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4V
i
(I + i)
I
2V
00
--3V
1
2
4Ω
Ω
6Ω
Ω
3/27/2006
3Ω
Ω
Circuits (© F.Robilliard)
1
Introduction:
Electrical circuits are ubiquitous in the modern world, and it is difficult to overstate
their importance. They range from simple direct current (DC) circuits, through
alternating current (AC) circuits, up to the more complex circuits of electronics,
communications and computing.
In all circuits, particles carrying electrical charge move around a closed path – the
circuit loop.
We will begin by defining the basic physical quantities needed to describe circuits.
These quantities include charge, current, potential difference, resistance, and
emf.
Then we will use these quantities to analyse DC circuits, in which the charge
carriers are usually electrons, and the electrons travel around the circuit in one
direction only.
Although we will consider only the simplest class of circuit, the principles and
methods we will use are fundamental to all circuits.
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Circuits (© F.Robilliard)
2
Charge:
We begin by defining some essential circuit quantities.
In the simple model, atoms are composed of three types of particle - protons,
neutrons, and electrons. Protons and neutrons are confined to a compact
central region of the atom called the nucleus. The electrons are distributed in
the space surrounding this nucleus.
Charge is a fundamental quantum property possessed by protons and electrons.
There are two types of charge - positive and negative. Protons carry a positive
charge. Electrons carry a negative charge. The magnitude of the charge on
the proton is exactly equal in magnitude to that on the electron. All protons and
electrons in the universe have precisely the same magnitude of charge.
Neutrons carry zero net charge.
Charge is, theoretically, the most fundamental of the electrical quantities.
SI unit of charge is the Coulomb = C
1 C is the total charge on 6.24x1018 electrons
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Circuits (© F.Robilliard)
3
Current:
Current is the time-rate of flow of charge through a designated surface.
dq
i≡
dt
dq in dt
where: i is the current flow, when charge-carriers convey a total
charge of dq through a designated surface, in a time interval of dt.
In materials, protons are generally confined to the nucleus, and cannot move.
However, the outer electrons of atoms, in metals, and some other materials, are
loosely held, and can readily move from one atom to another, thereby
constituting an electric current.
Free electrons can also move through a vacuum, as in the electron beam of a
cathode-ray tube
Current is proportional to the number of electrons that pass through a
designated cross-section of the conducting medium, per second.
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Circuits (© F.Robilliard)
4
Ampere:
SI unit of current is the Ampere = A
A current of 1 A flows, when 1 C of charge passes through a given
cross-section per second = 6.24x1018 electrons per second.
Although current is usually a flow of negative electrons, the flow of any
charged species, such as protons or free charged atoms ( ions ) will
constitute a current. By convention, we develop the theory of circuits on
the assumption that the charge-carriers are positive. Such a current is
called a “conventional current”.
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Circuits (© F.Robilliard)
5
Definition of the Ampere:
Although charge is the more fundamental quantity, in a theoretical sense, the
fundamental experimental unit of electrical quantities, in the SI system, is
chosen to be the Ampere.
The reason for this has to do with practicalities of the experimental definition
of the quantity. When a current flows, it generates a magnetic field, whose
strength depends on the magnitude of the current. Thus, the current can be
defined in terms of its associated magnetic field.
It turns out, that current can be more accurately defined by its magnetic
field, than charge can be defined by any known experimental means.
We use the magnetic force of attraction between two long, thin, straight
current-carrying conductors. For fixed separation d, the magnetic force, F,
increases with the current, i.
If the precise separation of the
i
wires, d = 1 m, and the magnetic
d force between the wires,
F F
F = 2 x 10-7 N, per metre length of
the wires, then the current, i, is
i
defined to be precisely 1 A.
