Electric Circuits

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Transcript Electric Circuits

ELECTRICAL CIRCUITS
The CELL
The cell stores chemical energy and transfers it to
electrical energy when a circuit is connected.
When two or more cells are
connected together we call this
a Battery.
The cells chemical energy is
used up pushing a current round
a circuit.
What is an electric current?
An electric current is a flow of microscopic particles
called electrons flowing through wires and
components.
+
-
In which direction does the current flow?
from the Negative terminal to the Positive terminal of a
cell.
simple circuits
Here is a simple electric circuit. It has a cell, a
lamp and a switch.
cell
wires
switch
lamp
To make the circuit, these components are connected
together with metal connecting wires.
simple circuits
When the switch is closed, the lamp lights up. This is
because there is a continuous path of metal for the
electric current to flow around.
If there were any breaks in the circuit, the current
could not flow.
circuit diagram
Scientists usually draw electric circuits using symbols;
cell
lamp
switch
wires
circuit diagrams
In circuit diagrams components are represented by
the following symbols;
cell
ammeter
battery
voltmeter
switch
motor
lamp
buzzer
resistor
variable
resistor
types of circuit
There are two types of electrical circuits;
SERIES CIRCUITS
PARALLEL CIRCUITS
SERIES CIRCUITS
The components are connected end-to-end, one
after the other.
They make a simple loop for the current to flow
round.
If one bulb ‘blows’ it breaks the whole circuit and
all the bulbs go out.
PARALLEL CIRCUITS
The components are connected side by side.
The current has a choice of routes.
If one bulb ‘blows’ there is still be a complete circuit to
the other bulb so it stays alight.
measuring current
Electric current is measured in amps (A) using
an ammeter connected in series in the circuit.
A
measuring current
This is how we draw an ammeter in a circuit.
A
A
SERIES CIRCUIT
PARALLEL CIRCUIT
measuring current
SERIES CIRCUIT
• current is the same
at all points in the
circuit.
2A
2A
2A
PARALLEL CIRCUIT
• current is shared
between the
components
2A
2A
1A
1A
copy the following circuits and fill in the
missing ammeter readings.
3A
?
4A
?
3A
1A
?
4A
?
4A
1A
1A
?
measuring voltage
The ‘electrical push’ which the cell gives to the current
is called the voltage. It is measured in volts (V) on a
voltmeter
V
measuring voltage
Different cells produce different voltages. The
bigger the voltage supplied by the cell, the bigger the
current.
Unlike an ammeter a voltmeter is connected across
the components
Scientist usually use the term Potential Difference
(pd) when they talk about voltage.
measuring voltage
This is how we draw a voltmeter in a circuit.
V
SERIES CIRCUIT
V
PARALLEL CIRCUIT
measuring voltage
V
V
V
V
series circuit
• voltage is shared between the components
3V
1.5V
1.5V
parallel circuit
• voltage is the same in all parts of the circuit.
3V
3V
3V
measuring current & voltage
copy the following circuits on the next two
slides.
complete the missing current and voltage
readings.
remember the rules for current and voltage
in series and parallel circuits.
measuring current & voltage
a)
6V
4A
A
V
V
A
measuring current & voltage
b)
4A
6V
A
V
A
V
A
answers
a)
b)
4A
6V
6V
4A
6V
4A
4A
3V
2A
3V
4A
6V
2A
Electric Circuits
In electricity, the concept of voltage will be
like pressure. Water flows from high
pressure to low pressure (this is consistent with
our previous analogy that Voltage is like height
since DP = rgh for fluids) ; electricity flows
from high voltage to low voltage.
But what flows in electricity? Charges!
How do we measure this flow? By Current:
current = I = Dq / Dt
UNITS: Amp(ere) = Coulomb / second
The rate at which electrons move
along field lines is called drift speed,
typically about 10-4 m/s
Electric current defined in
terms of the flow of positive
charge opposite the electrons is
called conventional current
Current will always be in the
same direction as the local
electric field
Voltage Sources:
batteries and power supplies
A battery or power supply supplies voltage. This is
analogous to what a pump does in a water system.
Question: Does a water pump supply water? If
you bought a water pump, and then plugged it in
(without any other connections), would water
come out of the pump?
Question: Does the battery or power supply
actually supply the charges that will flow
through the circuit?
• Charges move from
higher to lower potential
• For the process to
continue, charges that
have moved from a
higher to lower potential
must be raised back to a
higher potential again
• A battery is able to add
charges and raise the
charges to higher electric
potential
Symbol for a battery
Voltage Sources:
batteries and power supplies
Just like a water pump only pushes water (gives
energy to the water by raising the pressure of the
water), so the voltage source only pushes the
charges (gives energy to the charges by raising
the voltage of the charges).
