Download current electricity

CURRENT ELECTRICITY
Electrostatics is the study of charges that are stationary (static). Current electricity deals
with charges in motion.
Current is the net time-rate of flow of charge. Let charge q flow through a point in
time t . Then the average current at that point is
q
t
limit t  0 , this
q dq
I  lim

t 0 t
dt
I ave 
In the
becomes the instantaneous current at that point:
Current is measured in Coulombs/Second ( Cs 1 ), a unit called the Ampere (A). Thus, the
Ampere is the current that flows when a unit of charge or a Coulomb of charge passes
through a particular point.
Although current is a scalar, we associate a direction with it; that in which positive charge
would flow. If, for instance, the charges causing the current are electrons, the direction of
current is opposite the flow of the electrons.
Electromotive Force
To maintain the flow of current in a conductor, energy must be supplied continuously by
the source of electromotive force (emf). Electromotive force is not a force, but energy. It
is the amount of energy needed to push a unit charge through a source. Charge flows
externally from the positive pole of a cell to the negative cell. This indeed, is due to
electrons going from the negative terminal to the positive terminal. To make the circuit
complete, the electrons must move internally (inside the cell) from the positive to the
negative terminal. This they would ordinarily not do, as they should be attracted to the
positive terminal and repelled by the negative terminal. Therefore, some work needs to be
done to achieve the completion of the flow of electrons.
Emf is measured in Joules, the unit of work.
The emf of a cell is the amount of work done in moving a unit charge through the source.
That is, if the emf of a Lechlanche cell is 1.5 V, it means the amount of work needed to
push a unit (+ve) charge through the cell from the negative terminal to the positive one is
1.5 Joules.
Types of Charge Carriers
For current to flow in a material, such a material must have free charge carriers. For
example, insulators have no free charge carriers. The charge carriers in metals are free
electrons, those which have been detached from the outer atoms. A metal has a large
number of free electrons, and this is why it is a good conductor of electricity. In an
electrolyte, the charge carriers are positive and negative ions. In a plasma, a mixture of
electrons and positive ions, these are the charge carriers. In semiconductors, electrons and
1
holes (absence of electrons) are the charge carriers. Holes behave very much like
electrons with a positive charge and the same mass as electrons. As electrons move
opposite the external field, the holes move in the direction of the external field.
Charge Drift, Drift Velocity
Metals are unique in that they have a large number of free electrons. Consider the figure
below, a section of a metal in which the atomic sites are represented by dots. The motion
of an electron is indicated by the arrows. In the absence of an external electric field, the
free electrons move around due to their thermal energy. The direction of the motion of
the electron is random, necessitated by the absence of the external electric field and the
fact that it collides with atomic centers and changes direction.
Fig. 1: No external electric field
The electron has no preferred direction, and there is no net current.
Upon the application of an external electric field, the electron is attracted to the direction
in which a “positive external charge” might have caused the external field, Fig. 2. Such
an electric field could be due to the terminals of a DC voltage source.
E
vt
C
D
Fig. 2: External electric field engenders a preferred direction
There is, now, a preferred direction of flow of the electrons. This results in a flow of
charge, or we now say that there is a current. However, the electrons still collide with
atomic sites, slowing them down. Thus, we talk of the drift velocity of the electrons.
Let the cross sectional area of the conductor be S , the number of charge carriers per unit
volume of the material be n , and let the charge on each charge carrier be q . Then, the
amount of charge in the volume enclosed between the two vertical lines in Fig. 2 is
vt  S  (nq)
2
where vt is the distance moved by a charge carrier with velocity v in time t . The
volume of charge that flows through the plane CD is vt  S . In this volume is the amount
of charge vt  S  (nq) , since each carrier has charge q, n charge carriers carry nq amount
of charge.
The current through the surface is then,
Q
 Svnq
t
=I
If there are more than one type of charge carriers, then we can write,
m
I  S  ni v i q i
i 1
Example 1
Copper has a density of 8.94 g/cm³, an atomic weight 63.546 g/mol. Hence, there are
140685.5 mol/m³. There are 6.02×1023 atoms (Avogadro’s constant) in 1 mole of any
substance. Therefore in 1m³ of copper there are about 8.5×1028 atoms (6.02×1023 ×
140685.5 mol/m³). Since copper has one free electron per atom, n = 8.5×1028 electrons
per m³.
It can be shown that the drift current for copper is then,
v
I
 2.8 10  4 m / s
Snq
We can also define the current density as j 
I
S
in units of Amperes per square metre.
Hence, we can also write
j  (nq)v  charge per unit volume  drift velocity
We can write this equation in the vector form as,
j  nqv
It follows that j is in the direction of v if q  0 and opposite for q  0 .
Example 2
Find the current density in Example 1.
Ohm’s Law
Ohm's law states that, provided the temperature and pressure remain constant, the
potential difference across a metallic conductor is directly proportional to the current in