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Circuits (© F.Robilliard)
6
Potential Difference:
To cause a current to flow in a conductor, we set up an electric field through the
conductor. The electric forces which then act on the free electrons of the
conductor, cause those electrons to accelerate. As a consequence of their motion,
the electrons collide with the atoms of the conductor inelastically, thereby loosing
the energy they gained from the electric field, to the atoms. This causes the
conductor atoms to vibrate. We perceive this vibration of the atoms as heat, which
is transferred to the surroundings of the conductor, and lost to the system of
electrons and conductor atoms.
In summary, the electrons gain energy from the electric field, and lose it
through collisions with conductor atoms. Work is done by the field, to move the
electrons.
The potential difference ( PD ) between two points in a circuit, is defined as
the work that has to be done per unit charge, to move that charge from
one point to the other.
SI unit: Work/Charge = J/C = Volt = V
There is a PD of 1 V between two points, if 1 J of work must be done, to
move 1 C of charge, from the first point, to the second.
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Circuits (© F.Robilliard)
7
Ohm’s Law:
If we send a current, i, through a conductor, AB, a potential difference (which is a
potential loss), v, will occur between A & B. If we increase i, then v will also
increase. Say we make an experimental plot of the values of i, against v. It turns
out, that the shape of the i-v curve depends on the nature of the conductor.
+
A
i
conductor
B
-
For some conductors, we will get a linear relationship.
For other conductors, the relationship will be non-linear.
v
v
v
i
i
Ohmic conductor –
most metals, carbon.
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Non-Ohmic conductor –
semiconductor diode,
plasma tubes
Circuits (© F.Robilliard)
8
Resistance:
v
For Ohmic conductors, the potential across the conductor
is proportional to the current flowing through it,
v∝i
thus v = Ri
i
Ohm’s Law.
where R is a constant of proportionality that depends on the composition
and geometry of the particular conductor, and is called the resistance of
the conductor.
Note: Ohm’s law applies to a particular conductor.
SI unit for R: V/A = Ohm = Ω
Circuit
Symbol
Fixed R:
Variable R:
3/27/2006
( kΩ = 103 Ω; MΩ = 106 Ω ).
RΩ
or
RΩ
Circuits (© F.Robilliard)
RΩ
or
RΩ
9
A Simple Picture:
Ohm’s law is a property of some materials. It is not really a law of Physics. We
can get a simple picture of what is happening, as follows.
When an electric field is applied along a conductor, free electrons will move
(drift) along that conductor, collide with conductor atoms, and suffer energy
losses in those collisions. The gain in potential energy of electrons from the
field, will equal the losses in collisions.
As the field is increased, the potential losses are increased in proportion, by
more energetic collisions, occurring more frequently.
Because of the increase in the accelerating field, the average drift velocity of
electrons, between collisions also increases. Therefore the electrons move
more rapidly along the conductor, between collisions, which constitutes an
overall increase in current.
X
v
Y
In ohmic conductors (such as X, Y, Z, in the figure),
an increase in electric field strength, which
corresponds to an increase in potential difference
across the conductor, is proportional to the increase
in overall effective drift velocity of electrons (ie. to the
increase in current through the conductor).
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Circuits (© F.Robilliard)
Z
i
10
Resistors in Combination:
There are only two different ways to connect two resistors, R and r, in a circuit –
e
in
series
e
R
r
in
out
e
A
B
C
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R
e
out
r
In a series connection, an input
electron, e, has NO choice. If it
goes through R, it must go
through r. It can do nothing else.
Example:
parallel
In a parallel connection, an input
electron, e, has an exclusive-or
choice. It can go through either R or
r, but not both. And it has no other
choice.
A is not in series with B, since an
electron could go through A then C.
B is in parallel with C.
Circuits (© F.Robilliard)
11
Series:
Consider two resistors, R and r, connected in series. It is important to be able to find
the total equivalent resistance, ρ, of the combination.
equivalent resistance.
i
i
R
r
=
ρ
vR
vr
v
v
i = current through both R & r.
v = PD across both R & r.
vR = PD across R alone
vr = PD across r alone
For unit charge passing through R and r, by conservation of energy -
Energy
Energy
Energy
=
+
lost in ρ
lost in R
lost in r
v = vR + vr
i = iR + ir.......by Ohm' s law
thus ρ = R + r
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In series, resistors add directly.
Circuits (© F.Robilliard)
12
Parallel:
Two resistors, R and r, connected in parallel, have an equivalent single resistor ρ.
iR
i
i
R
r
ir
equivalent resistance.
ρ
=
v
v
Input current, i, splits into two currents, iR (through R) and ir (through r).
v = PD across the parallel combination.
By conservation of charge -
i = iR + ir
v v
+ .....by Ohm' s Law
R r
1 1 1
giving
= +
R r
thus
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v
=
In parallel, resistors add
by reciprocals.
Circuits (© F.Robilliard)
13
Example:
For the following circuit, find the total equivalent resistance between A and B
(all resistances in ohm).
2.6
A
Series:
r = 3.0 + 2.0 + 1.0 = 6.0
3.0
4.0
2.0
1.0
B
Redraw:
A
2.6
parallel
4.0
6.0
B
Redraw: A
Redraw:
2.6
2.4
Series:
r = 2.6 + 2.4 = 5.0
B
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1 1
1
=
+
r 4.0 6.0
r = 2.4
Circuits (© F.Robilliard)
A
5.0
B
Answer: 5.0 ohm
14
Electromotive Force (emf):
Electrons lose energy as they pass through the resistive elements in a circuit.
This energy must be replaced, for the electron current to continue.
This energy is supplied by a power supply (PS), such as a cell, battery,
generator, or electronic power supply.
The emf, ε, of a PS is the energy supplied per unit charge
that passes through the PS.
SI Unit for emf: (work)/(charge) = J/C = Volt = V
Note: A PS is an energy source, whereas a resistor is an energy sink, for
charge carriers in a circuit
When charge-carriers pass through a PS, they generally encounter
resistance along the internal path within the PS itself. This internal
resistance, Ri, can be regarded as being in series with the PS.
Ri is the energy lost per unit charge as it flows through the PS.
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Circuits (© F.Robilliard)
15
Circuit Symbol for emf :
ε
+
-
or
Ri
ε
+
ε = emf of the PS
-
Ri = internal resistance of the PS.
Ri
emf’s in Series & Parallel:
ε1
+
- +
Ri1
ε2
-
Ri2
≡
ε
+
Ri
-
where:
ε = ε1 + ε2
Ri = Ri1 + Ri2
In series, both the emf’s and the internal resistances are additive.
For the parallel connection, we will only consider the case where both
emf’s are the same, and their internal resistances are negligible.
+
ε1
Ri1=0
≡
ε1
+
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+
ε
Ri
-
where:
ε = ε1
Ri = 0
Ri2=0
Circuits (© F.Robilliard)
16
Switches, Fuses & Earths:
A
A
a
B
b
Double pole, double throw, switch:
A is switched to a, and B to b, simultaneously.
a
b
c
A
a
A
A
Single pole switch:
A is connected, or disconnected (switched) to a.
a
A
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E
or
Rotary switch:
A is switched to one only of a, b, or c.
Fuse, safety switch or circuit breaker:
A is disconnected from a, if the current
exceeds a set value.
Earth:
A is connected to earth (zero potential)
Circuits (© F.Robilliard)
17
The Simple Circuit:
By use of the rules for adding resistances, and emf’s, many circuits can be
reduced to a single emf, in series with a single resistance.
+
ε
Ri
i
R
Individual emf’s of the circuit have been
lumped into a single emf, ε, with a single
series internal resistance, Ri. Individual
resistances of the circuit have been lumped
into a single equivalent “external” resistance,
R, in series with the emf.
v
i = the current drawn from the emf = the current through the external resistance
v = the voltage across the terminals of the emf
= the voltage across the external resistance, R.
