Laboratory #3 Voltage Dividers

This experiment is mainly about making a electrical circuit which let the load part work in a certain range of voltage, so it could work properly.

 Before we set up the circuit, we did all the calculations that we need to get the sources. Reqmax=1k ohms, Reqmin=333.333 ohms. Because we want Vbus be the max so the resistance should be the max, so Vbus min need to use the Reqmin. The equation which i set up for Vbus max is [(Vs-5.25V)/Rs]*1000ohms=5.25V, Vbus min=[(Vs-4.75V)/Rs]*333.333ohms=4.75V. According to the two equation, we got Vs=5.54V, Rs=55.55 ohms. After we knew the resistance and voltage, we could know the Ibus, max=Vs/(Rs+Reqmin)=0.0142A, Ibus,min=Vs/(Rs+Reqmax)0.005248A. It's the time to set up the circuit which as the image that R1, R2, R3 were parallel and serious with Rs. We measured the resistors that were all 988ohms, the power supply was 6.07v, and the Resistor box was 56ohms. As what I read from the power supply and the resistor box, the maximum current were both within the capability.
The calculation of the power with two loads is P= Ubus*Ibus=5.35*10.89*10^-3=0.0583W. To figure out the actual percentage in load voltage variation, we need to first find out the Vaverage=(5.65V+5.05V)/2=5.35, so the %Diff=[(5.65-5.05)/5.35]*100%= 3.21%.
If we add the forth 1kohms in the load, we should get [5.54/(55.55+250)]*250=4.53, and we tested which i got closed number that was 4.95V.
To make the load voltage variation to 1%, we could just change the Rs to make it happen. We actually tested in the circuit. We just lower 23ohms from the Rs, so the Rs=33ohm, and the voltage variation is zero percent. The voltage was just 5.35V which is the Vaverage.