# KVL - Wrong solution?

1. Feb 24, 2013

### charlies1902

I attached the circuit. This is the solution that was posted on my course's website. I do not think it's right, but I want to make sure to avoid making myself look dumb.

Around loop A, KVL should be:
V1-R1*I1-R2*(I1-I2)=0
V1=R1*I1+R2*(I1-I2)

Is the answer supposed to be c?

never mind....I was being stupid, I1 and I2 aren't mesh currents so their answer is right.

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2. Feb 24, 2013

### charlies1902

I have another question regarding this circuit. Another question asked "Which of the following is needed for the diode to be forward biased?"
The options are:
1) V2<0
2) v1>0
3) V1+V2>0
4) V1-V2>0
5) V2-V1>0

This is my procedure:
The voltage on the right of the diode must be greater than the voltage on the left (Vd<0).
I found the voltage at the right node (Let's call this V_r) of the diode by doing KVl around the outer loop of the circuit. Not the whole loop, just until I got to that node. This gave:
V1-R1*I1-V2=V_r
Similarly, I used KVl to find the voltage at the node to the left of the diode (V_l):
V1-R1*I1-R2*(I1+I2)=V_l=0

So V_r>V_l=0
V1-R1*I1-V2>0
V1-V2>R1*I1

That's as far as I got. As you can see, this doesn't fit any of the answers.

3. Feb 24, 2013

### Staff: Mentor

Take each of the answers in turn and show that the diode being forward-biased is one of: T, F or maybe. Choose the TRUE one.

The question's wording could be improved. I suggest, "Which of the following ensures that the diode will be forward biased."

4. Feb 24, 2013

### charlies1902

Isn't #1 true as well? This is one of those questions where I can't understand what to do when I plug in each answer.

5. Feb 24, 2013

### Staff: Mentor

Suppose the resistors are of equal value, so by themselves they would halve V1.
Let's say V1 = 10V.

Now, take V2=9V. (This satisfies V1-V2>0, so you believe it is the answer.)

Will the diode be conducting?

6. Feb 24, 2013

### charlies1902

Oh I see what you're saying. Yes, the diode would then be forward biased as the voltage on the right of the diode will be greater than its left.
In my original post, I tried deriving the answer using KVL, but couldn't get it to work. Were you able to catch an error in my KVL equations?

Thanks for the help.

7. Feb 24, 2013

### Staff: Mentor

I chose those values so the diode will be reverse biased.

8. Feb 24, 2013

### charlies1902

Huh? I'm confused, did you choose the wrong values by accident or do those values make it reverse biased?

10-9=1>0 so it fits the answer that V1-V2>0 ensures forward bias.
But...since the resistors are of equal value, they will both drop 5V. So that means the right node of the diode is at -4V and its left is at 0V, which isn't right.

Is V1-V2 not right?

9. Feb 25, 2013

### Staff: Mentor

That.
That condition is not sufficient.

10. Feb 25, 2013

### charlies1902

I mean the condition V1-V2>0. That is what the solution said. But it looks like (from the 2 voltage values you gave), the condition V1-V2>0 does not ensure that it's forward biased.

11. Feb 26, 2013

### Staff: Mentor

That's the conclusion I formed. So what option are you now going to back?

12. Feb 26, 2013

### charlies1902

v2-v1>0.

13. Feb 26, 2013

### pyroknife

I don't think your conclusion is correct. 1. We don't know if the resistors are of equal value. 2. Even if they are of equal value, they will not have the same voltage drop (the current through R1 is not the same as the current through R2). Thus we can't calculate the voltage on the right of the diode.

I think the solution they gave is correct.
To have a forward biased diode, V_d (as shown in the schematic) must be less than 0 (polarities reversed).

14. Feb 26, 2013

### Staff: Mentor

That is equivalent to saying v2>v1. In this case, that is sufficient condition for ensuring that the diode is reverse biased.

15. Feb 26, 2013

### pyroknife

V1-V2>0 is correct. Your last few posts regarding that are inaccurate.