Just like a pump needs water coming into it in order
to pump water out, so the voltage source needs
charges coming into it (into the negative terminal)
in order to “pump” them out (of the positive
terminal).
Voltage Sources:
batteries and power supplies
Because of the “pumping” nature of voltage sources,
we need to have a complete circuit before
we have a current.
Circuit Elements
In this first part of the course we will consider
two of the common circuit elements:
capacitor
resistor
The capacitor is an element that stores
charge for use later (like a water tower).
The resistor is an element that “resists” the
flow of electricity.
Electrical Resistance
&
Ohms’ Law
The current established is directly proportional to
the voltage difference
Ohm’s Law: ΔV ∝ I
In a plot of ΔV vs I,
the slope is called the electrical resistance
Resistance
Current is somewhat like fluid flow. Recall
that it took a pressure difference to make
the fluid flow due to the viscosity of the
fluid and the size (area and length) of the
pipe. So to in electricity, it takes a voltage
difference to make electric current flow due
to the resistance in the circuit.
Resistance
By experiment we find that if we increase the
voltage, we increase the current: V is
proportional to I. The constant of
proportionality we call the resistance, R:
V = I*R
Ohm’s Law
UNITS: R = V/I so Ohm = Volt / Amp.
The symbol for resistance is
Resistance
Just as with fluid flow, the amount of
resistance does not depend on the voltage
(pressure) or the current (volume flow).
The formula V=IR relates voltage to
current. If you double the voltage, you will
double the current, not change the
resistance. The same applied to capacitance:
the capacitance did not depend on the charge and
voltage - the capacitance related the two.
As was the case in fluid flow and capacitance, the
amount of resistance depends on the
materials and shapes of the wires.
Resistance
The resistance depends on material and
geometry (shape). For a wire, we have:
R=rL/A
where r is called the resistivity (in Ohm-m)
and measures how hard it is for current to
flow through the material, L is the length of
the wire, and A is the cross-sectional area of
the wire. The second lab experiment deals with
Ohm’s Law and the above equation.
Electrical Power
The electrical potential energy of a charge is:
U = q*V .
Power is the change in energy with respect to
time:
Power = DU / Dt .
Putting these two concepts together we have:
Power = D(qV) / Dt = V(Dq) / Dt = I*V.
Electrical Power
Besides this basic equation for power:
P = I*V
remember we also have Ohm’s Law:
V = I*R .
Thus we can write the following equations for
power: P = I2*R = V2/R = I*V .
To see which one gives the most insight, we
need to understand what is being held
constant.
Example
When using batteries, the battery keeps the
voltage constant. Each D cell battery
supplies 1.5 volts, so four D cell batteries in
series (one after the other) will supply a
constant 6 volts.
When used with four D cell batteries, a light
bulb is designed to use 5 Watts of power.
What is the resistance of the light bulb?
Example
We know V = 6 volts, and P = 5 Watts; we’re
looking for R.
We have two equations:
P = I*V and V = I*R
which together have 4 quantities:
P, I, V & R..
We know two of these (P & V), so we should
be able to solve for the other two.
Example
Using the power equation we can solve for I:
P = I*V, so 5 Watts = I * (6 volts), or
I = 5 Watts / 6 volts = 0.833 amps.
Now we can use Ohm’s Law to solve for R:
V = I*R, so
R = V/I = 6 volts / 0.833 amps = 7.2 W .
Example extended
If we wanted a higher power light bulb,
should we have a bigger resistance or a
smaller resistance for the light bulb?
We have two relations for power that involve
resistance:
P=I*V; V=I*R; eliminating V gives: P = I2*R and
P=I*V; I=V/R; eliminating I gives:
P = V2 / R .
In the first case, Power goes up as R goes up; in the
second case, Power goes down as R goes up.
Which one do we use to answer the above question?
Example extended
Answer: In this case, the voltage is being held
constant due to the nature of the batteries. This
means that the current will change as we change
the resistance. Thus, the P = V2 / R would be the
most straight-forward equation to use. This means
that as R goes down, P goes up. (If we had used the
P = I2*R formula, as R goes up, I would decrease – so it
would not be clear what happened to power.)
The answer: for more power, lower the
resistance. This will allow more current to
flow at the same voltage, and hence allow
more power!
Kirchhoff’s Laws
• Junction Law: at a junction in a circuit, the
sum of the current entering the junction will
equal the sum of the current leaving.
ΣI= ΣI
in
out
• Loop Law: the sum of the potential drops
around any closed loop must add to
ΣV= 0
loop
Connecting Resistors
There are two basic ways of connecting two
resistors: series and parallel.
