In this experiment, we assume some value for resistance, power and permissible voltage drop for a load. We want to determine the permissible cable resistance so that the load can function normally. We would like to be able to determine the maximum distance the battery and the load can be separated if we use different types of cable. We will also determine the battery life and the efficiency.
Experiment
We first creates assumptions:
- Load is rated to consume 0.144W when 12V is supplied
- The voltage drop across load has to be more than 11V
- Battery voltage remain constant at 12V and has capacity of 0.8Ahr
Once the assumptions were made, the theoretical value for R can be calculated as follow:
R = V² / P
R = 12² / 0.144W
R = 1000Ω
After the theoretical resistor was calculated, a resistance box with power rating of 0.3W is used to become the representation of resistance of cable. The power supply rating have a maximum supply voltage of 12V and max current of 2A. The resistance box was also adjusted so that the voltage drop across the load is 11V.
The resistor used for the experiment are listed below:
|
Element
|
Nominal Value (Ω)
|
Measured Value (Ω)
|
Tolerance?
|
Wattage
|
|
Load
|
1000
|
980
|
Yes
|
0.25
|
|
Cable
|
88
|
87.3
|
Yes
|
0.3
|
The voltage of the power source was measured and comes out to be 12.09 ± 0.005 V
Once every element have been measured and calculated, the circuit was set up as follow:
 |
| Setup of the lab and the measurement of voltage drop across load |
As seen in the picture and also several measurement,some data are achieved:
V_load = 11.1 V
I_battery = 11.3 mA
R_cable = 87.3 Ω
After the data is taken, battery life can be determined by using:
Battery capacity = current*time
Time = Battery Capacity / Current
= 0.8 A-hr / 0.0113 A
= 70.80 hr
Power was also calculated in this lab and turns out as follow:
P = V²/R = I²R
Power to the load = 0.126 W
Power to the cable = 0.0111 W
Once the power is calculated, the efficiency is also calculated using the following formula:
η = [ P_out / (P_out + P_lost) ] * 100
η = 91.9%
Questions
- In one of the question, the total power is not exceeding the power capability of resistor box since the total power used is around 0.1W but never exceed the 0.3 W which is the power caps.
- Using the data taken, and with the resistance of AWG # 30 is 0.3451 Ω / m, we can find the maximum distance for the cable. L = 87.3 * 0.5 / 0.3451 = 126.49 m
- In the question about robosub project, AWG # 28 wire is used with resistance of 0.0764Ω / ft. This particular circuit have nominal rating of 2.6 - 5 V and 20 mA as provided in the question. To determine the maximum length of wire that can be used, we first must calculate the maximum resistance in cable R_cable = 2.4V / 0.02 A = 120 Ω. Once the resistor is calculated, the length of the cable can be determined as L = 120 Ω * 0.5 / 0.0764 Ω/ft = 785 ft.
- In the next question, we are sending 48 volts at 10A down the sub. 36 volts have to reach the sub and minimum cable gauge has to be determined for a 60 ft tether cable. Since there can only be a 12V drops, the resistance can be figured out to be R = V / I = 12V /10A =1.2Ω. Once the resistor is determined, the cable gauge can be tested and the minimal gauge required for 60 ft tether cable appears to be the AWG#24 since by using it L = 1.2 / 0.0302 = 39.7 ft.
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