Consider the circuit:
If we want to find the open-circuit voltage across the 5Ω resistor, first disconnect it from the circuit:
Find the equivalent resistance in loop 1 to find the current in loop 1. Use Ohm’s law with that current to find the potential drop across the resistance C. Note that since no current is flowing through resistor B, there is no potential drop across it, so it does not affect the open-circuit voltage.
The open-circuit voltage is the potential drop across the resistance C, which is: C C + A 100 V ∼ . {\textstyle {\frac {C}{C+A}}\ 100\ V_{\sim }\ .}
This is just an example. Many other ways can be used.2
"Open Circuits". www.learnabout-electronics.org. Archived from the original on 2009-01-29. https://web.archive.org/web/20090129201243/http://learnabout-electronics.org/resistors_18.php ↩
Thevenin's Theorem-One Independent Source, 11 January 2012, retrieved 2018-06-03 https://www.youtube.com/watch?v=LN7RoJ6GC9k ↩