Download 3 - MACscience

DC Electricity &
Capacitors
NCEA AS3.6
Text Chapters 12-13
Kirchhoff’s Laws
Current Law: The total current entering a
junction in a circuit equals the total current
leaving.
I1=2A
I3=5A
I2=3A
I1  I 2  I 3
Kirchhoff’s Laws
Voltage Law: The total of all the potential
differences around a closed loop in a circuit is
zero
V2
V1
V3
 V1  V2  V3  0
Kirchhoff’s Laws
For voltage sources:


Going from + to – represents a loss of energy
so potential difference is negative
Going from – to + represents a gain in energy
so potential difference is positive
+V
-V
Kirchhoff’s Laws
For resistors:


Passing in the same direction as the current
represents a loss of energy -V=-(IR)
Passing in the opposite direction to the
current represents a gain in energy +V=+(IR)
+V=+IR
-V=-IR
I
I
Kirchhoff’s Laws
To solve problems,
follow a loop around a
circuit…
a-b
0V
 b-c
+6V
 c-d
-8V (2Ax4Ω)
 d-a
+3V (3Ax1Ω) –V
Then add up the p.d’s.
0V+6V-8V+3V-V=0
V=1V!
a
b
2A
V?

6V
1Ω
3A
4Ω
d
c
Kirchhoff’s Laws
Harder….






a
a-b
0V
b-c
-0.75A x 4Ω = -3V
I1?
c-d
-I1x 2 = -2I1V
d-a
+5V
Adding: -3 -2I1 +5 = 0
b
4Ω
So I1 = 1A
Using the current law:


d
5V
2Ω
0.75A
I2?
1A = 0.75A + I2
So I2 = 0.25A
V?
e
c
8Ω
f
Kirchhoff’s Laws
Continued…






d
a
b-c
-0.75A x 4Ω = -3V
c-f
+0.25A x 8Ω = +2V
f-e
+V?
e-b
0V
Adding: -3V + 2V + V? = 0 b
So V = 1V
5V
I1?
2Ω
0.75A
c
4Ω
I2?
V?
e
8Ω
f
Internal Resistance
All components have some resistance,
including cells and meters
The potential difference (voltage)
measured across a cell when no current is
being drawn is called the E.M.F (electromotive force)
When current flows, the voltage measured
across the terminals of the cell will drop,
because of the cell’s internal resistance.
Internal Resistance
V (V)
EMF
r
EMF
r
A
V
V  EMF  Ir
I (A)
Electrical Meters
Galvanometer – very sensitive meter that
can be adapted to read either current or
voltage
Must know 2 things about the
galvanometer


The internal resistance
The current that causes full scale deflection of
the needle If.s.d
Electrical Meters
To make a voltmeter the galvanometer
must be connected in series with a large
resistor Rs
Small current
passes
through G
G
r
Most current continues through circuit
Rs
Electrical Meters
Example: Need to measure up to 10V with a
meter that has 200Ω of internal resistance and
a full scale deflection at 5mA.



V=IR
10V= 0.005A x (200 + Rs)
Rs=1800Ω
G
200Ω
5mA
10V
Rs
Electrical Meters
To make an ammeter the galvanometer
must be connected in parallel with a very
small resistor Rp
Small current
passes through G
G
r
Rp
Most current continues through circuit
Electrical Meters
Example: Need to measure up to 1A with a
meter that has 200Ω of internal resistance and
a full scale deflection at 5mA.



V (across meter) =IR = 0.005 x 200 = 1V
Current through Rp = 1A - 0.005A = 0.995A
Rp = V/I = 1V / 0.995A = 1Ω
G
200Ω
5mA
Rp
Capacitors
A capacitor is a device that can store
electric charge and release it later on.
They consist of two metal plates,
separated by an insulating material called
a dielectric.
Capacitors are used for



