Introduction to DC Circuits

Pictures will be uploaded soon

Introduction: 

This lab is done in order to represent real life scenario in which equipment may have to be tethered anywhere from hundreds to thousands of feet.  In this lab we have to create a circuit with a battery and a load. We will run multiple tests to the circuit and determine what problems can arise. 

 

Steps: 

Given Assumptions

• A "Load" is rated to consume .144w when supplied 12V

• A miminimum load of 11V is required

• Constant battery voltage of 12V(approx) with a capacity of 0.8Ahr

Firstly, the theoretical value of resistance must be calculated.  

R=V^2/P               R=12^2/.144           R=1000ohms 

We then set up the experiment, connecting the equipment into a series and using a DMM for the readings as shown in the diagram. We set the power supply to 12V, and slowly increasing the resistance on the resistance box until the voltage drops to 11V, which will determine the max permissible cable resistance in the wire. 

Next, measure components 

Components      Nominal Value      Measured Value     Within Tolerance     Wattage

Power Supply          12V                12.09V                         Yes           1/8watt

RLoad                     1000ohm         985ohm                        No(1%)           1watt    

Next we performed the experiment

Measurement:

VLoad = 11V

IBatt = 12.1mA

RCableTot = 98Ohm

Data Calculations:

a)Amp - hr = amps x time

  .8 = .0121 x t

  t = 66.66 hrs

b)P = (I^2)R     

  Pcable = (.0121^2)(98) = .0143Watt    

  Pload = (.0121^2)(985) = .144Watt

   Efficiency = (Pout)/(Pout + Plost)100 = (.144)/(.144+.0143)100 = 90.96% efficient

c) Are we exceeding the limits of the resitor box? No

d) Given that the resistance of AWG#30 wire is 0.3451Ω/m, determine the maximum distance between the battery and the load. 

The total length of the wire came out to be 283.97 meters, but usable distance is 142 meters.

e) In the case of the robo-sub project running #28 wire with a 60 foot tether at 20mA 5V. The minimum power required to run the application is 2.6V and 5V is the max the wire can handle.

5V-2.6V = 2.4V left for the wire

V=IR        2.4 = .02(R)     R=120 Ohms
120Ohms * 1foot/.0764ohms = 1570.68 feet maximum for the wire length

f) We are sending 48 volts at 10amps but need at least 36 volts to the sub. So a maximum of 12 Volts can be lost but still safe for regulation.

12V = 10R     R=1.2 ohms

1.2ohms/60ft = .02 ohm/foot minimum

Therefore a minimum of 22 Awg is required.