In the late 1870s, Thomas Edison and his team at Menlo Park set out to build an electric lamp that ordinary people could use. The idea of making a wire glow by passing current through it was already known — the trouble was that every wire they tried either melted within minutes or demanded so much current that no practical supply could feed it.
The team worked their way through material after material, carbonising cotton thread, cardboard and eventually strips of bamboo, sealing each one inside a glass bulb from which the air had been pumped out. What they were searching for was not simply a substance that glowed. They needed a filament that was thin and highly resistive, so that the current passing through it would dump a large amount of energy into a very small volume — enough to bring it to white heat — while the thick copper wires bringing that same current across the room stayed perfectly cool.
Late in 1879 a carbonised filament finally burned steadily for hours instead of seconds. The lamp that resulted lit streets and homes for the next century.
Diagram
Fig. 11.1 – An early carbon-filament bulb glowing in a laboratory, connected by thick copper leads to a cell, with the current path marked.
Behind that stubborn search lie the ideas of this chapter: what an electric current really is, how fast the charges actually move, why some materials resist their passage far more than others, and where the energy goes as they travel.
Each of these questions is answered somewhere in this chapter. Read on, and by the end you will be able to explain every one of them in your own words.
Every atom has electrons arranged in shells around its nucleus. The electrons in the outermost shell are called valence electrons, and they are the ones held most loosely by the nucleus.
Diagram
Fig. 11.2 – A lattice of positive metal ion cores with de-localized free electrons moving between them.
These de-localized electrons are called conduction electrons or free electrons. When a potential difference is applied across the two ends of the conductor, these free electrons are the particles that flow and constitute an electric current.
This idea — free charge carriers responding to an applied potential difference — is the foundation of the whole chapter. In the sections that follow, this picture is made quantitative: first by defining what “current” means, then by finding how fast these electrons actually drift, and finally by relating current to the potential difference that drives it.
To define electric current in the most general way, imagine a gas made up of both negatively and positively charged particles. This is an imaginary situation, but it is useful because it lets us treat every kind of conductor — metals, electrolytes, gases — with a single definition.
Now imagine a plane P drawn across the region through which these particles move, as shown in Fig. 11.3. The charged particles are moving randomly in all directions, so at any moment some of them cross P in the forward direction and some cross it in the backward direction.
Diagram
Fig. 11.3 – Positively and negatively charged particles moving randomly in various directions across a plane P, with the forward direction marked.
Let us watch the plane P for a time interval \(t\). In that time, let \(q^{+}\) be the amount of positive charge that crosses P in the forward direction, and let \(q^{-}\) be the amount of negative charge that crosses P in the forward direction.
Negative charge moving forward has the same electrical effect as positive charge moving backward. So the two contributions do not add — they oppose each other. The net charge transported in the forward direction is therefore the difference:
\[q = q^{+} - q^{-}\]
For a steady flow, the net charge \(q\) that crosses the plane keeps growing in direct proportion to the time \(t\) for which we watch. Twice the time gives twice the charge. This means the ratio \(q/t\) is a constant for a given steady flow — and it is this constant ratio, the rate at which charge crosses the plane, that we call the electric current \(I\):
\[I = \frac{q}{t} \qquad \text{--- (11.1)}\]
Here \(q\) is the net charge crossing the plane, measured in coulomb (C), and \(t\) is the time interval, measured in second (s). The SI unit of current \(I\) is therefore the ampere (A), where one ampere is one coulomb per second.
Equation (11.1) works only when the flow is steady, so that the charge builds up at a uniform rate. But in many real circuits — a camera flash, a switch being closed, a device being turned on — the current itself keeps changing from instant to instant. For such a case, the ratio \(q/t\) over a long interval would only give an average, not the current at a particular moment.