Tuning circuits eg radios
Flashing circuits eg camera flashes
Controlling alternating current
Capacitors
There are several types of capacitor:
Non-polar
Electrolytic
Variable
Capacitors
Capacitors are charged by
connecting them to a power
supply or battery
This causes an accumulation
of electrons on one plate
(negative) and removal of an
equal number of electrons
from the other (positive)
Capacitors
When a capacitor is fully charged:




the current flow in the circuit stops
both plates have equal but opposite amount
of charge on them
The voltage across the plates equals the
supply voltage
An electric field exists between the plates.
The strength of this field is given by: E=V/d
Capacitance C
The value of a capacitor is
a measure of how much
charge it can store per volt
applied across the plates
The unit for capacitance is
the farad F
1F=1CV-1
(Most capacitors are much less
than 1F in size, it’s a big unit!)
Q
C
V
Capacitance
The capacitance of a capacitor depends on 3
things:
 The area of the plates which overlap (Cα A
ie. The more area overlapping, the bigger
the C)
 The distance separating the plates (C α 1/d
ie. The closer the plates the bigger the C)
 What is used as the dielectric material.
Capacitance
These are summed
up in the capacitor
construction formula:
A
d
εo
εr
= Area of overlapping plates
= Distance separating plates
= Absolute permittivity of free
space (!!) – A constant
= 8.84x10-12 Fm-1
=Dielectric constant (no units,
it’s a factor)
Air
1
Oiled paper 2
Polystyrene 2.5
Glass
Water
6
80
Charging Capacitors
When the switch is
closed, charge begins
to build up on the
plates.
The process
continues until the
cap has the same
voltage across it as
the power source
Charging capacitors
It’s easy at first
because the plates
are empty, but as
they fill the stored
charge starts to
repel further charge
and the rate
decreases
V
Vmax
t
Charging Capacitors
As the rate of
charge movement
decreases, the
current flow in the
circuit decreases
I
Imax
t
Discharging Capacitors
The voltage will be
at a max to begin
with but will drop to
zero as charge
leaves the plates.
V
Vmax
t
Discharging Capacitors
The current will be
large at first, but
decrease as
potential left to
“push” decreases
I
Imax
t
Charge/Discharge Rates
The rate that a cap will charge or
discharge depends on two things:


The size of the capacitor. The bigger the cap,
the more charge it can store, so it will take
longer to fill/empty.
The initial current in the circuit. This is
determined by the resistance of the circuit.
More R means smaller current, so longer to
fill/empty.
Time Constant
This is defined as the time it takes the
voltage/current to rise to 63% of it’s max value or
for it to drop 63% from it’s max value
It is calculated using this formula:
  RC
A cap is considered fully charged or discharged
after 3 time constants
Capacitor in Series
Joining capacitors in
series:
1
1
1
 
CT C1 C2
Why?
The charge on each
cap is the same.
The voltage across
each cap adds to the
supply voltage
V=V1+V2
-Q
+Q
V1
V
-Q
+Q
V2
Q
C
V
1
V
 
C
Q
1 V1  V2
 
C
Q
1 V1 V2
 

C
Q Q
1
1
1
 

C
C1 C 2
Capacitor Networks
Joining capacitors in
parallel:
CT  C1  C2
Why?
The voltage across each cap
is the same as the supply.
The total charge stored is the
sum of the charge in each
cap. Q=Q1+ Q2
V
Q1
Q2
Q
C
V
Q1  Q2
C
V
Q1 Q2
C

V V
C  C1  C2
(NB. This is the opposite way round to resistors)
Energy in Capacitors
Capacitors store energy as electric
potential energy.
The amount they store is half of the
energy supplied by the battery. (The
other half is dissipated as heat in the
resistance of the circuit)
When a capacitor discharges, the
energy is dissipated in the resistance
of the circuit as heat, light etc.
Energy in Capacitors
For a cell: Each
coulomb of charge
gains V Joules of
energy
So the energy
provided by the cell
ΔE=VQ
V
ΔE= area
=VQ
Q
Energy in Capacitors
For a capacitor:
V
voltage increases
gradually as cap
charges until it
equals the cell
voltage
So the energy stored
by the capacitor
ΔE=1/2VQ
ΔE= area
= ½ VQ
Q
Energy in Capacitors
Some rearrangement and
substitution puts this
equation into a familiar
form…
c.f. the formula for kinetic
energy E=1/2mv2
c.f. the formula for elastic
potential energy E=1/2kx2
1
E p  QV
2
but
Q
C
V
so
1
2
E p  CV
2