it.
Thus,
V I
The constant of proportionality is R, the resistance of the conductor. So we can write,
V  IR
3
For a uniform conductor, the resistance can be written as R 
l
S
, where  is the
resistivity of the metal. That is, the resistance of a uniform conductor is directly
proportional to the length and inversely proportional to the cross-sectional area. It is now
clear why the pressure must be constant for Ohm’s law to hold: increasing the pressure
on the conductor reduces the cross-sectional area.
If we write I  jS and I 
jS 
S
l
S
l
V . Equating the two expressions,
V
Hence,
j
1V
 l
If we can write
j  E
V
E,
l
then, we can write
where   1 /  is the conductivity of the metal.
Vectorially, we can write,
j  E
Under what condition would E 
| E |
V
l
? In electrostatics, we have been taught that
dV
dx
Only if V is a linear function of
x,
or equivalently, when the electric field is uniform in
the direction of the length of the conductor. In that case, E 
V
l
V (x)
Slope = E = V/l
x
Fig. 3: Linear dependence of V on x
The reciprocal of resistance is conductance. The reciprocal of resistivity is conductivity.
Example 3
It is observed that a current of 1.2 A flows through a wire of cross-sectional area 0.55
mm2 when its length is 1 m, the resistance of the wire being 20  . Find the current
density when an electric field of 20 V/m is applied across a 10 m length of a uniform
conductor made of the material of the wire.
4
Solution
j  E
Find the value of  and take the reciprocal. Put this expression in the formula above. The
length of the conductor is immaterial in this question as conductivity has nothing to do
with the length of the material.
Effect of temperature
Earlier, we mentioned that Ohm’s law holds only when the temperature is kept constant.
This is because the graph of V versus I is linear only within what we would call the
linear region. We shall see more on this later.
For materials, we can write the resistivity at any temperature t (0C) as,
 (t )   (t 0 )[1   (t  t 0 )]
where  (t 0 ) is the resistivity ( m ) at the reference temperature (0C), and  (  1 m 1 or
S/m) is the temperature coefficient of resistivity. S (Siemens) is the reciprocal of the
Ohm, and is the unit of electrical conductance (reciprocal of electrical resistance).
Taking t 0  0 0 C ,
 (t )   (0)[1   t ]
Material
Silver
Copper
Calcium
Tungsten
Zinc
Nickel
Lithium
Iron
Platinum
Tin
Lead
Manganin
Constantan
Mercury
Carbon
Germanium
ρ [Ω·m] at 20 °C σ [S/m] at 20 °C
1.59×10−8
1.68×10−8
3.36×10−8
5.60×10−8
5.90×10−8
6.99×10−8
9.28×10−8
1.0×10−7
1.06×10−7
1.09×10−7
2.2×10−7
4.82×10−7
4.9×10−7
9.8×10−7
3.5×10−5
0.6
6.30×107
5.96×107
2.98×107
1.79×107
1.69×107
1.43×107
1.08×107
1.00×107
9.43×106
9.17×106
4.55×106
2.07×106
2.04×106
1.02×106
2.85×104
1.67
Temperature
coefficient K-1
0.0038
0.0039
0.0041
0.0045
0.0037
0.006
0.006
0.005
0.00392
0.0045
0.0039
0.000002
0.000008
0.0009
-0.0005
-0.048
Carbon and Germanium are semiconductors. Semiconductors have negative temperature
coefficient of resistivity.
5
Fig. 4: Resistivity versus temperature for a metallic conductor.
Notice the linear region within which Ohm’s law holds
Superconductivity
For some metals, the resistivity suddenly drops to zero as the zero of absolute
temperature is approached. Such metals are said to exhibit superconductivity. For
example, the resistivity of mercury disappears at 4 K.
Platinum does not exhibit this property, and is one of the reasons it is used for resistance
thermometers.
CURRENT-VOLTAGE CHARACTERISTICS
The current-voltage (I-V) characteristic of a load illustrates the way the current respond
to variations in the voltage across it. The set-up in the figure below can be used to
investigate the current-voltage characteristic of the load, L.
V
A
Fig. 5: Set-up for I-V characteristics
A voltmeter (high resistance) is connected across the load under investigation, while the
ammeter (low resistance) is connected in series with the load. We vary the voltage across
the load as the current flows in one direction. Thereafter, we reverse the terminals of the
battery. This ensures that current flows in the opposite direction. The voltage is increased
again to obtain the response to voltage in the other region.
6
i.
Metallic Conductor
For a metallic conductor, the I-V characteristic is linear within the Ohmic region. Outside
this region, the temperature is no longer constant, having increased. We expect, therefore,
that the material presents more resistance to the motion of current. Thus, the slope
becomes gentler as indicated in Fig.
ii.
Semiconductor Diode
For the semiconductor diode, current grows exponentially in forward bias. In the reverse
biased direction, the increase in current is low for a certain range of voltage, before the
breakdown voltage. At the breakdown voltage, a large current flows with little change in
the voltage, as shown in Fig.
Branch Circuit with Sources of Emf
Consider the branch circuit in the figure below.
AI
R1
 1 R2
2
R3
3
R4
4
R5
R6
B
If current flows from A to B, then the potential at A is higher than that at B. So, we can
write
VB  V A  IR1   1  IR2   2  IR3   3  IR4   4  IR5  IR6
The voltages that are positive tend to drive current in the direction of the current I.
Voltage sources that drive current in the opposite direction are negative.
Hence,
VB  V A   2   3   1   4  I ( R1  R2  R3  R4 )
This is the potential difference between the points A and B.
We can then write,
B
I
V A  VB   1   4   2   3
=
R1  R2  R3  R4
V AB   
A
B
R
A
This is the equivalent of Ohm’s law for a branch circuit. In the case where A = B, then,
V A  VB , and
7
B
I

