INTRODUCTION TO ELECTRONICS: Thevenin Voltage and Resistance Determination

Now that a conceptual understanding of Thevenin’s theorem has been established, we are ready to determine the Thevenin voltage ( V_TH ) and Thevenin resistance ( R_TH ) for an open circuit:

The equivalent voltage will represent a system in which the voltage source ( V_s ) has been replaced by a voltage that “ appears “ across the A-to-B terminals. Likewise, the current ( I ) that used to flow through the system is now stationary after having been redirected through a short across the source voltage. The introduction of the short enables us to eliminate V_s from the circuit:

Slight rearrangement of the circuit’s resistors clarifies that we now have a series-parallel circuit downstream of R₄:

From this point, we determine the equivalent voltage and resistance from the vantage point of the A-to-B terminal. Since the former current is now static charge at rest within the circuit, there is no voltage drop across the R₄ resistor!!! There are, however, voltage drops across R₁ and the circuit branch that contains R₂ and R₃, because they are situated parallel to the voltage drop across V_TH; thus, an observer at V_TH will see the voltage drop across R₂ and R₃ as its equivalent.

How must we go about solving V_TH? We must now treat the loop that contains R₁, R₂, and R₃ as a series circuit!!! The voltage drops of interest to us are those across R₂ and R₃ as a fraction of the net energy differential across the loop. Remember, an equivalent circuit is a static modification of an original circuit, so tweaks in the usual problem-solving protocols are justified by mathematics, measurements, experiments, and observation. We must now use the voltage-divider formula to solve for V_TH:

V_TH = V₂ + V₃

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

V_TH = [ 690 Ω / ( 1.69 kΩ ) ]( 10 V )

V_TH = 4.08 V

Fortunately, we use the familiar protocol of resistor addition to determine R_TH. From the vantage point of V_TH, R₄ is in series with two circuit branches, one of which contains the R₁ resistor whilst the other branch contains the R₂ and R₃ resistors:

R_TH = R₄ + ( R₁ )[ R₂ + R₃ / ( R₁ + R₂ + R₃ ) ]

Note: Although it may look somewhat confusing, the above fraction is the familiar double-reciprocal formula being applied to three resistors.

R_TH = 1.0 kΩ + ( 1.0 kΩ )( 690 Ω / 1.69 kΩ )

R_TH = 1.0 kΩ + ( 1.0 kΩ )( 690 Ω / 1.69 kΩ )

R_TH = 1.0 kΩ + 408 Ω

R_TH = 1.41 kΩ

Once Thevenin’s voltage and resistance values have been determined, the current that could be supplied by the bridge is determined ( how? ). For the time being, however, we should become familiar with the diagrammatic representation of equivalent circuits: