Introduction to DC Circuits

When we discuss electronic circuits, most often we refer to either direct current (DC) or alternating current (AC). The properties of a circuit can be expressed by Ohm's law.

V = IR

The most obvious of concerns when engineering simple circuits come from selecting the amount of current or voltage for the required source. Simple circuit diagrams explains perfectly well how to build circuits and describe the voltage, current, and resistance of each element. 

The equipment we wish to model has a load of .144W when supplied 12V. The battery will have a voltage of 12V with a capacity of 0.8Ahr.

 Obviously, the equipment that we will be powering will have a resistance as well, which we emulate by selecting a resistor with the same resistance of our equipment. Using Ohm's Law we find that the resistance of the load is 1000 Ohms.

A simple circuit diagram may very well be too "simple" by not explicitly stating the material, diameter, or even length of the wire. With considerable distances wires have considerable resistances. Thus, resistance should be as much of a concern as voltage and current when engineering circuits.

R = pL/A

Since we do not possess such long wires, we can emulate one by using a resistance box. A simple calculation using the expression above could give us an idea of what kind of current will run through the wire created by the voltage source.

We find the measurements to be:

Voltage of load: 11V

Current of battery: 12.4mA

Resistance of cable: 89 Ohms

With the specifications of our equipment, the time it would take for our battery to discharge is 64.5 hrs.

With the Joules Heating Law,

P = V^2/R = I^2R

we can find the power lost by the cable and consumed by the load which we calculated to be:

lost by the cable =  0.0137W

consumed by the load = 0.153W

efficiency of the circuit = 91.78%

Our resistance box has a power rating of 1W which can perfectly handle the power lost by the cable. Had we be handling AWG #30 wire which has a resistance of 0.3451 Ohms/meter, we could have a wire that is about 260 meters.

In the case of the robosub, using AWG #28 wire (resistance of .0764 Ohm/foot) and nominal rating is 0 -0.4 V for low and 2.6-5V for high. So, if we leave 2.6V for the load and 2.4V for the resistance of the cable, then we have sufficient voltage for our circuit. The current that is a 60 foot tether is 20mA. We can use Ohm's law to solve the problem for the maximum length that this wire can be to keep in proper communication with the sub.

2.4V = (20mA)(0.0764 Ohm/foot)(# of feet)

# of feet = 5/[(.02)(.0764)]

# of feet = 1570.68 ft

When sending 48V at 10A through a 60 ft tether the calculation for a proper guage for the wire required can also be solved using Ohm's laws,

48V = (10A)(60ft)(Resistance)

Resistance = 48/[(10)(60)]

Resistance = 0.08

We look this number up on a chart and find that the guage most appropriate is between #28 and #30 guage wire.