A
B
R
=
algebraic sum of the emfs
B
R
A
A
Thus, the branch circuit is equal to the algebraic sum of the emfs divided by the total
resistance in the branch. Note that there is no Ohm’s law for a branch circuit, only the
equivalent.
I-V Characteristic of a Cell
The conventional cell can be seen as a source of voltage with an internal resistance that
can be visualised as being a voltage source,  , in series with the internal resistance, r, of
the cell, as indicated in the figure below.

I
r
V
We know that V    Ir , since there is some voltage drop across the internal resistance,
r. That is, V is less than  .
Rewriting,
 V
I 
r r
This is in the form of the linear equation,
y  mx  c

1
where y  I , x  V , c  , and m   .
r
r
Drawing,

r
I
slope 
1
r
V
8
Suppose a cell supplies an Ohmic conductor, then the terminal voltage of the cell, and the
current flowing at this voltage could be determined by finding the points where the two
characteristics intercept. This is because the requirement for both the cell and the
conductor must be satisfied. For the Ohmic conductor whose characteristic is as shown in
the figure below, the linear line through the origin, with the cell with characteristic as
shown, the terminal voltage is V 0 , and the current through the conductor at this terminal
voltage is I 0 .
I
I0
V
V0
Note that this problem is equivalent to solving two simultaneous linear equations.
For a non-Ohmic conductor, the I-V characteristic might be given by a nonlinear
equation, e.g., a quadratic equation. In this case, the appropriate set of two equations is
solved, e.g., one linear, one quadratic, as in the example below.
Example
A cell of emf 6 V, and internal resistance 1 ohm is connected across a device whose
1
characteristic is given by V  I 2 . Calculate the current through the device and the
2
terminal potential difference of the cell at this current.
Solution
This problem can be solved graphically. We shall, however, solve the problem as
quadratic equations.
E  Ir  V : 6  I  V
V 
1 2
I
2
Hence,
6I 
1 2
I
2
Therefore,
I 2  2 I  12  0
 2  2 2  48  2  52

2
2
I  5.2111 or  9.2111
I
9
1 2 1
1
I  (5.2111) 2  13.5778 or (9.2111) 2  42.4222
2
2
2
Interpret these values of V.
V 
Combination of Resistors
Resistors can either be connected in series or in parallel. When in series connection, the
same current passes through all the resistors. In parallel connection, different currents
pass through different resistors.
Series Connection
Consider n resistors in series. The same current passes through them. The voltage drop
across them can be written as,
V  V1  V2  V3  ...  Vn
V  IR1  IR2  IR3  ...  IRn  I ( R1  R2  R3  ...  Rn )
Hence, we can write the voltage as
V  IReff
where Reff  R1  R2  R3  ...  Rn
Parallel Connection
In parallel connection, the current divides up, but the voltages across each resistor is the
same across every other resistor. The currents add up to the current supplying all the
branches.
I  I1  I 2  I 3  ...  I n
V V
V
V
V
I


 ... 

