In a previous exercise, we saw how the addition of a stiff voltage divider to a two-resistor series circuit lowers the voltage ( V ) drop across the lattermost resistor. We are now ready to examine this phenomena with a circuit that contains unequal resistor values:
Q: a. What is the unloaded output voltage?
b. What are the loaded output voltages when RL has a value of 10 kΩ
and 100kΩ?
A: We begin by determining the voltage across R2 using the familiar voltage-divider formula:
V2 = [ R2 / ( R1 + R2 ) ]( Vs )
V2 = ( 10 kΩ / 14.7 kΩ )( 5.0 V )
V2 = ( 0.68 )( 5.0 V )
V2 = 3.4 V
By attaching a load to the midpoint of R1 and R2, current ( I ) will be split up. Since less current will then travel across R2, the voltage drop across R2 will decrease in value:
V2before = IbeforeR
Ibefore > Iafter
Vafter < Vbefore
We must now use the double-reciprocal method to determine the equivalent resistance downstream of the midpoint:
R2∥RL = [ R2RL / ( R2 + RL ) ]
R2∥RL = [ ( 10 kΩ )( 10 kΩ ) / ( 20 kΩ ) ]
R2∥RL = 5.0 kΩ
Via usage of the voltage-divider formula, the voltage drop across V2 can be determined:
Vout ( loaded ) = [ R2∥RL / ( R1 + R2∥RL ) ]( Vs )
RTotal = 5.0 kΩ + 4.7 kΩ = 9.7 kΩ
Vout ( loaded ) = ( 5.0 kΩ / 9.7 kΩ )( 5.0 V )
Vout ( loaded ) = ( 0.52 )( 5.0 V )
Vout ( loaded ) = 2.6 V
If we increase the value of RL, electrons will have a more difficult time traveling across it. Thus, more electrons will travel across R2, and the voltage across R2 will increase:
R2∥RL = [ ( 10 kΩ )( 100 kΩ ) / ( 110 kΩ ) ]
R2∥RL = 9.1 kΩ
RTotal = 9.1 kΩ + 4.7 kΩ = 13.8 kΩ
V2 = ( 9.1 kΩ / 13.8 kΩ )( 5.0 V )
V2 = ( 0.66 )( 5.0 V )
V2 = 3.3 V
Bonus 1: What is the voltage drop across R1?
5.0 V – 3.3 V = 1.7 V
Also,
V1 = ( 4.7 kΩ / 13.8 kΩ )( 5.0 V )
V1 = ( 0.34 )( 5.0 V )
V1 = 1.7 V
Check:
Vs = V1 + V2
Vs = 1.7 V + 3.3 V = 5.0 V
Bonus 2: What is the voltage drop across R1 when RL = 10 Ω?
RENAISSANCE PHYSICS 