In series, we connect resistors together like
railroad cars; this is just like we have for
capacitors:
+
-
+
high V
-
low V
R1
R2
Formula for Series:
To see how resistors combine to give an
effective resistance when in series, we can
look either at
R1
I
V = I*R,
+
V1 V
R2
or at
Vbat
2
R = rL/A .
Formula for Series
Using V = I*R, we see that in series the
current must move through both resistors.
(Think of water flowing down two water falls in series.)
Thus Itotal = I1 = I2 .
Also, the voltage drop across the two resistors
add to give the total voltage drop:
(The total height that the water fell is the addition of the two
heights of the falls.)
Vtotal = (V1 + V2). Thus, Reff = Vtotal / Itotal =
(V1 + V2)/Itotal = V1/I1 + V2/I2 = R1 + R2.
Formula for Series
Using R = rL/A , we see that we have to go
over both lengths, so the lengths should add.
The distances are in the numerator, and so
the values should add.
This is just like in R = V/I (from V = IR)
where the V’s add and are in the numerator!
Note: this is the opposite of capacitors
when connected in series! Recall that
C = Q/V, where V is in the denominator!
Formula for Parallel Resistors
The result for the effective resistance for a
parallel connection is different, but we can
start from the same two places:
(Think of water in a river that splits with some water
flowing over one fall and the rest falling over the
other but all the water ending up joining back
together again.) V=I*R, or R = rL/A .
Itotal
+
Vbat
I1
-
R1 I2
R2
Formula for Parallel Resistors
V=I*R, or R = rL/A
For parallel, both resistors are across the same
voltage, so Vtotal = V1 = V2 . The current can go
through either resistor, so: Itotal = (I1 + I2 ) .
Since the I’s are in the denominator, we have:
R = Vtotal/Itotal = Vtotal/(I1+I2); or
1/Reff =
(I1+I2)/Vtotal = I1/V1 + I2/V2
= 1/R1 + 1/R2.
Formula for Parallel Resistors
If we start from R = rL/A , we can see that
parallel resistors are equivalent to one
resistor with more Area. But A is in the
denominator (just like I was in the previous
slide), so we need to add the inverses:
1/Reff = 1/R1 + 1/R2 .
Terminal Voltage
Terminal voltage, VT , is the potential difference
between the terminals of a battery.
Ideal voltage, VB , is determined by the chemistry
of the battery.
Internal resistance, ir : some charge will be los due
to the random thermal motion of the battery
Terminal voltage will be:
For recharging a battery:
VT = VB - ir
VT = VB + ir
Current Division
When current enters a junction, Kirchhoff’s first
law tells you the sum of the current entering
must equal the sum of the current leaving.
Example:
8 = I + 4I + 5I  I = 0.8 A
1/RP = 1/20 +1/5 +1/4 = ½  RP = 2Ω
V = IRP = 8 • 2 = 16V
16 = I1(20)  I1 = 0.8A
16 = I2(5)  I2 = 3.2A
16 = I3(4)  I3 = 4A
Simple Circuits
• A simple circuit is a connection of batteries
and resistors that meets 2 criteria
1. All batteries are in series
2. The equivalent resistance of the entire circuit
can be obtained by repeated use of just the
series and parallel equivalent resistance
formulas
VB = 60 – 18 = 42V
1/RP = 1/12 + 1/6 = ¼  RP = 4Ω
Requiv = 4 + 8 + 6 + 3 = 21Ω
42 – I(21) = 0  I = 2A
VT60 = 60 – 2 • 1 = 58V
VT18 = 18 – 2 • 2 = 22V
2 = I + 2I  I = 0.67A
Capacitance
We define capacitance as the amount of
charge stored per volt: C = Qstored / DV.
UNITS: Farad = Coulomb / Volt
Just as the capacity of a water tower depends
on the size and shape, so the capacitance of
a capacitor depends on its size and shape.
Just as a big water tower can contain more
water per foot (or per unit pressure), so a
big capacitor can store more charge per
volt.
Parallel Plate Capacitor
For a parallel plate capacitor, we can pull
charge from one plate (leaving a -Q on that
plate) and deposit it on the other plate
(leaving a +Q on that plate). Because of the
charge separation, we have a voltage
difference between the plates, DV. The
harder we pull (the more voltage across the two
plates), the more charge we pull: C = Q /DV.
Note that C is NOT CHANGED by either
Q or DV; C relates Q and DV!
Capacitance
• What happens when a water tower is overfilled? It can break due to the pressure of
the water pushing on the walls.
• What happens when an electric capacitor is
“over-filled” or equivalently a higher
voltage is placed across the capacitor than
the listed maximum voltage? It will
“break” by having the charge “escape”.
This escaping charge is like lightning - a
spark that usually destroys the capacitor.
