Intro to DC Circuits

PROBLEM
Today in lab, we constructed a circuit similar to that above.
The problem is to determine:
   - maximum cable resistance required for load to function normally
   - maximum distance batter and load can be separated using AWG #30 cable
   - distribution efficiency
   - approximate time before the battery discharges

METHOD
We first determined the theoretical value for Rload, given that the load consumes p = 0.144W when supplied   v = 12 V, where R = (v^2)/p = 1 kohm. We took measurements of all components before assembling the circuit, and recorded the data. To measure the current, we placed an ammeter in the circuit; a voltmeter was used to measure the voltage across the load. The circuit was modified:



The RcableTot resistance was adjusted until the voltage across RcableTot read 11V; this value was recorded. Values for voltage across the load, Vload, and current, Iamm.
  - Vload = 10.8 V
  - Ibatt = 11.4 mA
  - RcableTot = 97 ohms

QUESTIONS/CALCULATIONS
A)  Given an Amp-hour rating of 0.8Ahr for the battery, and that Ahr=amp*time >> time=Ahr/amp.
  - Time to Discharge: t=70 hrs

B) The power to the load(power out) and the power to the cable(power lost) need to be determined to find the distribution efficiency.
  - Pload = Pout = 0.1254 W
  - Pcable = Plost = 0.013 W
Efficiency = (Pout/(Pout+Plost))*100 = 91% EFFICIENCY

C) We are not exceeding the power capability of the resistor because our efficiency is reasonable. Accordingly, I can infer that the power capability of the resistor is not exceeded.

D) Resistance of AWG#30 wire is 0.3451 ohm/m. Max distance between battery and load,           dMax=R/p=97/0.3451= 281.1 m

E)

F)