To get the current at a particular instant, we must look at a very short interval. Let \(\Delta q\) be the net charge that flows across the plane P between time \(t\) and \(t + \Delta t\), that is, during the small time interval \(\Delta t\). The current at that instant is then
\[I(t) = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} \qquad \text{--- (11.2)}\]
In words, the instantaneous current is the limiting value of the ratio \(\Delta q / \Delta t\) as the time interval \(\Delta t\) is made smaller and smaller, tending to zero. Here \(\Delta q\) is in coulomb (C) and \(\Delta t\) is in second (s), so \(I\) is again in ampere (A). Shrinking the interval in this way removes the averaging, and what is left is the rate of flow of charge at that very moment.
Currents met with in practice cover an enormous range of values. The current during lightning could be as high as 10,000 A, while the current in a household circuit could be of the order of a few amperes. At the other extreme, currents of the order of a milliampere (mA), a microampere (µA) or a nanoampere (nA) are common in semiconductor devices.
Getting comfortable with these prefixes matters in numerical work: \(1\ \text{mA} = 10^{-3}\ \text{A}\), \(1\ \mu\text{A} = 10^{-6}\ \text{A}\), and \(1\ \text{nA} = 10^{-9}\ \text{A}\). A very common exam mistake is to substitute a current given in milliamperes straight into a formula as though it were in amperes — always convert to SI units first.
A current can be generated by positively charged particles, by negatively charged particles, or by both together. What matters is only that charge is being transported — not the sign of the charge doing the transporting.
The kind of carrier depends on the material. In an electrolyte — a conducting solution such as copper sulphate solution or the acid inside a car battery — both positively and negatively charged particles take part in the conduction. The positive ions drift one way and the negative ions drift the other way, and both movements add to the same net current.
In a metal, the situation is simpler. The positive ion cores are locked in their fixed positions in the lattice and cannot travel, so only the free electrons are responsible for conduction.
Left to themselves, the free electrons in a metal simply move about at random and transport no net charge. To get a current, an electric field must be applied inside the conductor. These electrons then flow and generate a net current under the action of the applied electric field.
The key point is that the field must be maintained. As long as a steady electric field exists inside the conductor, the electrons continue to flow in the form of a steady current. If the field disappears, the net flow stops almost at once. Such steady electric fields are generated by cells and batteries, which is why a circuit needs a source connected across it to keep a current going.
There is a subtlety here. In a metal wire, the particles that actually move are electrons, which are negative. But the direction we mark for the current on a circuit diagram is not the direction the electrons travel.
So in a metal, the electrons drift one way and the arrow for the current is drawn the other way, as shown in Fig. 11.4. This “conventional current” direction is what is used everywhere in circuit analysis, and it is the direction you must use when answering exam questions unless electron flow is asked for explicitly.
Diagram
Fig. 11.4 – A wire connected to a cell, showing electron drift in one direction and the conventional current arrow in the opposite direction.
Imagine a copper rod with no current flowing through it. Fig. 11.5 shows the schematic of such a conductor with the free electrons in random motion inside it.
Diagram
Fig. 11.5 – Free electrons in random motion inside the conductor, arrows pointing in all directions.
These electrons are moving fast, but they are moving in every possible direction. For every electron heading left there is another heading right, so there is no net motion of these electrons in any direction. That is why an isolated copper rod carries no current, even though it is packed with mobile charge.
Now suppose an electric field is applied along the length of the copper rod, and a current is set up in the rod. The random motion does not stop — the electrons still move randomly, colliding and changing direction as before. What changes is that superposed on this randomness, they now tend to drift slowly in one particular direction.
Because the electron carries a negative charge, the force on it is opposite to the field. So the direction of this drift is opposite to that of the applied electric field.
The numbers here are worth pausing on. The drift speed in a copper conductor is of the order of \(10^{-4}\) m/s to \(10^{-5}\) m/s, whereas the electron random speed is of the order of \(10^{6}\) m/s. The drift is therefore about ten billion times slower than the random motion — a slow, patient crawl riding on top of a furious jiggle.
How is the current through a conductor related to the drift speed of electrons? To answer this, Fig. 11.6 shows a part of a conducting wire with its free electrons having the drift speed \(V_d\) in the direction opposite to the electric field \(\vec{E}\).
