Day 2- Introduction to DC circuits

When doing theoretical questions on paper, we always assume that wires in the circuits are perfectly conductive, without any resistance. In reality, wire has resistance proportional to the ratio of its length and cross section area.

For example, if we have to operated a device that is connected with a wire which is thousands of feet in length, the resistance in the wire will add up into a significant number that will effect the circuits if we did not put it into consideration.

In this experiment, we will consider a circuits such as this

We will consider few assumptions
1. Load is rated to consume 0.144W
2. Minimum voltage across the load is 11V
3. The battery has a capacity of 0.8Ahr and will remain constant at 12V

The goal of this experiment is to determine
1. The maximum permissible cable resistance while the circuit still work functionally
2. The maximum distance the battery and load can be separated if we use AWG#30 cable
3. Distribution efficiency
4.Approximate time before the battery is out of power

Since we know the power rating on the load, we are able to determine the load resistance

P=V^2/R
and by rearranging the equation and solve for R
we will get a R value of R_load=1000Ω


Next, we will set up the experiment as shown in the schematic
 We will first set up the power supply to 12V, and by increasing the cable resistance until the voltage drop to 11V, that will be the maximum resistance of the wire

The yellow resistance box will represent the resistance of the cable and the part located in the bottom left corner of the picture will be the 1000Ω, which is the load

We will measure the voltage and the current at different location as shown below

Next, we will measure any componets before assembling the circuit and note the ratings

Parts	Nomial Value	Measured Value	Tolerance	Rating
power supply	12V	12.14		2A
R_load	1000	978	5%	0.25W

Next, we connect the circuit and measure the voltage across the battery, current, and the voltage acrosst he load.
we got

Measurement
V_Load	10.97V
I_batt	11.1mA
R_cable	101Ω

 After the measurement,  we disconnect the wiring place all the part to its original location


Few question that we are able to solve after do some calculation


a. The time to discharge the battery is given by the formula Amp-hr=amps*time
where we know he Amp-hr, and the amps, we can calculate the time, which is 72.07 hrs


b. To calculate the distribution efficiency, we need to find the power of the load, the cable, and the power that is supplied by the power supply
To find the efficiency is P-out/(P_out+P_lost)

P_out	0.1205W
P_in	0.0125W
efficiency	90.60%

C Are we exceeding the power capability of the resistor box, which is 0.3W
NO, becuase P=I^R=0.0124W

D Given that the resistance of AWG#30 wire is 0.3451Ω/m, determine the maximum distance b/w the battery and the load
Since we know the resistance of the cable and the linear resistance density, we can calculate the lenght will be 3.19m