Such circuits are called simple circuits, and can be analysed using
essentially the circuit methods that we have developed so far.
Some circuits consist of a network of emf’s and resistors, which cannot be
reduced to this simple form. These are called complex circuits, and
require more general methods for their analysis.
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Circuits (© F.Robilliard)
18
Basic equation of the Simple Circuit:
+
ε
Ri
i
R
This equation is the application of the
conservation of energy to the simple circuit
We assume that the circuit has reached a
steady-state condition.
We apply the conservation of energy to the
charge-carriers as they go round the circuit.
v
energy given per unit charge
energy lost per unit charge
to charge - carriers
= by charge - carriers
as they pass through the emf
in the external and internal resistances
= iR + iR i
= v + iR i
using Ohm' s Law
This equation, together with Ohms Law, and the rules for combining
resistors, and emf’s, are sufficient to analyse the simple circuit.
3/27/2006
Circuits (© F.Robilliard)
19
Potential around a Simple Circuit
-
ε
+
Ri
i
R
Say we earth the negative terminal of the cell,
which will set its potential to zero.
We then plot the potential of points around the circuit,
moving clockwise from this earthed point.
v
-
ε
+ Ri
R
ε
v = (ε – i Ri)
(ε – i Ri – i R) = 0
Potential rises, from zero, to ε, in the cell, drops by (i Ri) in the internal cell
resistance, and finally falls, by a further (i R), back to zero in the external resistance.
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Circuits (© F.Robilliard)
20
Solving Simple Circuit Problems:
Steps:
The general pattern is 1. use the rules for series and parallel to find the total external resistance
2. use ε = i R + i Ri to find the total current, i, drawn from the emf
3. use Ohm to find particular voltages, and currents
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Circuits (© F.Robilliard)
21
ε=10V
+
i
Example:
-
Ri=2.6Ω
i4
R4=4Ω
i6
(i) Total resistance, R:
(parallel resistors)
R6=6Ω
Fig.1
1
1
1
=
+
R R4 R 6
v
1 1
+
4 6
thus R = 2.4
=
ε=10V
+
For Fig.1, find (i) total external resistance, R.
(ii) total current, i, drawn from the PS.
(iii) voltage, v, across the 6 Ω.
(iv) current, i6, in the 6 Ω.
i
-
Ri=2.6Ω
v
R=2.4Ω
Fig.2: Equivalent Circuit
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(ii) Total current, i:
= iR + iR i
10
=
=2A
thus i =
R + R i 2.4 + 2.6
(iii) voltage, v:
Applying Ohm’s law to
the external resistance of
the equivalent circuit
(Fig.2) –
v = i R = 2.0 x 2.4 = 4.8 V
Circuits (© F.Robilliard)
(iv) current, i6:
Applying Ohm’s law
to the 6Ω resistance
of the original circuit
(Fig.1) –
i6 = v/R6 = 4.8/6
=0.8V
22
Measurement of ε and Ri for a PS:
+
ε
Ri
i
R
v
v
intercept=ε
grad=-Ri
i
(
= experimental point )
The emf, and internal resistance of a PS cannot be
measured directly, using a multimeter. The following
gives a precise method for their measurement.
Connect a variable external resistor, R, across the
terminals of a PS.
If we decrease R, from its maximum value, the current, i,
will increase, but the PD, v, across R, will simultaneously
decrease.
Plotting the values of v against the corresponding
values of i, gives a straight line.
ε = i R + i Ri
= v + i Ri
thus: v = -Ri i + ε
negative gradient
This has the form
y=mx+c
(a straight line)
intercept
By measuring the v-intercept and gradient of the plot, we can find ε and Ri.
3/27/2006
Circuits (© F.Robilliard)
23
Power In Circuit Elements:
So far, we have considered the total amounts of energy per unit charge, being
given to, or lost by, charge-carriers, as they pass through the emf’s and
resistors of a simple circuit. This takes no account of the time-rate of energy
gain or loss = power.
i
dq
P
R
dt
v
Let P = the power dissipated in a resistor R, when it
conducts a current, i, with an associated PD of v.
