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Electrical Circuits
They keep the lights on!!
Electrical Current is how fast
charge is moving in a unit of
time.
I is current.
Current is
measured in
amperes or amps
(A).
Remember it is charge
that is moving! Electrons
move very little!!
Physics knows its electron flow but
often use conventional current.
Direction of the current is taken to be
opposite of the electron motion.
Ohm’s Law is that different metals
have different resistance to electrical
flow independent of Voltage.
V = IR
V is voltage or how much electricity
(how much “push”) and is measured in
Volts (V)
I is current and is measured in Amps (A)
R is resistance to current and is
measured in Ohms (Ω)
Voltage is called “potential
difference” and change in
Voltage is called “drops”.
This means you have “more”
charge in one area and
“less” charge in another!
There must be a potential drop from
one end to another to have current.
potential difference = drop
V = IR
V is created by having more electrical
charge at one end and less electrical
charge at another end.
Deviation from Ohm’s Law
• Not all materials
obey Ohm’s Law.
• The filament of a
light bulb has
increasing
resistance as
current
increases.
• It only obeys
Ohm’s Law at low
temperatures.
Resistance slows electricity
down as it moves through a
circuit. It has a symbol of R
and is measured in Ohms (Ω).
Factors Affecting Resistance
R is directly proportional to length of the wire.
The longer the wire the more resistance.
R is inversely proportional to cross-sectional
area of the wire. The “thicker” the wire the
less resistance.
R is directly proportional to temperature. The
higher the temperature in the wire the more
resistance.
Electrical Power
• Either thermal energy or work
done by an electrical device.
• These equations come in
handy!!
Electrical Power
P = VI
P = RI2
V2
P=
R
If any two of
voltage,
current, and
resistance are
known then
power can be
calculated.
Electricity and Circuits
Property
•
•
•
•
Voltage (V)
•
Current (I )
•
Resistance (R) •
Power (P)
•
Units of Measure
Volts (V)
Amps (A)
Ohms (Ω)
Watts (W)
Simple Circuits
Here are examples of both a
series and parallel circuit.
Simple Circuits
The left circuit is in series, if one
bulb goes out the entire circuit is
disrupted.
Simple Circuits
The right circuit is in parallel. If
one bulb goes out the other will
remain on.
Simple Circuits
All circuits must be complete from
the positive to negative terminal
of the battery.
Symbols in Circuit Diagrams!
Battery
Conductor
Switch
Light Bulb or Lamp
Resistors
Resistance in Series
R1
R2
24 V
R3
0V
I
Rtotal = R1 + R2 + R3
Resistance on the circuit is the
sum of the individual resistors
R1
R2
24 V
R3
0V
I
Rtotal = R1 + R2 + R3
Resistance on the circuit is the
sum of the individual resistors
2Ω
4Ω
24 V
5Ω
0V
I
Rtotal = 2Ω + 4Ω + 5Ω = 11Ω
Current on the circuit follows
Ohm’s Law
2Ω
4Ω
24 V
5Ω
0V
I
V = IR
Current on the circuit follows
Ohm’s Law
2Ω
4Ω
24 V
5Ω
0V
I
24V = I 11Ω
Current on the circuit follows
Ohm’s Law
2Ω
4Ω
24 V
5Ω
0V
I
I = 2.2A
This is true across any resistor
as well.
R1 = 2Ω
4Ω
24 V
5Ω
0V
I
V = 2.2A 2Ω
There is a 4.4V drop at the first
resistor.
R1 = 2Ω
4Ω
24 V
5Ω
0V
I
V = 4.4V
At point A the voltage has
dropped.
R1 = 2Ω
4Ω
5Ω
A
24 V
0V
I
24V - 4.4V = 19.6V
Resistance in Parallel
R3
R2
R1
24 V
0V
1
1
I
1
Rtotal
=
R1
+
R2
+
1
R3
Resistance in Parallel
R3
R2
R1
24 V
0V
I
R1= 2Ω
1
Rtotal
=
1
R1
+
1
R2
+
1
R3
R2= 4Ω
R3= 5Ω
Resistance in Parallel
R3
R2
R1
24 V
0V
I
R1= 2Ω
1
Rtotal
=
1
2Ω
+
1
4Ω
+
1
5Ω
R2= 4Ω
R3= 5Ω
Resistance in Parallel
R3
R2
R1
24 V
0V
I
1
Rtotal
= 0.5 + 0.25 + 0.2
R1= 2Ω
R2= 4Ω
R3= 5Ω
Resistance in Parallel
R3
R2
R1
24 V
I
1
Rtotal
= 0.95
0V
R1= 2Ω
R2= 4Ω
R3= 5Ω
Resistance in Parallel
R3
R2
R1
24 V
I
Rtotal = 1.05 Ω
0V
R1= 2Ω
R2= 4Ω
R3= 5Ω
Resistance in Parallel
R3
R2
R1
24 V
0V
I
Remember the Golden Rule: In parallel
resistors resistance on the system must be
less than that of the lowest resistor!
