Transcript current+electricity+IV - PTEC107 AC/DC Electronics/Electrical

Circuit Examples
By observing and participating in this lesson, you
will be able to:
 Identify and describe
series and parallel
circuits.
 Simplify those circuits.
 Calculate the current
and/or voltage of each
part of a circuit.
1/81
Series and Parallel Circuits
Electrical components can be connected
in various ways. This drastically changes
the properties of the circuit.
2/81
Series Circuits
One simple way to arrange components in an
electrical circuit is to create one large continuous
loop with the components:
switch
2
batteries
Light
bulb
resistor
3/81
This is similar to a TV series where one episode
follows another.
Parallel Circuits
Another way to connect a circuit is in parallel. In
this arrangement, each component is connected
separately in its own “loop”.
2
batteries
3 resistors in parallel
4/81
Resistors in Parallel
To find the equivalent resistance of resistors
added in parallel:
1
1
1
1



...
RP R1 R 2 R 3
Parallel or total
combined
resistance, Ω
5/81
Individual
resistors, Ω
Parallel Resistor Example
Calculate the total effective resistance of two 10
Ohm resistors connected in parallel.
1
1
1


RP R1 R 2
1
1
1


R P 10Ω 10Ω
1
2

R P 10Ω
RP  5Ω
6/81
Combine, use
common
denominator if
needed
Take the
reciprocal of
each side of
equation
Parallel Resistor Observations
5 Ohms. Notice that the total overall resistance is
lower than either one of them individually!
This occurs because there are multiple paths for
the electrons to take, lowering their resistance.
7/81
Circuit Problems
This section will detail how to calculate
the various electrical quantities in a
circuit.
8/81
Series Circuit
R2=10Ω
Calculate the current and electric potential
difference for each component of the circuit
shown.
5V
R1=5Ω
A good first step is to simplify the circuit.
9/81
Circuit Simplification
5V
R1=5Ω
5V
R S  R1  R 2  R 3 ...
R S  5Ω  10Ω  15Ω
10/81
Rs=15Ω
R2=10Ω
Because this is a series circuit, to combine the
resistors and simplify the circuit, they are merely
added together.
5V
Rs=15Ω
Current in a Series Circuit
V  IR
V
I
R
5V
I
 .33A
15Ω
Use the voltage of the power supply and the total
resistance of the circuit to find the total current
flowing through the circuit.
Because the electron flow has no where else to go,
this amount is also the current flowing through
both resistors. I1 and I2 is that same .33 Amperes.
11/81
R2=10Ω
Voltage in a Series Circuit
5V
Since we know the current
flowing through each
resistor, we can use Ohm’s
law to find the potential
difference for each of those
resistors.
R1=5Ω
V1 IR
V 2  IR
V1 (.33A)(5Ω )
V2  (.33A)(10Ω )
V2  3.33V
V1  1.67V
12/81
Notice how the sum of the two voltages adds up
to the power supply for the circuit.
Parallel Circuit
Calculate the current and electric potential
difference for each component of the circuit
shown.
5V
R1=5Ω
R2=10Ω
Notice how this parallel circuit contains the
exact same components as the series circuit,
they are just arranged differently.
13/81
Again, a good first step is to simplify the circuit.
Circuit Simplification
5V
R1=5Ω
1
1
1


RP R1 R 2
1
3

R P 10Ω
1
1
1


R P 5Ω 10Ω
10
RP 
Ω  3.3Ω
3
1
2
1


R P 10Ω 10Ω
14/81
R2=10Ω
Notice this parallel
resistance is less than
either one individually.
Shortcut Formula
An equivalent formula can be used for two
resistors, R1 and R2, connected in parallel.
Sometimes this formula is easier to manipulate.
R 1R 2
RP 
R1  R 2
It may be easier to remember this formula as the
product over the sum for the two resistors.
15/81
Voltage in a Parallel Circuit
5V
R1=5Ω
R2=10Ω
The easy part about any parallel circuit is the
voltage applied to each item.
Since each item has its own independent
connection to the battery or power supply, each
item receives that potential.
In this case, V1 and V2 are each 5V.
16/81
Current in a Parallel Circuit
5V
R1=5Ω
R2=10Ω
Once you realize that the electric potential for each
resistor is 5V, finding the current is easy using
Ohm’s law, V=IR.
17/81
V1
I1 
R1
V2
I2 
R2
5V
I1 
 1A
5Ω
5V
I2 
 .5A
10Ω
Current Observations
5V
R1=5Ω
R2=10Ω
Notice the two currents add up to the same value as
the total current in the circuit. This is a good way to
check your work.
V
5V
IT 

 1.5A
R T 3.3Ω
IT  I1  I2  1A  .5A  1.5A
18/81
Conclusion
You should now be able to describe series and
parallel circuits with words and numbers!
Questions???
Homework:
Check with your instructor
19/81