# Voltage across open in this diode circuit with two batteries

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?

• Here's a similar, simpler question, i.imgur.com/yV5EAyX.jpg. Basically, what's the voltage at point 1? Is it: (current through R1) times R1 + (current through R2) times R2???
– Jack
Oct 17, 2017 at 1:52

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.