In this lab we are given a theoretical problem and asked to solve it, both analytically and experimentally. We are given a few assumptions:
Our load is rated to consume .144W at 12V.
The load will operate as long as the voltage is greater than 11V.
Battery voltage remains constant at approximately 12V and the battery has a capacity of .8Ahr.
We must determine the maximum cable resistance, the maximum distance the battery can be seperated if we use AWG #30 cable, the distribution efficiency, and the approximate time before the battery dies.
We begin by modeling the system with laboratory equipment.
From the given constraints we can find that the load resistance to be approximately 1000 ohms.
The resistor box is rated to be at 0.3W. Our power supply is rated to be at 12V and 2A max voltage and current, respectively. Our 1000 ohm resistor is rated to 1/8 W.
We find the resistance of our cable to be approximately 101 ohms and is within tolerance. our load resistance, which is being modeled by a 1000 ohm resistor is also within tolerance.
We measure our voltage battery to be 12.05 V.
We then build our circuit with all the circuit elements using alligator clips. Our final readings for load voltage is 10.99V, current across the battery is 11.10A, and equivalent resistance of the cable being 101 ohms.
Questions:
a. Our time to discharge is calculated to be Ahr/current = 72.1 hours.
b. The power to the load is current*voltage = .125W. The power to the cable is .013W. The efficiency is given by p_out/(p_out+p_lost) and results in an efficiency of roughly 90%.
c. No, the resistor box is rated to .3W and we are only absorbing .013W through the cable.
d. 101 ohm/(2*0.345 ohm/m)= 146m.
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