Diagram
Fig. 11.6 – A conducting wire of length L with applied electric field E, free electrons drifting with speed V_d opposite to E, and the conventional current I marked.
To keep the calculation manageable, two assumptions are made. It is assumed that all the electrons move with the same drift speed \(V_d\), and that the current \(I\) is the same throughout the cross section (\(A\)) of the wire.
Consider a length \(L\) of the wire. Let \(n\) be the number of free electrons per unit volume of the wire — this is called the number density, and its SI unit is electrons per cubic metre (m\(^{-3}\)).
The volume of this length of wire is \(AL\), so the total number of electrons in the length \(L\) of the conducting wire is \(nAL\). Multiplying this count by the charge on each electron gives the total charge in the length \(L\):
\[q = n\,A\,L\,e \qquad \text{--- (11.3)}\]
where \(e\) is the electron charge, in coulomb (C).
Now comes the central idea. This is the total charge that moves through any cross section of the wire in a certain time interval \(t\) — because in exactly the time an electron takes to cross the length \(L\), every electron in that length will have passed the far end. That time is the length divided by the drift speed:
\[t = \frac{L}{V_d} \qquad \text{--- (11.4)}\]
From Eq. (11.1) and Eq. (11.3), the current is this charge divided by this time:
\[I = \frac{q}{t} = \frac{n\,A\,L\,e}{L / V_d} = n\,A\,V_d\,e \qquad \text{--- (11.5)}\]
Notice how the length \(L\) cancels out. It had to — the current through the wire cannot depend on how long a piece of it we chose to look at.
Rearranging Eq. (11.5) gives the drift speed in terms of the current:
\[V_d = \frac{I}{nAe} = \frac{J}{ne} \qquad \text{--- (11.6)}\]
In the last step above, a new quantity has appeared. Here \(J = I/A\) is the current density. \(J\) is uniform over the cross sectional area \(A\) of the wire. Its unit is A/m\(^2\).
\[J = \frac{I}{A} \qquad \text{--- (11.7)}\]
Current density is a more local description than current. Two wires can carry the same current of 1 A, but if one is thin and the other thick, the thin one has a far higher current density — the same traffic squeezed through a narrower road.
From Eq. (11.6), the current density can be written in vector form:
\[\vec{J} = (ne)\vec{V}_d \qquad \text{--- (11.8)}\]
For electrons, \(ne\) is negative and \(\vec{J}\) and \(\vec{V}_d\) have opposite directions. Here \(\vec{V}_d\) is the drift velocity — the drift speed together with its direction.
This sign result is exactly the convention of Section 11.3 reappearing in vector language: the electrons drift one way, and the current (and hence \(\vec{J}\)) is taken the other way.
By now we know that a potential difference across a conductor sets up an electric field inside it, and that this field makes the free electrons drift, producing a current. The obvious next question is a quantitative one: if the potential difference is doubled, what happens to the current?
The relationship between the current through a conductor and the applied potential difference was first discovered by the German scientist George Simon Ohm in 1828 AD. This relationship is known as Ohm’s law.
It states that “The current \(I\) through a conductor is directly proportional to the potential difference \(V\) applied across its two ends provided the physical state of the conductor is unchanged”.
That last condition is not a decoration; it is the heart of the law. Heat a wire up and its behaviour changes, and the simple proportionality breaks. Any exam statement of Ohm’s law that leaves out “provided the physical state of the conductor is unchanged” is an incomplete statement.
If Ohm’s law holds, then a graph of current versus potential difference across the conductor must be a straight line, as shown in Fig. 11.7 — because that is what direct proportionality looks like on a graph.
Diagram
Fig. 11.7 – I-V curve for a conductor: a straight line for ideal Ohm’s law, with the slightly bending curve of a good conductor shown alongside.
In general, \(I \propto V\). Removing the proportionality sign requires a constant:
\[V = IR \qquad \text{or} \qquad R = \frac{V}{I} \qquad \text{--- (11.9)}\]
where \(R\) is a proportionality constant and is called the resistance of the conductor. Here \(V\) is the potential difference in volt (V), \(I\) is the current in ampere (A), and \(R\) is the resistance. The unit of resistance is the ohm (\(\Omega\)),
\[1\ \Omega = \frac{1\ \text{volt}}{1\ \text{ampere}}\]
If a potential difference of 1 volt across a conductor produces a current of 1 ampere through it, then the resistance of the conductor is 1 \(\Omega\).