R3
R2
R1
24 V
I
0V
This is due to the
property current has to
take the path of least
resistance
R1= 2Ω
R2= 4Ω
R3= 5Ω
2Ω
4Ω
I
2A
5Ω
Voltage across parallel lines is
always the same!
The resistance of this parallel
system is 1.05 Ω
2Ω
4Ω
I
2A
5Ω
The voltage drop can be found
using V = IR
V = (2 A)(1.05 Ω)
V = 2.1V
When we learn current splits we
will discover it is the same drop for
each resistor individually!
Rules to Remember
• In a parallel circuit system, the voltage is the
same in all parallel paths.
The Voltage
through the
4Ω and 8Ω
resistors are
the same!
Rules to Remember
• It helps to think of Voltage as the “push” through
a circuit, like an invisible hand pushing charge
along.
The Voltage
through the
4Ω and 8Ω
resistors are
the same!
Rules to Remember
• To find current on the circuit the resistance
across the entire circuit must be known as
well as voltage out of the battery.
The
Resistance is
3+10+5 = 18Ω
The Current is
V = IR
9V = I 18Ω
I = 0.5 A
Rules to Remember
• In a series circuit system, the voltage drops as
current goes through each resistor.
The Voltage is
9V at point 1
and 7.5V at
point 2.
V = IR
V = (0.5A)(3Ω)
V = 1.5V
Rules to Remember
• In a series circuit system, the voltage drops as
current goes through each resistor.
The Voltage is
7.5V at point 2
and 2.5V at
point 3.
V = IR
V = (0.5A)(10Ω)
V = 5V
Rules to Remember
• In a series circuit system, the voltage drops as
current goes through each resistor.
The Voltage is
2.5V at point 3
and 0 at point
4.
V = IR
V = (0.5A)(5Ω)
V = 2.5V
Rules to Remember
• To understand this it helps to think of voltage as
“push”. Some of this “push” is used up as each
resistor system is crossed.
The Voltage is
2.5V at point 3
and 0 at point
4.
V = IR
V = (0.5A)(5Ω)
V = 2.5V
Rules to Remember
• In a parallel circuit system, the current splits as it
enters and comes back together as it exits.
The Current
through the
4Ω and 8Ω
resistors are
the same
going in and
out!
Rules to Remember
• Current is always the “same in, same out”
More Current
goes through
the 4Ω
resistor than
the 8Ω
resistor!
Rules to Remember
• The 0.5 A will be split between the 4Ω and 8Ω
resistors!!!
8/12 of the
Current goes
through the
4Ω resistor!
0.5A (8/12) =
0.33 A
Rules to Remember
• The 0.5 A will be split between the 4Ω and 8Ω
resistors!!!
4/12 of the
Current goes
through the
8Ω resistor!
0.5A (4/12) =
0.17 A
Rules to Remember
• The current comes back together after going
through the resistors!
The Current to
the battery!
0.33A + 0.17A =
0.5 A
Rules to Remember
• When you have parallel systems mixed with
resistors in series, solve for the resistance of
the parallel system and then treat it as part of
the series.
• It is very common to have series and parallel
resistors mixed together on a circuit.
Example
2Ω
2Ω
4Ω
3Ω
6Ω
The 4Ω, 3Ω, 6Ω resistors are a parallel system in
series with the other resistors!!
Example
2Ω
2Ω
4Ω
3Ω
6Ω
First solve for the total resistance of the
parallel system!!
Example
2Ω
2Ω
1
Rtotal
4Ω
1
1
1
=
+
+
4
3
6
3Ω
6Ω
Example
2Ω
2Ω
1
Rtotal
4Ω
3Ω
3
4
2
9
=
+
+
=
12
12
12
12
6Ω
Example
2Ω
2Ω
1
Rtotal
9
=
12
4Ω
3Ω
6Ω
12
Rtotal =
= 1.33Ω
9
Example
2Ω
2Ω
4Ω
3Ω
6Ω
Now add the resistance of the parallel
system to the series!!!
Example
2Ω
2Ω
4Ω
3Ω
2Ω + 1.33Ω + 2Ω = 5.33Ω
6Ω
Current in Parallel
In = Itotal
Rtotal
Rn
In is the current across the individual resistor.
Itotal is the total current entering the parallel
system.
Rtotal is the total resistance of the parallel
system.
Rn is the resistance of the individual resistor.
