With these assumptions, we need to find the maximum distance the load and battery can be separated using the AWG #30 cable. From this, we need to find the resistance in the cable to ensure we meet the 11 V needed to operate this piece of machinery. In practice, we expect there to be a small drop in voltage and we must account for this, especially with the use of long cables.
The setup consists of a DC power supply measured at 12.10 V to represent the battery, a variable resistor box as the cable resistance, and a 1000 Ω resistor for the load, all hooked up in series. We included an ammeter in series to measure current and a voltmeter parallel to each of these elements to find the voltage.
The setup with the power supply, resistor, and variable resistor box. |
The equivalent circuit. |
Our ammeter measured a current of 12.4 mA. With this measurement, and the fact that the battery has a 0.8 Ahr capacity, we calculated that the time it takes to discharge the battery is 64.5 hours.
By using the formula, P = I2 R, we calculated the power in the load and the cable. Their values were 0.154 W and 0.0137 W, respectively. The measured power of 0.0137 W therefore does not exceed the power capability of the variable resistor box rated at 1 W (written on the side of the resistor box). With these values, the efficiency of the load comes out to be 91.8 %. This means that the remaining 8.2 % is lost to the cables.
Knowing that the maximum cable resistance can go as high as 89 Ω and assuming we are using AWG #30 wires with a resistance of 0.3451 Ω/m, the maximum distance from the battery to the load is 128.9 m in order to maintain the 11 V needed for the device to remain functional.
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