Physically, resistance measures how much the conductor opposes the flow of charge. A large \(R\) means a large potential difference is needed to push even a small current through.
It is sometimes more convenient to speak of how easily a conductor passes current rather than how strongly it opposes it. The reciprocal of resistance is called conductance.
\[C = \frac{1}{R} \qquad \text{--- (11.10)}\]
The unit of conductance is siemens or \((\Omega)^{-1}\).
Ohm’s law was discovered experimentally. But why should nature obey it? To see this, we return to the electrons themselves.
We know that electrical conduction in a conductor is due to the mobile charge carriers, the electrons. It is assumed that these conduction electrons are free to move inside the volume of the conductor. During their random motion, electrons collide with the ion cores within the conductor. It is assumed that electrons do not collide with each other. These random motions average to zero — over any reasonable time, the random part of the motion transports no charge at all.
On the application of an electric field \(\vec{E}\), the motion of the electrons is a combination of two things: the random motion of the electrons due to collisions, and the motion due to the electric field \(\vec{E}\). The electrons drift under the action of the field \(\vec{E}\) and move in a direction opposite to the direction of the field \(\vec{E}\).
Consider an electron of mass \(m\) subjected to an electric field \(\vec{E}\). The force experienced by the electron will be \(\vec{F} = e\vec{E}\). From Newton’s second law, force divided by mass gives acceleration, so the acceleration experienced by the electron will then be
\[\vec{a} = \frac{e\vec{E}}{m} \qquad \text{--- (11.11)}\]
Here \(e\) is the electron charge in coulomb (C), \(m\) is the electron mass in kilogram (kg), \(E\) is the field in volt per metre (V/m), and \(a\) is the acceleration in m/s\(^2\).
Equation (11.11) raises a problem. A constant acceleration should make the electron go faster and faster without limit — yet we observe a steady current, which means a steady drift speed. The collisions are what resolve this.
The type of collision the conduction electrons undergo is such that the drift velocity attained before the collision has nothing to do with the drift velocity after the collision. In other words, each collision wipes the slate clean. After the collision, the electron will move in a random direction, but will still drift in the direction opposite to \(\vec{E}\), because the field starts accelerating it all over again.
So the picture is: accelerate, collide and lose all memory of the drift, accelerate again, collide again. The drift speed therefore never builds up beyond what can be gained between two collisions.
Let \(\tau\) (tau) be the average time between two successive collisions, measured in second (s). This is called the relaxation time. Starting from zero drift and accelerating for a time \(\tau\), the electrons will acquire a drift speed \(V_d = a\tau\), where \(a\) is the acceleration given by Eq. (11.11). Also, at any given instant of time, the average drift speed of the electron will also be \(V_d = a\tau\).
From Eq. (11.11),
\[V_d = a\tau = \frac{eE\tau}{m} \qquad \text{--- (11.12)}\]
We now have two independent expressions for the same drift speed. From Eq. (11.6) and Eq. (11.12),
\[V_d = \frac{J}{ne} = \frac{eE\tau}{m} \qquad \text{--- (11.13)}\]
which gives, on solving for \(E\),
\[E = \left(\frac{m}{e^2 n \tau}\right) J \qquad \text{--- (11.14)}\]
or, \(E = \rho J\), where \(\rho\) is the resistivity of the material and
\[\rho = \frac{m}{ne^2\tau} \qquad \text{--- (11.15)}\]
This is the key result. For a given material, \(m\), \(n\), \(e^2\) and \(\tau\) will be constant, and so \(\rho\) will also be constant. In particular, \(\rho\) is independent of \(\vec{E}\), the externally applied electric field.
