Voltage Dividers

In this experiment, we will be trying figure out the V_s and R_s of a variable resistor box, so that V_bus is within 5.5V < V_Bus < 6.25V

Procedure:

We start off with some calculations of the circuit with R_1, R_2, R_3 = 1000 Ω.

Afterwards, we built the circuit with the measured values.

Color Code				Nominal Value	Measured Value	Wattage
Br	Blk	Blk	Br	1000 Ω	986 Ω	1/8 W
Br	Blk	Blk	Br	1000 Ω	985 Ω	1/8 W
Br	Blk	Blk	Br	1000 Ω	982 Ω	1/8 W
Resistor Box				43 Ω	45.3 Ω	1 W
Power Supply				6V	6.00 V	N/A

The max bus current is within the Power Supply.
The max current is withing the power capability of the resistor box.

We recorded values from testing different loads at the same time.

Config	R_eq	V_Bus	I_Bus	P_load
1 Load	981 Ω	5.48 V	6.19 mA	33.6 mW
2 Load	493 Ω	5.08 V	10.97 mA	55.7 mW
3 Load	329 Ω	4.55 V	14.61 mA	66.5 mW

Questions:
a. 2 Load calculation: P = VI = (10.97 mA)*(5.08 V) = 55.7 mW

b.

Load	Voltage (V)	Variation (%)
1 Load	5.48	-8.67
2 Load	5.08	-15.33
3 Load	4.55	-24.17

The voltages are not with in 5% because of the unregulated power supply. Before the experiment, the power supply was measured to give 6.00 V, but as the loads increase, V_s changes whichc hanges the values for V_Bus.

c. V_Bus,min = R_eq*V_s/(R_s + R_eq). New R_eq = 1/4 kΩ. V_s = 6.53 V, R_s = 45.5 Ω
V_Bus,min = 5.52 V
The new variation using V_s = 6V would be -7.92 %.

d. Variation by 1% = 5.94 V < V_s < 6.06 V
V_s = 6.12 V
R_s = 10.2 Ω