I'm suppose to calculate the voltages across the diodes.. but I was wondering in this situation, are both diodes conducting, or is only the first one conducting because there is more power in that, and it shuts off the other one.
in diode problems like this usually, you should check all cases to see which is the only consistent one. In this case, there are four cases both off, both on, D1 on only and D2 on only. Now I am not sure whether those diodes are ideal or you put 0.6V across them when they are ON. But the technique is the same. Remember what determines ON/OFF is the potential difference across each of them.
they both need 0.7 V, so I'm guessing the D1 will shut off D2
some one please help? :(
If your guess works, you did good.
Let me ask you:
why couldn't they both be turned ON?
why couldn't D1 be OFF and D2 be ON?
lol, its kind of a long explanation.. but even if the 5V turned on much before the 8 V did... but as soon as the 8V kicks in, there is 7.2 volts against the 5 volts.. and so it doesn't go through the diode.. I know its more complicated than that. but...
It's really not complicated though.
here's how i'd do the first case (why can't they both be on?)
call the node inbetween the diodes Vx
~the drop across D1 is 0.7, making Vx = 7.3V
~the drop across D2 is 0.7, making Vx = 4.3V
two different voltages for one node isn't allowed, so this is why they can't both be turned on.
Now try acting as if D1 is OFF and D2 is ON, and see what happens