And that is Ohm’s law, derived rather than assumed. Because \(\rho\) does not depend on \(E\), the relation \(E = \rho J\) says that the current density is simply proportional to the field — double the field, double the current density, double the current. Translated back into the language of a whole wire, that is \(V \propto I\).
Ohm’s law is obeyed by various materials and devices, but it is not a universal law of nature — it is a description of how some conductors behave, under some conditions. It is therefore important to know exactly where it applies and where it fails.
The devices for which the potential difference (\(V\)) versus current (\(I\)) curve is a straight line passing through the origin, inclined to the \(V\)-axis, are called linear devices or ohmic devices (Fig. 11.7). The resistance of these devices is constant. Several conductors obey Ohm’s law. They follow the linear \(I\)–\(V\) characteristic.
Two features of that definition carry weight. The line must be straight, so that the ratio \(V/I\) never changes; and it must pass through the origin, so that zero potential difference gives zero current. A metal wire at constant temperature is the standard example.
The devices for which the \(I\)–\(V\) curve is not a straight line, as shown in Fig. 11.8, are called non-ohmic devices.
Diagram
Fig. 11.8 – I-V curve for a non-Ohmic device: a curved, non-linear characteristic with current in arbitrary units against voltage in volts.
They do not obey Ohm’s law, and the resistance of these devices is a function of \(V\) or \(I\) — that is, the resistance itself changes as the operating conditions change, instead of staying fixed. Examples include liquid electrolytes, vacuum tubes, junction diodes, thermistors, etc.
For a non-ohmic device, the ratio \(V/I\) is different at every point on the curve, so we need a more careful definition. Instead of taking the ratio of the total voltage to the total current, we ask a local question: near a chosen operating voltage, how much does the current change when the voltage is changed slightly?
The resistance \(R\) for such non-linear devices at a particular value of the potential difference \(V\) is given by
\[R = \lim_{\Delta I \to 0} \frac{\Delta V}{\Delta I} = \frac{dV}{dI} \qquad \text{--- (11.16)}\]
where \(\Delta V\) is the potential difference between the two values of potential
\[V - \frac{\Delta V}{2} \quad \text{to} \quad V + \frac{\Delta V}{2},\]
and \(\Delta I\) is the corresponding change in the current. Here \(\Delta V\) is in volt (V), \(\Delta I\) is in ampere (A), and \(R\) is in ohm (\(\Omega\)).
Geometrically, \(dV/dI\) is the slope of the \(V\)–\(I\) curve at that point. So for a non-ohmic device the resistance is read from the steepness of the curve at the operating point, not from the ratio of the coordinates of that point. For an ohmic device the two happen to give the same answer, because a straight line through the origin has the same slope everywhere — which is another way of seeing why \(R\) is constant there.
Consider a resistor AB connected to a cell in a circuit as shown in Fig. 11.9, with current flowing from A to B. The cell maintains a potential difference \(V\) between the two terminals of the resistor, with the higher potential at A and the lower at B.
Diagram
Fig. 11.9 – A simple circuit with a cell and a resistor AB, current I marked, and the potential difference V shown across the resistor.
Let \(Q\) be the charge flowing in time \(\Delta t\) through the resistor from A to B. Now recall what potential difference means: the potential difference \(V\) between the two points A and B is equal to the amount of work \(W\) done to carry a unit positive charge from A to B. It is given by
\[V = \frac{W}{Q}, \qquad W = VQ \qquad \text{--- (11.17)}\]
Here \(W\) is the work in joule (J), \(Q\) is the charge in coulomb (C), and \(V\) is the potential difference in volt (V). One volt is therefore one joule per coulomb.
The cell provides this energy through the charge \(Q\), to the resistor AB where the work is performed.
When the charge \(Q\) flows from the higher potential point A to the lower potential point B — that is, through a decrease in potential of value \(V\) — its potential energy decreases by an amount
\[\Delta U = QV = I\,\Delta t\,V \qquad \text{--- (11.18)}\]
where \(I\) is the current due to the charge \(Q\) flowing in time \(\Delta t\). The second form simply uses \(Q = I\Delta t\) from the definition of current, Eq. (11.1).
