Lab 1: Introduction to DC Circuits

Introduction:
The purpose of this experiment is to study the basic elements in a DC circuit. This experiment is designed for students to determine the maximum resistance of the cables so that the voltage of the power supply can be remained greater than the requirement of the "Load" when it is delivered to the "Load". Students will theoretical calculate the value for the resistance of the "Load" and theoretical maximum resistance of the cable. Then students will build a circuit to experimental measure the maximum resistance of the cable in which the voltage across the "Load" is still greater than than the requirement. The "Load" can operate properly when the voltage across it is greater than 11V. The power supply is remaining constant at 12V and has capacity of 0.8Ahr. The "Load" is rated to consume 0.144W when supplied 12V.

Procedure:

Using the the rated values of the "Load", and the equation: P = V²/R, we calculated the theoretical value for R_load is R =V²/P  = (12)²/0.144 = 1000 Ω.

Set up the circuit similar to the diagram. We have determined the maximum permissible cable resistance before the voltage across the "Load" equals 11V is 94 Ω.

Measure the components of the circuit:

Resistor box power rating: 1W

Max supply voltage: 12 V

Max supply current: 2A.

Color code	Nominal value	Measured value	Within Tolerance	Wattage
brown black black brown brown	1000 Ω	980 +/- 2 Ω	no	1/8 W
Resistance box	94 Ω	94 Ω +/- 1 Ω	yes	1 W

 Connect the the ammeter and voltmeter to the circuit and make measurement. Adjusted the resistance box until the voltmeter read about 11V.

V_load = 11.00 +/- 0.02 V

I_batt = 11.44 +/- 0.02 mA

R_cable = 84 Ω

Calculation:

a)

Time to discharge = (Amp-hr) / (measured Amp) = (0.8 A-hr)/(11.44 mA) = 69.93 hr

Uncertainty:

  u = (0.02 mA / 11.44 mA) * 69.93 hr = 0.12 hr

Time to discharge = 69.93 +/- 0.12 hr.

b)

Power to the load:   P_out  = I*V = (11.44mA)*(11.00 V) = 0.1254 +/- 0.0005 W

Power to the cable:  P_lost  = I²R = ((11.44mA)^2) * (84Ω) = 0.011 +/- 0.0004 W

efficiency: n = ( P_out )*100/(P_out  + P_lost ) = (0.1254 * 100)/ (0.1254 + 0.011) = 91.9% 

c)

 No, we were not exceeding the power capacity of the resistor box. The resistance box power rating is 1W although we calculated the power to the resistance box in this experiment is only 0.011 W.

d)

Given the resistance of AWG #30 wire is 0.3451 Ω/m ; in this experiment, we measured that the maximum resistance of the wire is 84 Ω. Therefore, the length of the AWG #30 wire can be:

l = (84 Ω) / (0.3451Ω/m) = 243.4 m.

The wire consists of a forward and a return conductor, so the maximum distance between the battery and the load is: (243.4 m)/2 = 121.7 m.

e) 
According to the reference the composite video, the TTL signal does not reach to the nominal high rating. Therefore, we need to keep the signal at the low rating which is 0-0.4V. The signals through the tether are 20mA 5V, and the low rating is 0 - 0.4V. Thus, the maximum voltage can be lost in the cable is :
= 5V - 0.4V = 4.6 V.
The maximum resistance of the cable is = V/I = (4.6V)/(20mA) = 230Ω  
The AWG # 28 has 0.0764 Ω/ft, the longest length is = (230Ω)/(0.0764Ω/ft) = 3010 ft 

f)
We are sending 48V at 10A and we need to insure that at least 36V reaches the sub. Thus, the maximum voltages can be lost in the cable is = 48V - 36V = 12V.
The maximum resistance = (12V)/(10A) = 1.2 Ω.
The length of the tether is 60 ft. Therefore, R= (1.2 Ω)/(60 ft) = 0.02 Ω/ft.Therefore, the minimum cable gauge we can use is AWG 22 has resistance 0.0190 Ω/ft.