Say a charge dq passes through the resistor, R, in a time dt.
From the definition of PD, the energy, dW, lost by the
charge dq, as it passes through the resistor is -
dW = v dq
since v ≡
dW
dq
Dividing both sides by dt we get
dW
dq
= v
dt
dt
Thus P = v i ...................(1)
dW
dq
)
(since P ≡
and i ≡
dt
dt
3/27/2006
Circuits (© F.Robilliard)
Power = voltage x current
24
i
dq
P
R
dt
v
Power Dissipated in Resistance:
P = v i …….(1)
By use of Ohm’s law, we get two other useful versions of (1).
From Ohm v = iR ..........(2)
v
and
i = ..........(3)
R
v2
(2) → (1) : P =
R
(3) → (1) : P = i 2 R
Summary: The power, P, dissipated in a resistor, R, which has a current, i,
flowing through it, with a PD of v, across it, is given by -
v2 2
P = vi =
=i R
R
An example: In power transmission lines, the i, is kept small, to minimise
power losses = i2R.
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Circuits (© F.Robilliard)
25
Example:
Find the current drawn by a 40 W lamp, running on a 240 V supply.
i
P = vi
P 40 1
= = 0.167A
thus i = =
v 240 6
P=40W
R
v=240 V
Example:
Find: (i) the power dissipated in the external 8 ohm resistor of the following circuit
(ii) the total power supplied, by the emf, to the circuit.
+
ε=20V
(i) First find the current:
-
thus i =
Ri=2Ω
i
R + Ri
=
20
=2A
8+ 2
Thus Power in 8 ohm = i2 R = 22 8 = 32 W
R=8Ω
v
3/27/2006
= iR + iR i
(ii) Total power supplied
Circuits (© F.Robilliard)
= power lost in both R and Ri
= i2 ( R + Ri )
= 22 (8 +2) = 32 + 8 = 40 W
26
Kirchoff’s Rules:
If a circuit cannot be reduced to an equivalent simple circuit (a single emf in
series with a single external resistance), then the methods we have used so
far will fail. For such complex circuits, we need a more general method.
Kirchoff’s Rules give such a general method, and allow for the solution
of any DC circuit.
Furthermore, Kirchoff gives the bases for the analysis of more general AC, and
electronic circuits
Kirchoff realised that any network of circuit elements consists of two basic elements:
1. Node = a point in a circuit where 3 or more conducting paths meet.
2. Loop = any closed conducting path in a circuit.
There are two Kirchoff rules.
The first rule deals with nodes; the second rule deals with loops.
Kirchoff assumes that the circuit is in a steady state.
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Circuits (© F.Robilliard)
27
Kirchoff Rule 1:
Rule 1 (K1): Conservation of charge at a node.
The sum of currents into a node = the sum of currents out of that node
i1
i in =
i2
i4
N
i out
2A
6A
1A
i3
N
5A
2+5=6+1
i1 + i3 = i2 + i4
In a steady-state circuit, there is no gain, nor loss, of charge at a node.
If we don’t know the physical direction of a current, we assume a direction,
and proceed consistently on that basis. If the numerical solution for that
current is positive, we guessed correctly; if the solution is negative, the
current direction is the reverse to that assumed.
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Circuits (© F.Robilliard)
28
Kirchoff Rule 2:
Rule 2 (K2): Conservation of energy around a loop.
algebraic sum of potential drops across
Algebraic sum of emf' s
=
the conductors around the same loop
around a loop
loop
( ) = (ir )
loop
We assume that each point in the loop has a fixed potential (is in steady state).
If a positive test charge traverses a simple circuit loop, its potential will increase
in the emf’s, and decrease in the resistances. When it has completed the loop,
its potential must return to the value it had at its starting point. The gains must
equal the losses!
