Introduction to DC Circuits

In this lab, we used the relationship between voltage, resistance, and current (V=IR) to solve a hypothetical problem involving a battery, load, and unknown length of wire.

The problem

We have a battery that supplies a constant voltage to the circuit for a limited amount of time, and a load in the circuit with a constant resistance. In any real situation, the wires necessary to make the circuit also add resistance to the circuit.  As the total resistance in the series circuit increases, voltage across the load decreases in order to maintain a constant current (current is constant across series elements and our battery voltage is constant). There is a minimum voltage for which this load will operate, and we're trying to find the maximum added cable resistance that corresponds to this minimum voltage.

The model

Our setup

We modeled the battery with a constant-voltage power supply, the load with a resistor whose value was known, and the unknown length of cable with a variable-resistance box. These three things, along with an ammeter, were set up in series; a voltmeter was set up in parallel with the load to catch the operating voltage corresponding to any given resistor box value.

What we found

Values:

Actual load resistance: 975 ohm

Resistor box power rating: 0.3 W

Max supply voltage (Vbatt): 12.09 V

Max supply current: 2 A

V across the load: 10.96 V

I through battery: 11.45 mA

R of cable: 88 ohms

We tweaked the variable-resistance box until the voltage across the load dropped to exactly 11v (our minimum). The added resistance that caused this was 88 ohms, and the resulting current was 11.45 mA.

Using the capacity of the battery (0.8 Amp-hours) and the current flowing through it, we were able to calculate that the battery should sufficiently power the load for 69.87 hours.

Using P=IV and its other forms, P=(V^2)/R and P=(I^2)R we calculated the power going into the load and the "wire" (resistor box), respectively.

P to load: 0.123 W

P to cable: 0.0115

We could then calculate the efficiency of this setup, which was 91.4%. This means that 91.4% of the power goes into the load, and the remaining power goes into the wire.

Based on the resistance per unit length of AWG #30 wire, we were able to calculate the length that corresponds to the 88 ohms, and thus the maximum distance the load can be placed from the source while still maintaining a functional voltage.

Max distance: 128 meters