Introduction to DC circuits

Design Problem: A piece of electronic equipment is to operate at a considerable distance from its power source. The power will be transferred through a length of cable, with a maximum length to be determined.

Load is rated to consume .144 watts at 12 volts.
Load will continue to operate as long as the voltage does not drop below 11 volts.
The battery supplying the load has a charge of .8Ahr.

Question:
Determine...
Maximum possible cable resistance
maximum length of cable is using 30 AWG cable
distribution efficiency
approximate time before the battery discharges
The circuit in question can be represented as a battery, attached two a load with two resistors. Like so.

 It is easier to analyze this circuit if we combine like elements in series. Thus cable one and two can be represented as a single resistor.

In order to model this circuit, we used a power supply and a decade resistor box, seen below.

Based on the initial conditions, the load will be represented by a 1kOhm resistor. The banana cables, ammeter and resistor box had internal resistance of 1.8 Ohms, so this total was added to the final measurement. The power supply had a maximum voltage rating of 12 Volts and a maximum current rating of 2 Amps. 

As the power supply is non-ideal, we measured its output voltage and it was found to be 12.11 V. 

The decade resistor box was switched to read 84 Ohms, and the voltage across the supply was found to be 11.04 Volts. The true value of our 'load' resistor was found to be 975 Ohms. In this configuration, the ammeter read 18.37 milliamps. Based on this data, we determined that it should take 40.8 Hours for the battery to discharge. (In reality the battery would lose voltage as the charge was transferred, so these results are merely theoretical.)

Surprisingly,  these readings indicate that the load is consuming .201 watts, which is greater than its consumption at 12 Volts. However, this calculation neglects the current leeched by the parallel Volt meter. An ideal resistor of 975 Ohms dissipates .125 Watts at 11.04 Volts. Thus the Volt meter must be responsible for .076 watts of the calculated value. Additionally, 12.11 Volts and 18.36 mA corresponds to a functional resistance of 653 Ohms for the entire circuit.  Working backward, this indicates that the shunt resistance in the volt meter is approximately 1350 Ohms. This resistance is comparable to the resistance of our load, and thus including the voltmeter in the circuit skews the results.

 

According to Ohm's law, a 975Ohm resistor should pull .1128 A and this corresponds to a cable resistance of 98.4 Ohms. The inclusion of the Voltmeter changed the results of the experiment by 12.8 percent.

Using the theoretical results, the efficiency is 90.7%, and the length of cable equal to 142.56 meters.

Optional questions)

e)Communication through the cable is Transistor Transistor Logic, and this requires a  0-.4 Volt logic low and a 2.6-5.0 Volt logic high. If the load device requires 20 mA to operate and the signals are 5 volts at the input, how long can the cable be before the signal will be unable to trigger TTL at the other end? Additionally, the communication wire is approximately 28 awg.

Using the same schematic from earlier, the load is 20 mA at 5 Volts, which corresponds to a load resistance of 250 Ohms. Again, the cable is reduced to a single resistance element. If the current remains at 20 mA, and the load can function down to 2.6 V then the voltage drop in the cable can be as high as 2.4 volts. This corresponds to a resistance of 120 Ohms. Using the table for resistance as a function of wire gauge:

(more info can be found here http://hyperphysics.phy-astr.gsu.edu/hbase/tables/wirega.htm)

and a resistance multiplier for the step from 24 awg to 28 awg of 2.53, we find that the resistance of 28 awg is 213.7 mOhms per meter. This corresponds to a maximum cable length of 280 meters before TTL communication can no longer be transferred.

f) Using the same configuration, an electric ROV (remote operated vehicle) can get enough power to run its motors. Assuming that the ROV needs 10 A at a minimum of 36 V in order to operate, what is the maximum cable gauge that can be used with a 48V power supply and a 60 foot cable?

Repeating the calculations from the previous problem it is determined that the resistance per foot needs to be on the order of .001 Ohms. This corresponds to approximately 10AWG cable.