadiansmith]

In conclusion D2 is in fact off. Thank you very much. I got very confused with the negative potentials. But I realize now that they cannot make the potential at any point higher than the original potential.

 

[gneill]

The important thing is that you got there in the end

 

[phinds]

good work ... sounds like you learned something worth learning

 

[canadiansmith]

Thank you both very much for your help. I didn't think I would ever see the light at the end of the tunnel. Thanks again