However, in complex circuits, the situation is a little more involved than this.
Note: that we are talking here about the change in a scalar quantity, the
potential of a test charge. An increase is a positive change, a decrease is a
negative change. This is nothing to do with direction in the vector sense.
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Circuits (© F.Robilliard)
29
K2 Sign Convention:
In a simple circuit, test charges, +dq, can always be taken
around the loop in the direction of the loop current, so that
the test charge will gain potential in emf’s, and lose
potential in resistances. This is illustrated in Fig.1
+dq
ε
+
+
Fig.1
R
-
i
However, in a complex circuit, we may be forced to move our test charge
through emf’s and resistors, in directions opposite to those of Fig.1. In this case,
the test charge will lose potential in emf’s, and gain potential in resistors, the
opposite of Fig.1.
To keep correct an account of the potential, we need a sign convention
The directions of Fig.1 are taken as positive, and the reverse directions, as negative.
+ direction of (iR) is the direction of the current, i.
+ direction of ε is from the negative terminal to the positive.
+
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+
R
i
+
Circuits (© F.Robilliard)
- +
ε
30
K2 Example:
I=2A
A
R=5Ω
+
e=15V
r=4Ω
D
i=5A
Consider a loop ABCD, which is part of some network.
B
Let us write K2 for this loop.
K2 :
E=5V
C
K2 Algebraically:
-E + e = -IR + ir
loop
( ) = (ir )
loop
Traverse the loop clockwise (in direction ABCD).
We will call this the positive loop direction
( indicated by the blue circular arrow).
The current directions are shown (by the mauve
arrows); the positive directions for the two emf’s
are indicated (by the red arrows).
K2 Numerically:
-5 + 15 = -(2)(5) + (5)(4)
thus +10 = +10
If the direction of a current, or an emf is in the positive loop direction,
it is written as positive; if opposite to the loop direction it is negative.
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Circuits (© F.Robilliard)
31
Steps in Network Problems:
In the usual situation, we are given the emf’s, and must find all currents.
1. Label directions of known emf’s with an arrow, on the circuit diagram.
2. Assume directions for unknown currents and label on the diagram..
3. For each circuit loop, choose a positive loop directions (clockwise, or
anticlockwise), and label it.
4. Apply K1 to nodes (this can be done on the diagram, in simple cases).
5. Apply K2 to loops.
6. Solve node and loop equations for unknowns.
7. Interpret your numerical answers as physical directions on a circuit diagram.
Note: If we get a negative solution for a current ( or emf, if unknown ) we
guessed the wrong direction at the start of the problem – the physical
direction is the reverse of that assumed!
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Circuits (© F.Robilliard)
32
Example:
4V
i
I
2V
(I + i)
Label unknown currents in 4V & 2V cells.
(We need to guess these directions.)
Use K1 to label third current.
Label positive directions for emf’s.
3V
1
2
4Ω
Ω
6Ω
Ω
Find the magnitudes and
directions of the currents
delivered by the three cells of the
network illustrated.
Take clockwise as positive for both loops.
Write loop equations (K2) for both loops:
3Ω
Ω
loop
( ) = (ir )
loop
Loop 1: +4 -3 = ( I + i ) 4 + (i) 6
1 = 4 I + 10 i ……………(1)
Loop 2: -2 +3 = -(I) 3 - ( I + i ) 4
1 = -7 I - 4 i ………….…(2)
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Circuits (© F.Robilliard)
33
Example:
1 = 4 I + 10 i ……………(1)
1 = -7 I - 4 i …….…….…(2)
0.20 A
4V
i
(1)x2:
2 = 8 I + 20i …….….(3)
(2)x5:
5 = -35I – 20i..…….….(4)
(3) + (4): 7 = -27I
I = - 7/27 = - 0.26 A…(5)
(5)->(1): 1 = - 28/27 + 10 i
10 i =1 + 28/27 = 55/27
hence
i = +0.20 A
The physical directions for I & i need to
be interpreted by labelling on the circuit
diagram. Since I is negative, its physical
direction (red arrow) will be opposite to
the assumed direction (purple arrow).