Where will this energy go? Energy cannot simply disappear. By the principle of conservation of energy, it is converted into some other form of energy.
Energy delivered is one thing; the rate at which it is delivered is another, and in practice it is usually the more useful quantity. In the limit as \(\Delta t \to 0\), Eq. (11.18) gives
\[\frac{dU}{dt} = I.V \qquad \text{--- (11.19)}\]
Here \(dU/dt\) is power, the time rate of transfer of energy, and is given by
\[P = \frac{dU}{dt} = I.V \qquad \text{--- (11.20)}\]
Here \(P\) is the power in watt (W), \(I\) is the current in ampere (A), and \(V\) is the potential difference in volt (V). One watt is one joule per second.
We can also say that this power is transferred by the cell to the resistor or to any other device placed in place of the resistor, such as a motor or a rechargeable battery. In a motor the energy becomes mechanical work; in a rechargeable battery it becomes stored chemical energy; in a resistor, as we now see, it becomes heat.
Because of the presence of an electric field, the free electrons move across a resistor and there would be an increase in their kinetic energy as they move. When the electrons collide with the ion cores, the energy gained by them is shared among the ion cores. Consequently, vibrations of the ions increase, resulting in heating up of the resistor.
Thus, some amount of energy is dissipated in the form of heat in a resistor. The energy dissipated in the time interval \(\Delta t\) is given by Eq. (11.18), and the energy dissipated per unit time is actually the power dissipated, given by Eq. (11.20).
This connects the whole chapter together. The very collisions that gave us Ohm’s law in Section 11.5 — the collisions that stop the electrons from accelerating forever — are the same collisions that dump energy into the lattice and heat the resistor up.
Using Eq. (11.20), and using Ohm’s law, \(V = IR\):
\[\therefore P = \frac{V^2}{R} = I^2 R \qquad \text{--- (11.21)}\]
So the power dissipated in a resistor can be written in three equivalent ways: \(P = IV\), \(P = I^2R\), and \(P = V^2/R\). They are the same statement, and which one to use depends on which two quantities are known.
There is a trap here that costs marks in exams. \(P = I^2R\) suggests power grows with resistance, while \(P = V^2/R\) suggests it falls with resistance — and both are correct, because they assume different things are being held fixed. If the current through the resistor is fixed, use \(P = I^2R\); if the voltage across it is fixed, use \(P = V^2/R\). Deciding which quantity is common to the elements in a circuit is therefore the first step in any power problem.
It is the power dissipation across a resistor which is responsible for heating it up. For example, the filament of an electric bulb heats up to incandescence, radiating out heat and light.
Every real circuit needs a way to control how much current flows where. Resistors are used to limit the current flowing through a particular path of a circuit. By choosing a suitable resistance, a designer fixes the current in that branch to a safe and useful value, because by Ohm’s law a larger resistance means a smaller current for the same potential difference.
Commercially available resistors are mainly of two types: carbon resistors and wire wound resistors.
High value resistors are mostly carbon resistors. They are small and inexpensive, which is why they appear in such large numbers on the circuit boards inside everyday electronic equipment.
A carbon resistor is far too small to print a number such as “47000 ohm” on. Instead, the values of these resistors are colour coded to mark their values in ohms. The colour coding is standardized by the Electronic Industries Association (EIA), so that a resistor made anywhere in the world can be read anywhere in the world. One such resistor is shown in Fig. 11.10.
Diagram
Fig. 11.10 – A carbon composition resistor showing four coloured bands labelled 1st digit, 2nd digit, multiplier and tolerance.
| Colours | 1st digit | 2nd digit | Multiplier | Tolerance |
|---|---|---|---|---|
| Black | 0 | 0 | ×10⁰ | |
| Brown | 1 | 1 | ×10¹ | ±1% |
| Red | 2 | 2 | ×10² | ±2% |
| Orange | 3 | 3 | ×10³ | |
| Yellow | 4 | 4 | ×10⁴ | |
| Green | 5 | 5 | ×10⁵ | |
| Blue | 6 | 6 | ×10⁶ | |
| Violet | 7 | 7 | ×10⁷ | |
| Gray | 8 | 8 | ×10⁸ | |
| White | 9 | 9 | ×10⁹ | |
| For Gold | ×10⁻¹ | ±5% | ||
| For Silver | ×10⁻² | ±10% | ||
| No colour | - | ±20% |
In the four band resistor colour code illustrated in the above table, the first three bands (which are placed closest together) indicate the value in ohms. The first two bands indicate two numbers, and the third band is often called the decimal multiplier.