V or DV ?
When we deal with height, h, we usually refer
to the change in height, Dh, between the
base and the top. Sometimes we do refer to
the height as measured from some reference
point. It is usually clear from the context
whether h refers to an actual h or a Dh.
With voltage, the same thing applies. We
often just use V to really mean DV. You
should be able to determine whether we
really mean V or DV when we say V.
Parallel Plate Capacitor
For this parallel plate capacitor, the
capacitance is related to charge and voltage
(C = Q/V), but the actual capacitance
depends on the size and shape:
C plate ∝ A / d
A is the area of each plate, d is the distance
between the plates
Energy Storage
If a capacitor stores charge and carries
voltage, it also stores the energy it took to
separate the charge. The formula for this is:
Estored = (1/2)QV = (1/2)CV2 ,
where in the second equation we have used
the relation: C = Q/V .
Energy Storage
Note that previously we had:
U = q*V ,
and now for a capacitor we have:
Ucap = ½ QV = ½ CV2 = ½ Q2/C
Energy Storage
The reason is that in charging a capacitor, the
first bit of charge is transferred while there
is very little voltage on the capacitor (recall
that the charge separation creates the
voltage!). Only the last bit of charge is
moved across the full voltage. Thus, on
average, the full charge moves across
only half the voltage!
Hooking Capacitors Together
Instead of making and storing all sizes of
capacitors, we can make and store just
certain values of capacitors. When we need
a non-standard size capacitor, we can make
it by hooking two or more standard size
capacitors together to make an effective
capacitor of the value we need.
Two basic ways
There are two basic ways of connecting two
capacitors: series and parallel.
In series, we connect capacitors together like
railroad cars; using parallel plate capacitors
it would look like this:
+
-
+
high V
-
low V
C1
C2
Series
If we include a battery as the voltage source,
the series circuit would look like this:
C1
+
Vbat
C2
Note that there is only one way around the
circuit, and you have to jump BOTH
capacitors in making the circuit - no choice!
Parallel
In a parallel hook-up, there is a branch point
that allows you to complete the circuit by
jumping over either one capacitor or the
other: you have a choice!
High V
C1
C2
Low V
Parallel Circuit
If we include a battery, the parallel circuit
would look like this:
+
Vbat
+
C1
+
C2
Formula for Series:
To see how capacitors combine to give an
effective capacitance when in series, we
can look either at C = Q/V, or at
C plate = A /d
Formula for Series
Using C = Q/V, we see that in series the
charge moved from capacitor 2’s negative
plate must be moved through the battery to
capacitor 1’s positive plate.
C1
+
+Q
Vbat
C2
-
-Q
(  +Qtotal)
Formula for Series
But the positive charge on the left plate of C1 will
attract a negative charge on the right plate, and the
negative charge on the bottom plate of C2 will
attract a positive charge on the top plate - just
what is needed to give the negative charge on the
right plate of C1. Thus Qtotal = Q1 = Q2 .
(+Q1  )
C1
+
+Q1
-Q 1
+Q2
Vbat
C2
-
-Q2
(  +Qtotal)
Formula for Series
Also, the voltage drop across the two
capacitors add to give the total voltage drop:
Vtotal = (V1 + V2).
Thus, Ceff = Qtotal / Vtotal = Qtotal / (V1 + V2),
or (with Qtotal = Q1 = Q2)
[1/Ceff] = (V1 + V2) / Qtotal = V1/Q1 + V2/Q2 =
1/C1 + 1/C2 = 1/Ceffective
Parallel Circuit
For parallel, both plates are across the same voltage,
so Vtotal = V1 = V2 . The charge can accumulate
on either plate, so: Qtotal = (Q1 + Q2).
Since the Q’s are in the numerator, we have:
Ceff = C1 + C2.
+
Vbat
+Q1
C1
-Q1
+Q2
C2
+Q1 
 +Qtotal = (Q1+Q2)
 +Q2
-Q2
Review of Formulas
Electric Current
I= ΔQ/Δt
Resistor Voltage Drop
V = IR
Resistivity
R = rL/A
Electric Power
P = IV
Resistance Power P = I2R = V2/R
Junction Law
SI = SI
in
Loop Law
out
SΔV = 0
loop
Series Resistors
Rs = R1 + R2 +…
Parallel Resistors 1/RP = 1/R1 + 1/R2 +…
Terminal Voltage VT = VB ± IR
Capacitance
C = Q/V
Parallel Plate Capacitor C plate ∝ A / d
Capacitors in Series
1/Cs = 1/C1 + 1/C2 +…
Capacitors in Parallel
CP = C1 + C2 +…
Energy in Capacitor
U = ½ Q2 /C = ½ CV2 = ½ QV