Diodes question is this correct

With diode analysis, one thing you have to be completely aware of is that it's a guessing game. Yes, I said guessing. How well you guess will depend on your experience, so even if you are the worst guesser ever, its ok, because the math will tell you.

When Vd >= 0.7, it conducts. When Vd < 0.7, it does not conduct. Vd is the is the difference between anode and cathode.

Now to the guessing game.

There are 4 cases to test.

• Case 1: D1 = OFF | D2 = OFF
• Case 2: D1 = ON | D2 = OFF
• Case 3: D1 = OFF | D2 = ON
• Case 4: D1 = ON | D2 = ON

Your initial guess was that D2 conducts, and D1 does not. You didn't specify what your reasons were for that selection, but even if it was a blind guess, it does not matter because you'll need to prove your guess was correct. Don't ever guess, and not verify. Always verify your guess.

Case 3: D1 = OFF and D2 = ON

$$ -10V - 10k\Omega I + 0.7V - 5k\Omega I - 10V = 0 $$ $$ I = \frac{20V - 0.7V}{15k\Omega} = 1.28667mA $$

Lets look at the anode of D2 $$ 10V - (1.29mA)(10k\Omega) = -2.86667V $$

So let's look at the voltages at node between the two diodes.

$$ 5k\Omega I -10V = -3.6V$$

So now the main question, do these values hold true for our assumptions.

For D2 to be on, \$V_{D2} >=0.7\$ and \$V_{D2} = -2.9V - (-3.6) = 0.7\$

For D1, to be off \$V_{D1} < 0.7 \$ and \$V_{D2} = 0-(-3.6V) = 3.6V \$

But.. D1 is supposed to be off...

Repeat the process for all cases until you can confirm that the math holds true with your assumptions.

But your initial guess that D2 is on, and D1 is off is wrong.

Assuming all the diodes take 0.7V.

VR1 is parallel over D2 = 0.7V

VR2 = 20V- 2* 0.7V = 18.6V

With these values known you can calculate the currents