The fourth band is separated by a space from the three value bands — this gap is deliberate, so that you know which end to start reading from — and it indicates the tolerance of the resistor.
Tolerance matters because manufacturing is never perfect. A 47 kΩ resistor marked ±5% is guaranteed only to lie somewhere between 5% below and 5% above 47 kΩ; the manufacturer does not promise the exact figure.
i. Colour code of a resistor is:
| Yellow | Violet | Orange | Gold |
|---|---|---|---|
| Value: 4 | 7 | 10³ | ±5% |
i.e. \(47 \times 10^{3} = 47000\ \Omega = 47\ \text{k}\Omega \pm 5\%\)
The value of the resistor is \(47\ \text{k}\Omega \pm 5\%\).
The reasoning is: the first two bands give the digits 4 and 7, which are written side by side as 47; the third band tells us to multiply by \(10^3\); and the gold band fixes the tolerance at ±5%.
ii. From the given value of a resistor, find the colour bands of this resistor:
\[330\ \Omega = 33 \times 10\]
Here the working is reversed. The value is first written as two digits followed by a power of ten:
| 3 | 3 | 10¹ |
|---|---|---|
| Orange | Orange | Brown |
followed by the tolerance band.
So the bands are Orange, Orange, Brown, and then the tolerance band.
A fixed resistor has one value and one value only. But many situations call for a resistance that can be changed while the circuit is running. A rheostat, shown in Fig. 11.11, is an adjustable resistor used in applications that require adjustment of current or resistance in an electric circuit.
Diagram
Fig. 11.11 – A rheostat: a coil of wire wound on a cylinder, with terminal A, terminal B, terminal C and a sliding contact marked.
The working idea is simple. The resistance wire is wound as a long coil, and a sliding contact can be moved along it, as seen in Fig. 11.11. Moving the slider changes how much of the wire is actually included in the circuit — a longer included length means a larger resistance, and hence a smaller current.
The rheostat can be used to adjust the potential difference between two points in a circuit, to change the intensity of lights, and to control the speed of motors, among other uses. In each case the underlying action is the same: change the resistance, and by Ohm’s law the current changes with it.
Its resistive element can be a metal wire or a ribbon, carbon films, or a conducting liquid, depending upon the application. In hi-fi equipment, rheostats are used for volume control.
Circuits rarely contain a single resistor. When several are used together, we usually want to replace the whole group by one equivalent resistance — a single resistor that would draw exactly the same current from the same supply. There are two basic ways of connecting them.
In a series combination of resistors, these are connected in a single electrical path, as shown in Fig. 11.12. Hence the same electric current flows through each resistor in a series combination — the charge has nowhere else to go, so whatever passes through the first resistor must pass through the second.
Diagram
Fig. 11.12 – Series combination of two resistors R₁ and R₂ connected to a cell, with V₁ and V₂ marked across them and current I in the single path.
Because of the series combination, the supply voltage between the two resistors \(R_1\) and \(R_2\) is divided into \(V_1\) and \(V_2\) respectively, while the same current \(I\) flows through the resistor \(R_1\) and the resistor \(R_2\).
In short: in a series combination, the supply voltage is divided, and the current remains the same in all the resistors. This one sentence is the key to solving almost every series problem.
