A Wheatstone bridge circuit has the following voltage ( Vs ) and resistor ( R ) values:
Q: What is the value for the voltage ( VL ) drop and current ( I ) across the load resistor ( RL )?
A: We begin by removing the load resistor and marking the new terminals of the opening with the letter “ A “ and “ B “:
The Thevenin equivalent voltage ( VTH ) is the energy differential between points A and B:
VTH = VA – VB
VTH = ( Vs )[ R2 / ( R1 + R2 ) ] – ( Vs )[ R4 / ( R3 + R4 ) ]
VTH = ( 24 V )[ 680 Ω / ( 330 Ω + 680 Ω ) ] – ( 24 V )[ 560 Ω / ( 680 Ω + 560 Ω ) ]
VTH = ( 24 V )[ 680 Ω / ( 1,010 Ω ) ] – ( 24 V )[ 560 Ω / ( 1,240 Ω ) ]
VTH = ( 24 V )( 0.67 ) – ( 24 V )( 0.45 )
VTH = 16.16 – 10.80
VTH = 5.36 V
Note: If VTH had a negative value, what assumption could be made about the direction in which current travels across the load resistor?
The next step requires us to replace the source voltage with a short:
We now have two parallel circuits in series with one another. If the reasoning behind this step isn’t apparent, please review the lecture material that immediately precedes this entry. The Thevenin equivalent resistance ( RTH ) is determined by adding the aforementioned circuits to one another:
RTH = [ R1R2 / ( R1 + R2 ) ] + [ R3R4 / ( R3 + R4 ) ]
RTH = [ ( 330 Ω )( 680 Ω ) / ( 330 Ω + 680 Ω ) ] + [ ( 680 Ω )( 560 Ω ) / ( 680 Ω + 560 Ω ) ]
RTH = [ ( 330 Ω )( 680 Ω ) / ( 1,010 Ω ) ] + [ ( 680 Ω )( 560 Ω ) / ( 1,240 Ω ) ]
RTH = 222 Ω + 307 Ω
RTH = 529 Ω
The Thevenin equivalent circuit can now be diagrammed as a series circuit:
Reattaching the load resistor to the A and B terminals gives us the following circuit:
The voltage drop across the load is determined via usage of the voltage-divider formula:
VL = ( VTH )[ RL / ( RL + RTH ) ]
VL = ( 5.36 V )[ 1.0 kΩ / ( 1.0 kΩ + 529 Ω ) ]
VL = ( 5.36 V )[ 1.0 kΩ / ( 1.529 Ω ) ]
VL = ( 5.36 V )( 0.65 )
VL = 3.51 V
Finally, the current across the load resistor is derived from the Ohm’s law equation:
IL = ( VL / RL )
IL = ( 3.51 V / 1.0 kΩ )
IL = 3.51 mA
RENAISSANCE PHYSICS 