# This circuit is two parallel resistances of 10Ω. Convince me otherwise

I'm trying to find the Thévenin equivalent resistance of a circuit - one that resembles a Wheatstone bridge, but is not at balance. The load in question is where the galvanometer would appear.

Anyway, I have disconnected the load, removed the (ideal) sources, and redrawn the circuit as below.

To me, the resistance seen by the load on AB is two series combinations, both of which evaluate to 10, in parallel - an equivalent resistance of 5Ω.

This is apparently incorrect. My textbook calculates the Thévenin resistances as:

$$R_L = \frac{3 \times 7}{3+7} + \frac{4 \times 6}{4+6} = 2.1 + 2.4 = 4.5Ω$$

simulate this circuit – Schematic created using CircuitLab

TIA

• Please edit your question to show how you calculated 10 Ω for the series combinations. Maybe then we can help you. Commented Aug 5, 2018 at 12:41
• So you say 3 Ohm + 7 Ohm in series as well as 4 Ohm and 6 Ohm in series? Commented Aug 5, 2018 at 12:41
• Yes. Why is this incorrect? Commented Aug 5, 2018 at 12:42
• Because 3 Ω and 7 Ω are in parallel as are the 4 Ω and 6 Ω. Commented Aug 5, 2018 at 12:42
• The 3+7 series may not both start and end at point A. And the 4+6 series cannot both start and end in point B. Commented Aug 5, 2018 at 12:43

So this is what you drew

simulate this circuit – Schematic created using CircuitLab

Let me slightly redraw this:

simulate this circuit

it is clear there is two series connected parallel branches.

( 3||7 ) + ( 4 || 6 ) which equates to 4.5R

• This is a perfect explanation. It won't let me accept the answer for another 7 minutes, but be assured I will. Thank you for your help! Commented Aug 5, 2018 at 12:43