Required Concepts from Electronics
Electric Circuit Symbols and Terminology
Here is an electric circuit diagram:
Notice these symbols:
 The symbol for a voltage source
(this could be for example a battery). Voltage sources generate a force or "pressure"
which causes electric currents to flow in the circuit. The voltage source shown is rated
at 10 volts (abbreviation V). This means that the voltage or "pressure" is 10 volts higher
at the + or wideline side of the battery than at the  or narrowline side of the battery.
As a result, electric charges are given a force in the upward direction
(from the  side to the + side).
 The symbol for a resistor.
Resistors are devices that impede the flow of electric current. The resistor at the lower
right has a resistance of 55 ohms (abbrev ).
 The symbol for a wire. A wire is assumed
to have no resistance.
Principal Nodes or Junctions and Branches
In this diagram we define:
 principal nodes or junctions  These are points where 3 or more wires meet.
This circuit contains 4 of them, denoted N1, ..., N4. Notice the
symbol for a principal node.
 branches  A branch is any path in the circuit that has a node at each end and
contains at least one voltage source or resistor but contains no other nodes. This circuit
contains 6 branches, denoted B1, ..., B6.
(Note: If branch B4 did not contain a resistor then it could be deleted and nodes
N2 and N3 could be considered one and the same node.)
Electric Current
Electric charge flowing in a branch in a circuit is analogous to water flowing in a pipe.
The rate of flow of charge is called the current. It is measured in coulombs/second
or amperes (abbreviation A) just as the flow rate of water is measured in litres/second.
Water is incompressible, which means that if 1 litre of water enters one end of a length of pipe
then 1 litre must exit from the other end. The situation is the same with electric current.
If the current is 1A at a certain point in a branch then it is 1A everywhere else in that branch.
Kirchhoff's Current Law
An immediate consequence of this is Kirchhoff's Current Law. Kirchhoff's current law states
that the sum of the currents flowing into a node equals the sum of the currents flowing out of the node.
Here is an example:
This diagram also shows how we draw an arrow on the branch to indicate the current flowing
in the branch.
Electric Voltage
Electric current is the flow of electric charges. Electric voltage is the force that causes
this flow. Just as a pump pushes a "plug" of water through a pipe by creating a pressure
difference between its ends, so a battery pushes charge through a resistor by creating a voltage
difference between the two ends of the resistor. The picture shows the analogy:
This diagram also shows how we draw an arrow beside a resistor or any other device to indicate
a voltage difference between the two ends of that device. The arrow head is drawn pointing
to the higher voltage end.
Ohm's Law
We have just seen that a voltage difference between the two ends of a resistor causes a current
to flow through the resistor. For many substances the voltage and current are proportional.
This is expressed by the formula:
V = I R,
This equation is called Ohm's law and any device that obeys it is called a resistor.
V is the difference in voltage between the two ends of the resistor measured in volts,
I is the current through the resistor measured in amperes, and the proportionality
constant R is the resistance of the resistor measured in ohms. Given any two of these quantities,
Ohm's law can be used to find the third.
Kirchhoff's Voltage Law
Just as the water pressure drops in a garden hose the farther one moves away from the tap,
so the voltage changes as one moves around a circuit away from a voltage source.
Kirchhoff's Voltage Law states that:
Around any closed path in an electric circuit, the sum of the voltage drops through
the resistors equals the sum of the voltage rises through the voltage sources.
A closed path is a path through a circuit that ends where it starts.
Problem: Resistors in Series. Use Kirchhoff's voltage law and Ohm's law to find the value
of the unknown resistor R if it is known that a 2 ampere current flows in the circuit.
Solution: Let's follow the current as it flows clockwise around the circuit. If we start
at A and assume the voltage there is 0 then at B the voltage must be 10 volts because
the battery behaves like a pump that creates a higher pressure at the + side than the  side.
At C the voltage is still 10 volts but it drops going to D through resistor R,
and drops again going to E through the 2 ohm resistor. In fact it must return to 0 volts
since A and E are at the same voltage (voltage does not change along an ideal wire
that has no resistance).
Using Ohm's Law in the form V = I R we find that the I R (voltage) drop across the 2
resistor is (2 A) * (2
) = 4 V. Then by Kirchhoff ' s Voltage Law the
I R drop across the unknown resistor is 10 V  4 V = 6 V.
Again using I = 2 A, Ohm's law in the form R = V / I gives R = 3
.
The results are shown to the right.
Notice the directions of the voltage arrows across each of the devices.
Also notice that the voltage drops across the two resistors are proportional to their resistances.
This is called the Voltage Divider Rule. This rule is useful in many situations.
Suppose that we replaced the above circuit by the one shown to the right and didn't know what
was inside the "black box" but did know that the current flowing into the black box was 2 A and
that the voltage across it was 10 V. Then Ohm's law, R = V / I, would tell us that the black
box had a resistance of 5 .
Notice that this is exactly the sum of the two resistances in the original circuit.
This is true in general: two resistors R_{1} and R_{2} in series
may be replaced by a single equivalent resistor R_{eq} whose resistance is
the sum of the two resistances:
R_{eq} = R_{1} + R_{2}.
Problem: Resistors in Parallel. Find the value of the total current I_{T}
flowing into the parallel circuit.
Solution: Since there is no voltage drop along a wire, 20 V appears across each resistor.
Using Ohm's Law in the form I = V / R we find:
I_{1} = 20 V / 5 = 4 A,
and:
I_{2} = 20 V / 10 = 2 A,
and by Kirchhoff ' s Current Law I_{T} = 4 A + 2 A = 6 A.
Notes:
Suppose that we replaced the above circuit by the one shown to the right and didn't know what
was inside the "black box" but did know that the current flowing into the black box was
I_{T} = 6 A and that the voltage across it was 20 V. Let's call its resistance
R_{eq}. Ohm's law in the form I=V/R says that:
But when we could see inside we had:
Comparing these two expressions for I_{T} gives (after cancelling out the common
factor of 20 V):
This formula is true in general: two resistors R_{1} and R_{2} in
parallel may be replaced by a single equivalent resistor R_{eq} given by the formula:
Return to Linear Algebra and Electricity
