Introduction to DC Circuits

The goal in this experiment is to measure the resistance of a hypothetical couple of cables that are represented as a variable resistor box. A fixed resistor will act as a resistor load in the circuit. The resistor box will be adjusted to get a value for the cable resistance and then it will be compared to the following assumptions:

P = 0.144W at 12V

Proper operation at V>11V

Voltage source = 12V with capacity of 0.8A-hr

In this experiment we set up a circuit consisting of a power supply set in DC mode. We were provided with a composition fixed resistor of 1000 ohms which represented the load resistance. The resistor was then connected in series to a resistor box (cable resistance).

Circuit Diagram

Circuit Layout

Modeled cable resistance

 The theoretical value of the load resistor according to the assumption values is:

R = V^2 / P = 12^2 / 0.144 = 1000 ohms

Once we had the circuit set up, we placed an ammerter in series with the cable resistance to measure the current. We also put a voltmeter across the resistance load to measure the voltage through the load resistor.

The power rating for the resistor box was: 1 Watt with 1% accuracy.
For the power supply we got  a maximum voltage supply of 12V at a current of 2A.

The colors of the fixed blue resistor were: Brown - Black - Red which was the color code for 10*10^2 = 1000 ohms. We then measured the actual resistance of the fixed resistor using a multimeter to get 0.988+/-0.001 Kohms. The last color code was Brown which was a 1% tolerance (not within the regular 5% tolerance). The wattage was found to be 1/8 W.

Once the circuit layout was approved by the professor, we turned on the power supply.

Using a voltmeter we found the voltage across the load (fixed resistor) to be:
Vload = 10.99 +/- 0.01V

The current running through the circuit was:
Ibatt = 11.22+/-0.01A

The total cable resistance (resistor box) was modified to be 105 ohms

Using the current measured we now can estimate the amount of time it takes for the battery to supply the load before being discharged.

0.8A-hr = (11.22+/-0.01A) t
t =  0.0713 +/- .000064hrs or about 4.28 minutes.

To find the power to the load, that is the power out we take the voltage through the load and the load resistance;

P out= V^2 / R = (10.99+/-0.01V)^2 / (988+/-1ohms) =  0.1223+/-0.00020 Watts

The power to the cable (power lost) is obtained by multiplying the current out of the power supply and the resistance from the resistor box;

Plost = (I^2)*(Rtotcable) = ( 11.22+/-0.01A)^2 * (105ohm) =  13220+/-17 Watts

The efficiency is found by computing:

n = [Pout/(Pout + Plost)] = 0.1223+/-0.00020 / (0.1223+/-0.00020 + 13220+/-17)     =0.000925+/-0.0000019%

Given that the power capability of the resistor box is  1%, we can say that we are not exceeding the power capability of the resistor box.

If the resistance of AWG #30 wire is 0.3451ohm/m, determine the maximum distance between the battery and the load.

(988+/-1 ohms)/(0.3451ohm/m) =  2863+/-2.9m.