Download IV. Electrical Engineering

IV.
Electrical Engineering (circuits problem)
Electrical engineering is a broad description involving electrical power generation, distribution,
and use. All electrical circuits can be mathematically modeled using one or more circuit
elements including resistance, inductance, capacitance, voltage, and current sources. Let’s talk
about some of these elements.
Current (symbol i French word for intensite) is the flow of electrical charge which requires work
to move it from one point to another. We define current in the direction we do because the effect
was perceived sooner than the nature(i.e. electron flow). The unit of measure is Amperes (A).
Voltage (symbol V) is the change in electrical potential between two points(flow between tow
differently charged bodies). The unit of measure is volts (v).
Resistance (symbol R) impedes electron flow and is voltage divided by current. The unit of
measure is ohms ().
In 1827 Georg Ohm observed the relationship between voltage, current and resistance. He found
that voltage was proportional to current if the resistance remained constant.
Ohms’ Law: V = iR
What does this mean? When current flows in an electric circuit, free electrons collide not only
with each other but also with bound electrons in nonmoving atoms of the circuit, hence
resistance.
Aside Question: If a lamp is your resistance, why does it heat up?
Answer: Kinetic energy from the electrons bouncing off each other.
Aside Question: Why is shorting out an electric outlet dangerous? Hint, the body is 300
and the outlet is 110v.
Answer: Our low resistance drives a high current through our body (110v/300 =
36mA) which is enough to stop the heart beating (bentricular fibrillation). Only a few mA can
cause this to happen.
Work (symbol W) in electric circuits is the expenditure of energy.
W = Vit
Power (symbol P) is the rate at which work is done, the product of voltage and current. The unit
of measure is watts (W).
Or
P = Vi
P = i2 R
NOTES:
Now, let’s relate this to something you are familiar with . . .
radio
current flow
+
electron flow
DC battery
Battery / Resistor Circuit
Battery +
(V)
i
Resistance
(R)
Circuit Diagram of Above Battery / Resistor Circuit
Kirchhoff’s Voltage Law:
The sum of the voltage rises around a closed loop in a circuit must
equal the sum of the voltage drops.
Kirchhoff’s Current Law: The sum of the currents into a junction (mode) must equal the sum
of all currents flowing away from the junction.
NOTES:
SERIES CIRCUITS
We have three resistors connected in series with a DC power source (as shown in the following
diagram). The circuit has only one voltage rise, the battery with voltage V. The circuit also has
three voltage drops, the product of the current and each resistor in order to satisfy Kirchoff’s
voltage law. Expressing this situation in an equation yields
V  iR1  iR2  iR3
R1
V
Diagram of Resistors in SeriesR R2
R3
Diagram of Resistors Connected in Series
The current flow i is the same for each resistor, so we could rearrange the equation as
V  i ( R1  R2  R3 )  iReq
where Req is the single equivalent resistance that could replace all three individual resistances.
An equivalent resistance in series is equal to the sum of the individual resistances.
R eq   R i
(symbol  denotes summation
i
of subscripted variable Ri)
In series circuits the voltage drop across each resistor will vary according to the value of R.
NOTES:
PARALLEL CIRCUITS
For resistors connected in parallel (as shown in the following diagram), the voltage drop across
each resistor is the same and equal to the battery voltage. The current, however, varies for each
resistor in order to satisfy Kirchhoff’s Current Law.
V
R1
R2
Diagram of Two Resistors Connected in Parallel
From Kirchhoff’s Voltage Law, we can state the following equation.
V  i1R1  i2 R2  i3 R3
I haven’t been able to load MS
equation on my computer. I think a
two parallel resistor circuit should be
introduced first.
Using Kirchhoff’s Current Law for junctions x and y what are the currents?
i = i1 + i2
Let’s solve for i1 and i2.
We know that current leaving the battery equals the sum of the individual resistor currents. So,
individual currents may be found by applying Ohm’s Law (V = iR) to each resistor.
NOTE: All of these will first be introduced with just R1 and R2
i1 
V
R1
i2 
V
R2
i3 
V
R3
Utilizing the substitution property,
ix 
V V V


R1 R2 R3
And finally rearranging the terms,
ix  V (
1
1
1

 )
R1 R2 R3
The term in parenthesis can be re-stated as equivalent resistance:
1
1
1
1



Req R1 R2 R3
= R2 + R1
R1R2
So, Req = R1R2
R2 + R1
Now, solving for i1 and i2
i1 = V = i Req = i R1R2 . =
R1
R1
R1 (R2 + R1 )
i R2 .
(R2 + R1 )
i2 = V = i Req = i R1R2 . =
R2
R2
R2 (R2 + R1 )
i R1 .
(R2 + R1 )
Let’s look at a three resistor parallel circuit. What is the current at nodes x and y?
V
R1
R2
R3
Diagram of Three Resistors Connected in Parallel
ix  i1  i y
and
i y  i2  i3
therefore, combining the above two equations
ix  i1  i2  i3
Again, we can solve for i1, i2 and i3:
i1 = V = i Req
R1
R1
i2= V = i Req
R2
R2
i3 = V = i Req
R3
R3
Algebraically, solve for Req in the space below for the three parallel resistors:
Classwork Example
Find the equivalent resistance of the circuit shown below.
5
15 
10 
20 
Homework Problem Set 1-3
1.
The filament in a flashlight bulb has a resistance of 40 . The battery voltage is 6 V.
Determine the current flow.
2.
The maximum current flow from a 1.5-V battery is 45 mA. What is the minimum size
resistor that can be connected to it?
3.
If four 5- resistors are wired in series, what is the total resistance of the circuit?
4.
Many home lighting circuits have 15-A circuit breakers with a voltage supply of 110 V.
How many 100-W light bulbs may be placed in parallel in the circuit before the breaker
trips?
5.
You are using a 1250-W hairdryer on a 15-A, 110-V circuit. Your sister comes in and
turns on a 350-W stereo and a 100-W light, also in parallel on the same circuit. Does the
circuit breaker trip?
ANSWERS TO HOMEWORK SET 1-3
1.
2.
3.
4.
5.