Voltage across open in this diode circuit with two batteries

Question

Trying to understand how voltage \$ V_{D1} \$ is found.

Here's the original circuit with solution:

Here's how I think it can be redrawn, basically the grounds are just one node:

So to find voltage \$ V_{d1} \$ one first considers what's voltage at point 1, positive side of \$V_{d1}\$ right?

Well, at point 1, voltage is what? \$ -10K \times 1.33mA? \$

How about voltage across 5k resistor? Point 1 is between 10k and 5k resistors...

For point 2 it's 0, since the node of that point is connected to ground, right?

Answer 1

For \$V_{D1}\$ you can consider two KVL loops—they should both give you the same answer.

At point 1, you could start from the leftmost source (start from the ground) and say:

$$ \text{10V}-i_{D2}(10\text{k})-V_{D1}=0$$

^{simulate this circuit – Schematic created using CircuitLab}

You already have the current then solving for \$V_{D1}\$ results in \$V_{D1}=-3.3\text{V}\$

You may also use another KVL loop:

^{simulate this circuit}

$$ V_{D1}-i_{D2}(5\text{k})-(-10\text{V})=0$$

And solving for \$V_{D1}\$ will result in the same.