- #1

- 807

- 8

## Homework Statement

The problem and its solution are attached.

## Homework Equations

"Steps to follow for Thevenin’s Theorem:

(1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.

(2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.

(3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.

(4) Analyze voltage and current for the load resistor following the rules for series circuits."

- https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/

## The Attempt at a Solution

If one replaces the voltage sources with a short circuit, wouldn't that mean that the current ignores the 6 kΩ resistor and that R_ab = 12*16/(12+16) = 6.85714285714285714286 kΩ, instead of 3.2 kΩ?

Could someone please let me know whether I'm right or wrong? If I'm wrong, I'd appreciate having an explanation as to why I'm wrong, as well.