We did the lab to measure how many power loss in the cable. Before we did set this up, we measured the voltage of the power supply, the resistant of the load and recorded the data for the calculation at the later part which were 12.07V and 975 ohms. Then, we built the circuit as the image and read the voltage and the current of the circuit which the number were 11.96V and 12.5mA. According to the requirement, we changed the resistant of the cable to make the voltage of the road resistant drop 1 volt; then, we recorded the data again which were 88ohms, 10.96V and 11.45 mA.
After them all, we could do the calculation. Time of discharge: 0.8 Amp-hr= 0.01145A*t so the t=69.869 hours. Power of the load: P=V^2/R=10.96^2/975=0.123202W
Power of the Cable: P=I^2R= (0.01145^2)*88=0.011537W
The Efficiency= (Pout/Pout+P lost)*100%=0.123202/0.123202+0.11577*100%=91.43%
If the AWG #30 wire is 0.3451 ohms/m, so the maximum distance should be 88ohms/(2*0.3451)=127.499 meters. The power lost which could lost from the heat of the wires or the room temperature may consider.
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