The voltage ( V ) drops across parallel-circuit resistors ( R ) are equal in magnitude; conversely, the currents ( I ) traveling through parallel branches may or may not be the same. For this reason, parallel circuits are sometimes referred to as being current dividers. Take the following diagram into consideration:
Since the R1 and R2 values are equal, the total current ( It ) will be split equally between each path. If R1 had been greater than R2, the current across R2 would have been greater than that across R1 ( and vice-versa ); thus, the rate at which coulombs of charge deposit energy into R1 and R2 would differ, but the joules of energy carried per second ( W ) by each coulomb of charge is the same. The effect of resistance upon current flow in parallel circuits is modeled by the structure of the two-branch current-divider formula:
( I1 ) = ( It )[ R2 / ( R1 + R2 ) ]
( I2 ) = ( It )[ R1 / ( R1 + R2 ) ]
The question to consider now is as follows: If R1 and R2 were split into two paths that are still parallel to one another, would the voltage drops across the four resistors still have a symmetrical relationship? Let’s split R1 into R1a and R1b, and let’s split R2 into R2a and R2b to examine things a little more closely:
Furthermore, let’s assume the resistors have the following values:
R1a = 3.0 Ω
R1b = 7.0 Ω
R2a = 7.0 Ω
R2b = 3.0 Ω
Whether current travels along path 1 or path 2, a net 10 Ω of resistance must be overcome along each respective branch; however, the instantaneous energy value of a coulomb of charge at the midpoint of each branch will differ. The voltage-divider formula explicitly shows this to be the case:
V1a = ( Vs )[ R1a / ( R1a + R1b ) ]
V1a = ( 10 V )[ 3.0 Ω / ( 3.0 Ω + 7.0 Ω ) ]
V1a = 3.0 V
V1b = ( Vs )[ R1b / ( R1a + R1b ) ]
V1b = ( 10 V )[ 7.0 Ω / ( 3.0 Ω + 7.0 Ω ) ]
V1b = 7.0 V
V2a = ( Vs )[ R2a / ( R2a + R2b ) ]
V2a = ( 10 V )[ 7.0 Ω / ( 7.0 Ω + 3.0 Ω ) ]
V2a = 7.0 V
V2b = ( Vs )[ R2b / ( R2a + R2b ) ]
V2b = ( 10 V )[ 3.0 Ω / ( 7.0 Ω + 3.0 Ω) ]
V2b = 3.0 V
Let’s now measure the potential difference between the midpoints of R1a / R1b and the midpoint of R2a / R2b:
The new resistor ( Rwb ) is akin to that of a Wheatstone bridge’s voltage-measuring terminal. In order for no current to flow across this bridge, the energy status of the charges within each of the connected regions must be the same. Since the first voltage drop at the midpoint along path 1 is 3.0 V, while that along path 2 is 7.0 V, there is clearly a voltage imbalance between the two once-isolated regions. Since energy travels downward across energy gradients, a current would move from the midpoint of path 2 to the midpoint of path 1 if the two midpoint regions are connected. If the 7.0 V potential was instead located at the midpoint of path 1 with a 3.0 V potential now located at the midpoint of path 2, the current directions would reverse. In order for voltage gradients between the midpoint regions to be as they were prior to splitting R1 and R2 apart, the voltage drops across R1a and R1b must be proportional to the voltage drops across R2a and R2b. Expressed in ratio form, the following relationship can be established:
( R1a / R1b ) = ( R2a / R2b )
The net series resistance for branches 1 and 2 above were the same ( 10 Ω ), but this need not be the case to establish balance. Let’s now assume that R1a and R1b are 10 Ω resistors, and R2a and R2b are both 20 Ω resistors. Even though less current will travel across path 2, the voltage-divider formula confirms that a 50% voltage drop will still occur in both branches. Coulombs of charge travel at a faster rate across path 1, but at each path’s midpoint region, no energy differential exists:
V1a = ( Vs )[ R1a / ( R1a + R1b ) ]
V1a = ( 10 V )[ 10 Ω / ( 10 Ω + 10 Ω ) ]
V1a = 5.0 V
V2a = ( Vs )[ R1b / ( R1a + R1b ) ]
V2a = ( 10 V )[ 20 Ω / ( 20 Ω + 20 Ω ) ]
V2a = 5.0 V
Thus, the balanced Wheatstone bridge is established by fixing the R2a and R2b resistors along path 2 and allowing the R1b resistor to be a variable resistor:
Rx = ( Rv )( R2a / R2b )
The R2 to R4 ratio is called the scale factor of the circuit. Once a scale factor of known value is established, a variable resistor is adjusted until a zero-volt ( Vout ) reading is established across the Wheatstone bridge resistance Rwb. With Rv established, the Rx resistor value for a balanced Wheatstone-bridge circuit may be determined.
Q: A Wheatstone bridge has two fixed resistors, a variable resistor, and a resistor of unknown value as shown:
A voltage reading of zero is established when the variable resistor ( R3 ) is set at 7.5 Ω. What is the resistance value of Rx?
A: Rx = ( Rv )( R2 / R4 )
Rx = ( 7.5 Ω )( 42 Ω / 7.0 Ω )
Rx = ( 7.5 Ω )( 6.0 )
Rx = 42 Ω
Q: What is the scale factor in the example above?
A: 6.0
RENAISSANCE PHYSICS 