R1 R2 R3
Rn Reff
Notice that in parallel connection, the branch currents are inversely proportional to the
branch resistances. The largest current flows through the least resistance, and vice versa:
V
V
I1 
, I2 
, and so on.
R1
R2
Hence,
1
1
1
1
1



 ... 
Reff R1 R2 R3
Rn
For two resistors in parallel,
1
1
1


Reff R1 R2
Hence,
RR
Reff  1 2
R1  R2
In the case where R1  R2 , then R1  R2  R1
10
R1 R2
 R2
R1
Thus, the effective resistance in this case is approximately equal to the smaller resistance.
Indeed, it is slightly less.
Reff 
Exercise
Calculate the branch currents in the following circuit.
1
2V
2
3
2
3
11
KIRCHHOFF’S LAWS
So far, we have dealt with cases in which there is only one source of emf. In a situation
where there are more than one, we might not be able to tell apriori, in which direction
current flows in a particular branch. Kirchhoff’s laws can help solve the problem. There
are two of these:
i.
Node Law
This states that the algebraic sum of currents at a node is zero. In other words, the amount
of current entering a node is equal to the sum of currents leaving it. This laws is that of
conservation of charge, that charge cannot accumulate at any node.
Consider the figure below.
I1
I2
I5
I4
I3
Kirchhoff’s node law requires that I1  I 4  I 5  I 2  I 3  0 .
Put in another way, I1  I 4  I 5  I 2  I 3 .
Kirchhoff’s Loop Law
This states that the algebraic sum of emf in a loop is equal to the algebraic potential drop
round the loop, taken in the same sense.
   IR
I1
I2
R1
R2
1
2
+
5
3
I5
I3
R5
R3
I4
R4
12
Convention: We shall take currents flowing clockwise in the circuit as positive.
Anticlockwise sense of flow of current shall be taken as negative. The same convention
shall be used for the emf’s. Emf sources that tend to drive current clockwise shall be
taken as positive; the ones that tend to drive current anticlockwise shall be taken as
negative.
Thus,   1   2   3   5   I1 R1  I 2 R2  I 3 R3  I 4 R4  I 5 R5
The two laws hold whether the currents are steady or unsteady.
Application of Kirchhoff’s Laws
Consider the circuit below.
I1
I2
R1
R2
R6
1
2
I6
6
5
+
3
I5
I3
R5
R3
I4
R4
  1   2   3   5   I1 R1  I 2 R2  I 3 R3  I 4 R4  I 5 R5
The two laws hold whether the currents are steady or unsteady.
Application of Kirchhoff’s Laws
Consider the circuit below.
2
I1
I2
R1
1
1
5
R2
R6
L1
6
I8 R
8
I5
8
R5
5
2
I6 L 2
6
I9
L4
I4
R4
13
I7
R7
7
L3
R9
4
3
3
I3
R3
+
We shall need 9 equations: 4 loops and 5 nodes. The other node is not independent of the
5.
I 8  I1  I 8  0
I1  I 2  I 3  0
I2  I3  I7  0
I3  I4  I9  0
Node 5: I 4  I 5  0
Node 1:
Node 2:
Node 3:
Node 4:
Loop 1:
Loop 2:
Loop 3:
Loop 4:
 8   1   6  I 6 R6  I1 R1  I 8 R8
 2   6   7  I 2 R2  I 6 R6  I 7 R7
 3   7  I 7 R7  I 3 R3  I 9 R9
  5   8  I 5 R5  I 8 R8  I 9 R9  I 4 R4
It could be quite tedious solving nine equations. So, we use loop currents instead. This
leads to only four equations, as there are only 4 loops (see the figure below).
I1
R1
I2
1
i1
I5
R5
2
I 6 i2
6
I8 R
8
5
R2
R6
I7
8
i4
I9
I4
R4
14
R7
i3
R9
7
I3
R3
Loop 1:
Loop 2:
Loop 3:
Loop 4:
 8   1   6  i1 Ri  (i1  i2 ) R6  (i1  i4 ) R8
 2   6   7  i2 R2  (i2  i1 ) R6  (i2  i3 ) R7
 3   7  i3 R3  (i3  i2 ) R7  (i3  i4 ) R9
  5   8  i4 ( R4  R5 )  (i4  i1 ) R8  (i4  i3 ) R9
How do we recover the various currents from the loop currents?
I1  i1 ; I 2  i2 ; I 3  i3 ; I 4  I 5  i4
I 6  i1  i2 ; I 7  i3  i2 ; I 8  i4  i1 ; I 9  i4  i3
Numerical Example
Determine the branch currents in the circuit shown below.
1V
2
A
B
2V
C
i2
1
D
2
F
E
2V
Solution
Loop 1: 2  1  2i1  1(i1  i2 )
Loop 2:  2  2  2i2  1(i2  i1 )
(i)
(ii)
From equation (i),
3i1  i2  1
From equation (ii),
 i1  3i2  4
(iii)
(iv)
Multiplying equation (iv) by 3 and adding to equation (iii),
8i2  11
11
i2  
8
Substituting i 2 into equation (iv),
 33  32
1
 11 
i1  3i2  4  3    4 

