- #1

Firestrider

- 104

- 0

## Homework Statement

Code:

```
|------|----a
| |^ +
| ||
R_eq I_s V_0
| |
| | -
|------|----b
R_eq = 3kOhm
I_s = 6mA
Find V_0.
```

## Homework Equations

Ohm's law: KVL, KCL

## The Attempt at a Solution

Should this work just like any other circuit where V = IR? So V = (6 x 10^-3)(3 x 10^3) = 18? Or does it work differently since V_0 is an open circuit.

Here is the original problem: