INTRODUCTION TO ELECTRONICS: Thevenizing a Wheatstone Bridge Circuit

Many simple circuits can be categorized as being either a series circuit, parallel circuit, or a combination series-parallel circuit. To the contrary, analysis of Wheatstone bridge circuits is comparatively difficult, because no clear cut series-parallel relationship exists between its component resistors:

Thevenin’s theorem enables us to analyze the circuit with convenience via removal of the load resistor. Once the resistor has been removed, the now-familiar method of determining the Thevenin equivalent voltage ( V_TH ) and resistance ( R_TH ) can be applied to the system:

Removal of R_L gives rise to a possible potential difference between the A and B terminals. For convenience, the voltage differential ( V_s ) between the source and ground potential can be diagrammed with the voltage source located to the side of the Wheatstone bridge. Thus, the initial physics of the system remains the same, but the circuit becomes less difficult to examine:

If we wish, we can rearrange the circuit resistors to make things look familiar to us:

We must now realize that the voltage at points A and B are a fraction of the total voltage drop along each of the two branches. What separates the voltage potential at nodes A and B from ground are the voltages that exist across R₂ and R₄. As a consequence, the voltage-divider formula is applied to each branch as follows:

V_A = ( V_s )[ R₂ / ( R₁ + R₂ ) ]

V_B = ( V_s )[ R₄ / ( R₃ + R₄ ) ]

The Thevenin voltage is the energy differential ( if any ) between points A and B: 

V_TH = V_A – V_B 

V_TH = ( V_s )[ R₂ / ( R₁ + R₂ ) ] – ( V_s )[ R₄ / ( R₃ + R₄ ) ]

We are now ready to determine the Thevenin resistance of the circuit. Please recall that, for all practical purposes, the internal resistance of the source is zero; therefore, we replace the source voltage with a conductive wire that essentially shorts the system:

At this point, we must realize that introducing the short changed how resistor R₁ relates to R₂, as well as how R₃ relates to R₄. A hypothetical charge at point A has two paths to ground. The same can be said for a charge located at point B. For this reason, the R₁ resistor is parallel to R₂, and the R₃ resistor is parallel to R₄:

Notice how each parallel circuit shares attachments to the node above R₁ and the node beneath R₂. If we wish, we can again simplify the diagram with yet another redrawing of the circuit:

It is now easier to see that the Thevenin resistance between points A and B are the sum of two parallel circuits in series with one another:

R_TH = R₁∥R₂ + R₃∥R₄

The final step involves placing the Thevenin voltage in series with the Thevenin resistance prior to reconnecting the load resistor to the circuit:

The voltage across the load resistor is found via usage of the voltage-divider formula. Subsequently, Ohm’s law is used to determine what current flows across the load:

V_L = ( V_TH )[ R_L / ( R_L + R_TH ) ]

I_L = ( V_L / R_L )

In a subsequent problem, we will skip most of the steps above and deal solely with the mathematical procedures needed to solve the problem at hand.