electronics

[evaporade]

[TonyHem]

my guess is zero

[/0]

I think is 6V and the rest are 0V, but not sure.

[Damo17]

I would say 6v across v3 as well and 0v across v1 and v2.
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit. 

[TrueTears]

I think is 6V and the rest are 0V, but not sure.

was saying the same as that with tonyhem on msn.

But I'm not 100% sure...

[/0]

I would say 6v across v3 as well and 0v across v1 and v2.
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit. 

Ah that's why

My teacher was explaining to the class that there was no current in the circuit. But how would this account for the voltage drop across the reverse diode?
So there must be current in the circuit after all.

[TrueTears]

I would say 6v across v3 as well and 0v across v1 and v2.
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit. 

Ah that's why

My teacher was explaining to the class that there was no current in the circuit. But how would this account for the voltage drop across the reverse diode?
So there must be current in the circuit after all.

in Jarcaranda it says the circuit has very LITTLE current, so I assume there is still current, but just very very small...

[Damo17]

yeah there is still current, I think I read that it was as small as micro amps.
If a reverse-biasing voltage is applied across the P-N junction, the depletion region expands, further resisting any current through it. Key word here is 'resisting', it does not completely stop all current just makes it harder for current to flow as the depletion region gets larger.

[naved_s9994]

It will be zero for all for SURE... Guys, its a SERIES?? !!

[Mao]

Hello again evaporade...

[kurrymuncher]

im confused

[Mao]

http://vcenotes.com/forum/index.php/topic,2754.0.html

[TonyHem]

Lol@argument!!!

[TrueTears]

(Image removed from quote.)

I've confirmed.

V1 is 0
V2 is 0
V3 is 6V

basically the reverse bias allows no current in the circuit (There might be some, but let's assume it is a perfect diode and blocks out all current)

using Ohms law on the resistor we have V = IR when I is 0 there is 0 V

for the forward bias diode we can not use ohms law as it does not have a linear relationship.

Looking at the graph of a diode, we are only dealing with the positive half of the graph (Since forward bias implies positive and reverse bias implies negative)

when I is 0 we see that V = 0 Hence V2 = 0

However at V3 we see its a reverse bias diode, again looking back at diode graph, we are only dealing with the negative half. Since the battery is 6V and when I = 0, the voltage can be anything for the reverse diode, so the potential difference is 6 V.

[Mao]

TT: the only way to confirm something like this is to construct the circuit and experimentally test it out.

A diode has a forward voltage as well as a reverse voltage (usually excluded as it almost always lead to permanent damage to it). I'm not sure if 6V is enough to kill the diode in reverse bias, but if it is, there will be some interesting complications.