8
8
 8
1
11
units of current flow from B to A, current
units of current flow from F to E
8
8
1  11  5
and       units of current flow from D to C.
8  8 4
Thus,
15
RC CIRCUITS
The circuit shown in Fig. shows a voltage source V 0 connected to a series connection of a
capacitor of capacitance C and a resistor of resistance R. Applying Kirchhoff’s loop law
when the switch in position 1,
1
2
V0
i

–
R
vR
C
vC
V0  vC  iR
Hence,
q
C
dq
But i 
dt
dq q
R
  V0
dt C
Dividing through by R ,
V
dq
q

 0
dt RC R
Multiplying through by e t / RC , gives
dq
q t / RC V0 t / RC
e t / RC

e
 e
dt RC
R
iR  V0 
The two terms on the left can be written

differentiation.
Therefore,
V
d
qe t / RC  0 e t / RC
dt
R




Hence,
V
d
qe t / RC  0 e t / RC
dt
R
V
d qe t / RC  0 e t / RC dt
R


d
qe t / RC , courtesy of the product rule of
dt

16
 d qe
  qe
V0 t / RC
V
V
e dt  0  e t / RC dt  0 RCe t / RC  k  V0 Cet / RC  k
R
R
R
where k is an arbitrary constant of integration.
q  V0 C  ket / RC
t / RC
t / RC

At t = 0, q  0 , as the capacitor is initially uncharged, and 0  CV0  k , or k =  CV0 .
So,
q  CV0  CV0 e t / RC  CV0 (1  e t / RC )
As t   , the capacitor is fully charged and q  q0  CV0 .
Finally,
q  q0 (1  e t / RC )
The current is found from,
dq q0 t / RC
i

e
 i0 e t / RC
dt RC
q
where i0  0 .
RC
q
q0
t
i
i0
t
RC
17
In this position, the charge builds up on the capacitor until it is fully charged as t tends to
infinity. The current decreases from i 0 to zero as t tends to infinity.
The time taken for the current to reduce to 1 / e of its original value is called the time
constant of the RC circuit. This can be obtained by setting i0 e t / RC  i0 e 1 , resulting in
  RC . This can also be seen as the time it would have taken the current to reduce to
zero had decay continued at the initial rate:
i
di
d
1
 i0 e t / RC

i0 e t / RC
  0 (see figure)
dt t 0 dt
RC
RC
t 0
t 0
The larger the time constant, the more slowly the current decays and the more slowly the
charge builds up on the capacitor. The time constant can be seen as the time taken the
current to decay to 0.37 of its initial value or the time taken for the charge to build up to
0.63 of its fully charged level. i0 e 1  0.37i0 , q0 (1  e 1 )  0.63q0 .
When the switch is position 2, the voltage source is in open circuit. It supplies no current.
The capacitor discharges through the resistor. Hence, the direction of current reverses in
the latter. Notice now that vc is now positive.
vC  iR  0
q
iR   0
C
dq
q

0
dt RC
dq
1

dt
q
RC
t
ln q  
k
RC
 t 


 t 


e ln q  e  RC  e k  De  RC 
where D  e k
e ln q  q
q  De t / RC
At t = 0, q  q0
q0  D
Therefore,
q  q 0 e t / RC
The current is given by
q
dq
i
  0 e t / RC  i0 e t / RC
dt
RC
`
18
q
q0
t
t
 i0
With the switch oscillating between positions 1 and 2, we obtain the figure below:
19