Since i is positive, its physical direction
will be in the same direction as that
guessed (purple arrow).
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0.26 A
I
2V
0.06 A
(I + i)
3V
4Ω
Ω
6Ω
Ω
3Ω
Ω
The third current can be found by (I + i) = (-0.26 + 0.20) = -0.06A
This is also labelled.
Circuits (© F.Robilliard)
34
Capacitance:
So far, we have discussed DC networks, which contain only emf’s and resistances.
Circuits in general, may contain other circuit elements, among which are capacitors,
inductors, diodes, transistors, gates, and a huge array of integrated circuit chips for
various functions.
+q A
Here we will look at capacitance, and capacitors.
Electrical capacitance is a property of a system. It is essentially a
v
measure of the system’s ability to store energy in an electric field.
(-q)
B
Consider a system of two bodies, A & B.
Say we transfer charge carriers, carrying a total charge of (+q), from B to A. This will
give A a positive charge of (+q), and leave a residual negative charge of (-q) on B.
Because of the separation of charge, an electric field, and
consequent potential difference, v, will exist between A and B.
As more charge is transferred to A, the PD, v, between A and B
will increase, in proportion to the total charge, q, transferred.
v
q
The two bodies, A & B constitute a capacitor.
The gradient of the v-q line is used to define the capacitance of this capacitor.
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35
Definition of Capacitance:
The capacitance, C, of a capacitor, is defined as the ratio of the
charge, q, on either conductor to the magnitude of the potential
difference, v, between them
q
C≡
v
+q
A
v
(-q)
B
The capacitance of a capacitor is a measure of its ability to store charge, and
electrical potential energy. Capacitance will depend on the geometry, size and
composition of a system.
Capacitance is numerically equal to the charge, q, that must be
moved (from B to A) to raise the potential difference, v, by one volt.
SI Unit: Coulomb/Volt = C/V = Farad = F
(Common units used are µF, nF and pF)
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Parallel Plated Capacitor:
This is a common geometry for a capacitor, and consists of two parallel
conducting plates separated by an insulator.
d
To charge the capacitor, an emf is connected
across the plates. This causes a charge of (+q) to
be transferred from one plate to the other, producing
a potential difference of v across the plates.
A
-q
circuit
symbol
q
0 A
C≡ =
v
d
v
ε
+
v
(+q)
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+q
The capacitance, C, of a parallel plated capacitor
depends on the plate area, A, and on plate separation, d,
and is given by -
A = area of each plate
d = separation of the plates
ε0 = constant = permittivity of vacuum
-12
2 -1
-2
CN m
= 8.8542 x 10
(We will derive this result in the next section - electric fields.)
where:
Circuits (© F.Robilliard)
37
Example:
d
A
Find the capacitance of a parallel plate capacitor
with plates 10 cm square, separated by 1 mm, in
air.
C=
0
(
A
d
)
0.10 x 0.10
= 8.85x10
10 −3
= 8.85x10−11 F
= 0.885 pF
−12
Note: To make the capacitance, C, large, we can increase the plate
area, A, or reduce the plate separation, d. There are practical limitations
on the size of the plates, and when d is reduced below a certain
separation, short circuiting, or arcing, will take place between the plates.
Thus practical capacitors in electronics are typically no larger than µF.
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Capacitances in Parallel:
In circuits, capacitors are often wired together in series or in parallel. We
need to be able to compute the overall equivalent capacitance of such
combinations. An equivalent capacitor is one that could completely replace
the combination in a circuit. When the same potential difference exists across
both the equivalent capacitance and the original combination, both would be
storing the same total charge.
Consider two capacitors, C1 & C2, in parallel.