Example of Parallel Current
4Ω
I = 6A
I = 6A
6Ω
8Ω
1
Rtotal
=
1
4Ω
+
1
6Ω
+
1
8Ω
Example of Parallel Current
4Ω
I = 6A
I = 6A
6Ω
8Ω
1
Rtotal
=
6
24
+
4
24
+
3
24
=
13
24
Example of Parallel Current
4Ω
I = 6A
I = 6A
6Ω
8Ω
Rtotal =
24
= 1.85Ω
13
Now we can solve for the current through each
resistor!
In = Itotal
Rtotal
Rn
1.85Ω
I4 = 6A
4Ω
I4 = 2.77A
In = Itotal
Rtotal
Rn
1.85Ω
I6 = 6A
6Ω
I6 = 1.85A
In = Itotal
Rtotal
Rn
1.85Ω
I8 = 6A
8Ω
I8 = 1.38A
Example of Parallel Current
I = 6A
4Ω
2.77A
I = 6A
6Ω 1.85A
8Ω
1.38A
2.77A + 1.85A + 1.38A = 6A
Ammeters
• They are built into
circuits to give the
current level at that
point.
• They have as little
resistance as
possible.
• Ammeters are put in
series so they do not
change the current of
the circuit
• Remember current
splits in parallel!!!
Galvanometers read current
through magnetic fields and
are obsolete.
Voltmeters
• Voltmeters are built into
circuits and give voltage at
that point.
• They are very good for giving
the “voltage drop” of a
particular device.
• Voltmeters are put in parallel
to the devices they read.
• Remember the voltage
across any two parallel lines
are the same!
Digital Multimeters
• They can read voltage,
current, resistance, and
even temperature.
• They range from less
than $10 to hundreds of
dollars.
• They are making
ammeters and
voltmeters obsolete!!
Fuses
A fuse is designed to melt and
disrupt or break a circuit before
current levels get dangerous.
This is why computers and
LCD televisions should be
plugged into a “surge
protector” instead of straight
into a wall. Modern
electronics need more
protection from temporarily
high current levels or
“surges”.
Fuses
A fuse is rated by the amp level at
which it melts, such as a 10 amp
fuse.
The power of devices is
rated at a specific voltage
level. A 300 W device at
110 V will have a
different power at a
different voltage.
Calculating Power
Calculating Power
• When using the P = VI equation the voltage
used is the “voltage drop” or push “used up”
at that particular resistor.
• The current used is the current through the
resistor.
• Same with resistance, the Ω value of that
resistor.
• Only exception is when asked the power on
the circuit.
A Power
Example
4Ω
4Ω
2Ω
20 V
0V
I
What is the resistance of the two 4Ω in
parallel?
Rp = 2Ω
A Power
Example
4Ω
4Ω
2Ω
20 V
0V
I
What is the resistance on the circuit?
Rt = 4Ω
A Power
Example
4Ω
4Ω
2Ω
20 V
0V
I
What is the current on the circuit?
I = 5A
A Power
Example
4Ω
4Ω
2Ω
20 V
0V
I = 5A
What is the voltage drop on the 2Ω
resistors?
V = 10V
A Power
Example
4Ω
4Ω
2Ω
V = 10V
20 V
0V
I = 5A
What is the voltage drop on the parallel
4Ω resistors?
V = 10V
A Power
Example
4Ω
5A
4Ω
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
What is the current on the 2Ω resistor?
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
What is the current on the parallel 4Ω
resistors?
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = VI
P for 2Ω = 10V 5A = 50W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = VI
P for 4Ω = 10V 2.5A = 25W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = RI2
P for 2Ω = 2Ω (52)A = 50W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = RI2
P for 4Ω = 4Ω (2.52)A = 25W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = (V2) /R
P for 2Ω = (102)V / 2Ω = 50W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
Now calculate Power using P = (V2) /R
P for 4Ω = (102)V / 4Ω = 25W
A Power
Example
4Ω 2.5A
5A
4Ω 2.5A
V = 10V
2Ω
V = 10V
20 V
0V
I = 5A
See how they are all the same?
Power Changes
An electrical device is rated at 500 W at 220 V.
What will the power of the device be if it is
switched to a 110 V power source? Assume that
its resistance is not variable.
Resistance can be found with the equation
P = V2/R.
500W = 2202V/R.
R= 48400V/500W.
R= 96.8Ω.
Power Changes
Now we need to find the current when the device
is plugged in to the 110V source.
Current can be found with the equation
V = I R.
110V = I 96.8Ω.
I = 110V / 96.8Ω
I = 1.14A.
Power Changes
Now we can find the power of the device plugged
into the 110V outlet.
Current can be found with the equation
P=VI
P = 110V 1.14A
P = 125W
The End