Introduction to DC circuits

For this experiment, our goal is to model a tethered remote operated vehicle (ROV) to find:

1) The maximum permissible cable resistance required for the load to function properly.

2) The maximum distance between the battery and load using AWG #30 cable.

3) The distribution efficiency.

4) The approximate time before the battery discharges.

We will consider this circuit model with the following assumptions:

1) Load is rated to consume 0.114W when supplied 12V.

2) Load's minimum voltage is 11V.

3) The battery will be at a constant of 12V and has a capacity of 0.8 Ahr.

By using P = V^2/R, we found the R_Load = 1000 Ω.

As for R_CableTot, a variable resistor box of a power rating of 1 W was used.

For the power supply, V_Batt was set to 12 V, but the measured value was 12.34 V with a max supply current of 2 A.

Color Code				Nominal Value	Measured Value	Within Tolerance	Wattage
Br	Blk	Blk	Br	1000 Ω	989 Ω	No	1/8 W

Next, the schematic was built using a black resistor box as the variable resistor.

 

By connecting the DMM to the load in parallel, we were able to change the variable resistance until V_Load was close to 11 V. To finish the experiment, voltage, current, and the variable resistor was measured, and the circuit was disassembled and put back in its original place.

V_Load = 10.99 V

I_Batt = 11.22 mA

R_CableTot = 105 Ω

As for data calculation,

a) Since Amp-hr = amps x time for a 0.8 Ahr battery,

Time to Discharge = 71.2 hr

b) By finding the power to the load, and cable, we can find power efficiency using the formula 

P_out/(P_out+P­_lost).

P_out = 0.1222 W

P­_lost = 0.0132 W

Efficiency = 90.25% 

c) We are not exceeding the power capability of the resistor box of 1W because the power is 0.0132W.

d) Given the resistance of AWG #30 wire is 0.3451 Ω/m.

The total wire length is 105 Ω * 1m/0.3451Ω = 304 m.

Since the ROV needs a forward and return conductor,

304/2 m = 152 m is the maximum distance between the battery and load.

e) For the robo-sub using AWG #28 with a 60 foot tether. Since the signals through the tether are 20mA 5V and high is 2.6-5V, we are left with 5-2.6 = 2.4V. Then the resistance of the wire is 2.4/20 V/mA = 120 Ω. Given AWG #28 has a 0.0649 Ω/ft, then the length of the wire is 1849 ft.
The longest distance is 1849/2 = 924.5 ft.

f)Given 48V at 10A and ensuring the sub receives 36V, only 48-36 = 12V remain. Then the resistance = 12/10 = 1.2 Ω. Assuming that the tether is 60 ft, then 1.2/60 = 0.02 Ω/ft. Therefore, the minimum cable should be AWG #22.