C1 (+q1)
The equivalent capacitor is C.
If the same potential difference, v, exists
across each, then both must store the
C (+q)
(+q
)
2
C2
same total charge.
v
v
Thus
q = q1 + q2
From definition of capacitance
C v = C1 v + C2 v
Thus
C = C1 + C2
In parallel, capacitors add directly.
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Capacitors in Series:
Consider two capacitors, C1 & C2, in series.
The equivalent capacitor is C.
(-q) (+q)
(-q) (+q)
C1
C2
v1
v2
(+q)
C
v
v
To charge the combination, we move a
charge (+q) from the left plate of C1, to
the right plate of C2. The central wire
connecting the two capacitors contains
free charges. The residual (-q) on the left
plate of C1, will attract an equal but
opposite charge, (+q), to its right plate,
from the central wire. Similarly, C2 will
attract a residual charge (-q) to its left
plate, from the central wire.
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Thus both the series capacitors
carry the same charge, q.
However, because their
capacitances are different, they will
have different individual PD’s, v1 &
v2.
From conservation of energy
v = v1 + v 2
From definition of capacitance
q q
q
=
+
C C1 C 2
thus
1 1
1
=
+
C C1 C 2
Series capacitors add by reciprocals.
Circuits (© F.Robilliard)
40
Example:
In series, thus
1 1 1
= +
C 3 2
2+3 5
=
=
6
6
6
thus C = = 1.2 nF
5
Find the total capacitance between A and B of 3.0 nF
A
2.0 nF
B
0.8 nF
In parallel, thus
C = 1.2 + 0.8 = 2.0 nF
Redraw:
1.2 nF
A
B
0.8 nF
Thus total capacitance
between A & B is 2.0 nF.
A
B
2.0 nF
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Energy Stored in a Capacitor:
When a capacitor is charged, work is done in separating charges within the
capacitor. This energy comes from the emf that separates the charges.
If the capacitor is discharged through a resistance, this stored energy is
transferred to the resistance, where it is dissipated as heat.
Say a capacitance, C, has separated charge, q, and potential difference, v.
ε
+
v
(-q)
C
(+q)
We separate a small additional amount of
charge dq. To do this we need to do a small
amount of work dW.
Because the change in charge is small, the
potential difference, v, will remain essentially
constant.
From the definition of potential difference, the
work done, dW, is given by dW = v dq…………..(1)
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Energy Stored in a Capacitor
dW = v dq…………..(1)
Using definition of capacitance
q
q
C = → v = ......................(2)
v
C
(2) → (1) : dW = q dq.........(3)
C
Using the definition of capacitance
C=
q = Cv → (4) :
The total energy, E, stored in the
capacitor is the sum of all the
elements of work, dW, done in
storing the total charge, q.
E
Total work = dW =
0
1 q2
→E=
.............(4 )
2 C
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1
qv
2
1
E = Cv 2
2
E=
Summary:
q2 1
1
E=
= qv = Cv 2
2C 2
2
q
q
dq
C
0
q
→ (4 ) :
v
The energy stored in a capacitor is
proportional to the square of the
charge separated, or to the square
of the PD across the capacitor.
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43
Example:
An electric car needs to output 20 kW of power for 3 hours, before being
recharged. If this energy were to be stored in a 1 mF capacitor, what would the
PD across the capacitor terminals need to be?
Total energy to be stored = E = Power x time = (20 x 103) x (3 x 60 x 60) = 216 MJ
But
E = 1/2 C v2
thus
(216 x 106) = 1/2 (1 x 10-3 ) v2
which gives
v2 = (216 x 106) x 103 x 2 = 43.6 x 1010
hence
v = 6.60 x 105 V = 0.66 MV
This is clearly a lethally high voltage, and hence impractical!
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4V
i
(I + i)
I
2V
00
--3V
1
2
4Ω
Ω
6Ω
Ω
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3Ω
Ω
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