Applications of systems of linear equations
Circuits, part 1 (motivation)
This is a simple electric circuit:
It consists of
• A battery (symbolized on the drawing by
or
and in equations by the variable V ), which is a source of
electric potential. Electric potential (or potential difference) is measured in Volts (V ).
• A resistor (symbolized on the drawing by
in Ohms (Ω).
, and in equations by the variable R), with resistance measured
• A current (symbolized by an arrow, and in equations by the variable I) flows through the circuit. Typically,
you’ll need to sketch in the current, and make a guess about what direction it is flowing in. The convention your
book uses is that current flows OUT the long side of the battery symbol. Current is measured in Amperes (A).
These quantities are related by Ohm’s Law, which states
V = IR
If the battery is a 9V battery, and the resistor is a 2Ω resistor, what is the current in that circuit?
A more complicated circuit: Let V = 15V , R1 = 3Ω, R2 = 5Ω, R3 = 2Ω.
At each junction in the circuit (labels B and D), the current divides off. In a circuit like this, we may have to guess
at the direction the current takes on each branch. (It’ll turn out to be OK if we guess wrong – the problem corrects
itself). Circuits with junctions obey Kirchoff ’s Laws:
• All the current flowing into a junction must flow out of it.
• The sum of the IR terms around a closed path (i.e., the voltage across each resistor) must equal the total voltage
(i.e, that of the battery) in the path.
By applying Kirchoff’s Rules, we can set up a system of equations that can be solved for the unknown currents. Do
the following:
Assign directions to the currents I1 , I2 , and I3 .
Set up the junction rules (look at each junction – sum of currents in must equal sum of currents out):
Pick a closed path to walk around. As long as you are walking in the direction of the current, the product IR is
positive on the left hand side of the equation. If you walk against the current, it goes in as a negative. On the right
hand side, if you walk across a battery from short to long, it goes in as a positive. If you walk across it ‘backwards’,
it goes in as a negative.
• Path ABDA:
• Path BCDB:
• Path ABCDA:
You now have a system of equations. At the moment, you’ve got 5 equations total (although it’s immediately evident
that the two junction equations are really the same equation), and three unknown quantities. The questions are now
(1) how many equations do we really need? and (2) HOW DO WE SOLVE THIS THING?