-.--.-.-.--.

 

Like this?

    10V
     |
     o--------o
     |        |
    .-.      .-.
    | |      | |
    | |10k   | |10k
    '-'      '-'
     |        |
Vout-o---|<---o
     |        |
    .-.      .-.
    | |      | |
    | |10k   | |1k
    '-'      '-'
     |        |
     o--------o
              |
             ===
             GND

 

-.-.-.-.-.-.-.-.

 

yes, but there is also a separate input to the top right at +10V. and another ground at the bottom left, the two sides of the H are not connected.

This would be electrically the same as the schematic niftydog posted.

What you must do for the first question is replace the diode with a .6 volt source (such as a battery; positive on the right) and solve the network for the voltage at the left middle node. This assumes that the diode voltage can be assumed to be .6 volts independent of the current in it.

The question also asks for the value of R where the diode would stop conducting.

In this case, find the value of R such that the voltage at the right middle node is .6 volts greater than the left middle node.

 

-.--.-.-.-.--

 

Connect the 10 volt supplies together, and the grounds together so you only need 3 loops.

But, for the second question, take the diode (and battery) out before you solve. Just solve for R such that the voltage from right middle node to left middle node is just .6 volts.

 

--.-.-.-.-.-.

 

The main problem is with your I2. The currents in mesh analyses must be loops. To just say that I2 is going through the diode right to left doesn't define a loop; where else is it going?.

Try this. Let I1 and I3 be as you've defined them. Now let the top of the right and left legs, which are connected to a 10 volt source, be connected instead to an explicit 10 volt battery. Put this battery over on the right side of the H. Let this battery be shown as connected from the top of the H to ground, with the positive end connected to the top of the H. Now define a current loop I2 going upward through the two 10k resistors in the left leg, and then downward through the 10 volt battery and back to the bottom of the left leg.

Another problem is that you haven't written the I1 and I3 loop equations correctly. In your first and third equation, 10 (for the 10 volt supply) shouldn't even appear because those current loops don't pass THROUGH the 10 volt supply. I1 touches it, but doesn't pass through it. Also, the current I2 doesn't appear anywhere in your three equations, and it just won't work out if you don't do something with I2. And don't forget that your resistors are 10k ohms, not 10 ohms.

So you have 3 current loops.

I1 goes through the top two 10k resistors and the .6 volt battery.

I3 goes through the bottom 10k and 1k resistors and the .6 volt battery, and the direction through the .6 volt battery is opposite to the direction I1 goes.

I2 as I defined it above.

As a hint, your first equation should be:

10000 I1 + 10000 I2 + 10000 I1 +.6 = 0

starting at the lower left corner of the I1 loop.

If you don't see where that comes from, go back and re-read the part of your text on mesh analysis.

 

This would be electrically the same as the schematic niftydog posted.

What you must do for the first question is replace the diode with a .6 volt source (such as a battery; positive on the right) and solve the network for the voltage at the left middle node. This assumes that the diode voltage can be assumed to be .6 volts independent of the current in it.



In this case, find the value of R such that the voltage at the right middle node is .6 volts greater than the left middle node.

Am I going nuts? The circuit I see has the diode reverse biased. If this is the case, Vout=5 volts.

 

.--.-.-.-.-.-.-.-

 

No you are correct, sorry. I have been working on the same circuit with a 20k resistor instead of 1k after figuring out that part.

For the 1k I got Vout = 5 volts and after a little help I got Vout = 5.715 volts when there is a 20k resistor in the place of the 1k resistor. Can someone verify that?

OK, but my question was primarily meant for The Electrician.

 

Doh!!

Funny how you see what you want to see!

I just assumed that the instructor would have given the case where the diode conducts, since that's the difficult and therefore more interesting case.

For that case, just turn the diode around in the original circuit, and let the .6 volt battery have its positive end on the left. Then the voltage on the left node will be 2.046 volts and on the right node, 1.446 volts.

If you replace the 1k with a 20k resistor and have the diode facing to the left, the voltage at the left node will be 5.457 volts.

electo101, did your problem as given really have a 1k and the diode facing left? Trick question.

 

Ron, look at his Vout in your Quote of his post. He has Vout = 5.715 volts. But if you go back to that post, he's edited it to say Vout = 5.46 volts!

You must have been working on your post between 7:31PM and 7:54, have captured your quote of his post before he edited it, but he edited it before you submitted your post!

 

-.-.--.-.-.-

 

i couldn't get Vout = 5.46 volts, i use the previous loops but don't come up with the answer...