A series circuit that contains two equal-value resistors ( R ) will split the amount of work ( J ) done by the charges equally:
Prior to arrival at R1, a coulomb of charged particles ( I ) will contain 10.0 J of energy available to perform work. After passing through R1, the charges will have 5.0 J of energy remaining. This energy is then consumed by R2, and the charged species flow to ground potential elsewhere within the circuit. What if, however, a resistive element were attached to the region between R1 and R2? What effect would this have upon the voltage ( V ) measured across R2? As shown in the diagram below, the circuit would be transformed into a series-parallel circuit:
Note: Please recall that, in reality, electrons have negative charges.
Prior to attachment of RL, the voltage at point A would be 5.0 V; however, with the introduction of a new circuit branch, the system’s resistance downstream of point A will lessen. Since the current now has two possible routes of travel, R2 is no longer as much of an energy barrier as it was beforehand, and the voltage reading across it will be less than 5.0 V, but by how much? We will first attach a 1.0 kΩ resistive load to point A and subsequently attach a 10.0 kΩ resistor in its place. In this way, we can see how much the voltage across R2 is affected by the loading process:
R2 ∥ RL = [ R2RL / ( R2 + RL ) ]
R2 ∥ RL = [ ( 1.0 kΩ )( 1.0 kΩ ) / 2.0 kΩ ]
R2 ∥ RL = 0.5 kΩ
We now have an equivalent resistance that is in series with R1:
The voltage across R2 ∥ RL may now be determined using the voltage-divider formula:
Vout = [ R2 ∥ RL / ( R1 + R2 ∥ RL ) ]( V )
Vout = [ 0.5 kΩ / ( 1.0 kΩ + 0.5 kΩ ) ]( 10.0 V )
Vout = ( 0.5 kΩ / 1.5 kΩ )( 10.0 V )
Vout = ( 0.33 )( 10.0 V )
Vout = 3.3 V
Percent Decrease = [ ( Initial Value – Final Value ) / Initial Value ] x 100%
Percent Decrease = [ ( 5.0 V – 3.3 V ) / 5.0 V ] x 100%
Percent Decrease = ( 1.7 V / 5.0 V ) x 100%
Percent Decrease = 34.0 %
Let’s now compare this result to one in which a 10.0 kΩ load has been attached to point A:
R2 ∥ RL = [ R2RL / ( R2 + RL ) ]
R2 ∥ RL = [ ( 1.0 kΩ )( 10.0 kΩ ) / 11.0 kΩ ]
R2 ∥ RL = 0.9 kΩ
Vout = [ R2 ∥ RL / ( R1 + R2 ∥ RL ) ]( V )
Vout = ( 0.9 kΩ / 1.9 kΩ )( 10.0 V )
Vout = ( 0.47 )( 10.0 V )
Vout = 4.7 V
Percent Decrease = ( 0.3 V / 5.0 V ) x 100%
Percent Decrease = 6.0 %
As a general rule, a stiff voltage divider is one that is at least ten times greater than the resistor parallel to it. And as we’ll see in part 2, R1 and R2 need not have equal values within the system above.
RENAISSANCE PHYSICS 