According to Ohm’s law, applied separately to each resistor,
\[R_1 = \frac{V_1}{I}, \qquad R_2 = \frac{V_2}{I} \qquad \text{--- (11.22)}\]
which can also be written as \(V_1 = IR_1\) and \(V_2 = IR_2\). The two voltage drops must add up to the total supply voltage, since going from one end of the combination to the other means falling through both drops one after the other:
\[V = V_1 + V_2 \qquad \text{--- (11.23)}\]
From equations (11.22) and (11.23) we write
\[V = I(R_1 + R_2) \qquad \text{--- (11.24)}\]
\[\therefore V = I R_s \qquad \text{--- (11.25)}\]
Comparing the last two lines, the equivalent resistance of the series circuit is
\[R_s = R_1 + R_2\]
When a number of resistors are connected in series, the equivalent resistance is equal to the sum of the individual resistances.
For \(n\) number of resistors,
\[R_s = R_1 + R_2 + R_3 + \dots + R_n = \sum_{i=1}^{i=n} R_i \qquad \text{--- (11.26)}\]
Note what this means: adding a resistor in series always increases the total resistance. The equivalent resistance of a series combination is therefore larger than the largest individual resistance in it — a useful check on your answer.
In the parallel combination, the resistors are connected in such a way that the same voltage is applied across each resistor.
A number of resistors are said to be connected in parallel if all of them are connected between the same two electrical points, each having an individual path, as shown in Fig. 11.13. Because both ends of every resistor are tied to the same two points, every resistor must experience the same potential difference.
Diagram
Fig. 11.13 – Two resistors R₁ and R₂ in parallel combination across a cell, with total current I splitting into I₁ and I₂.
In parallel combination, the total current \(I\) is divided into \(I_1\) and \(I_2\) as shown in the circuit diagram Fig. 11.13, whereas the voltage \(V\) across them remains the same.
\[I = I_1 + I_2 \qquad \text{--- (11.27)}\]
where \(I_1\) is the current flowing through \(R_1\) and \(I_2\) is the current flowing through \(R_2\). This is simply conservation of charge: the current that arrives at the junction must equal the total current leaving it.
When Ohm’s law is applied to \(R_1\):
\[V = I_1 R_1 \qquad \text{i.e.} \qquad I_1 = \frac{V}{R_1} \qquad \text{--- (11.28a)}\]
Ohm’s law applied to \(R_2\):
\[V = I_2 R_2 \qquad \text{i.e.} \qquad I_2 = \frac{V}{R_2} \qquad \text{--- (11.28b)}\]
From Eq. (11.27) and Eq. (11.28),
\[\therefore I = \frac{V}{R_1} + \frac{V}{R_2}\]
Now let \(R_p\) be the single resistance that would draw the same total current from the same supply. Then
\[\text{If,} \qquad I = \frac{V}{R_p},\]
\[\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2}\]
Since \(V\) is common to every term, it cancels out, leaving
\[\therefore \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \qquad \text{--- (11.29)}\]
where \(R_p\) is the equivalent resistance in the parallel combination.
If \(n\) resistors \(R_1, R_2, R_3, \dots, R_n\) are connected in parallel, the equivalent resistance of the combination is given by
\[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} = \sum_{i=1}^{n} \frac{1}{R_i} \qquad \text{--- (11.30)}\]
Thus, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.
Adding a resistor in parallel always decreases the total resistance, because it opens one more path for the current. So \(R_p\) is always smaller than the smallest individual resistance in the combination — again, a quick check on any answer. A very common exam mistake is to stop at \(1/R_p\) and forget to take the reciprocal at the end.
Resistance is not a fixed property handed to a wire once and for all. At a particular temperature, the resistance of a given conductor is observed to depend on three things: the nature of the material of the conductor, the area of its cross-section, and its length.
It is found by experiment that the resistance \(R\) of a conductor of uniform cross section is
i. directly proportional to its length \(l\), i.e. \(R \propto l\)
ii. inversely proportional to its area of cross section \(A\), i.e. \(R \propto \dfrac{1}{A}\)
Both results make physical sense, as Fig. 11.15 suggests. A longer wire means the drifting electrons must run the gauntlet of more collisions before emerging at the far end, so the opposition is greater. A thicker wire means more parallel routes through which the same charge can travel, so the opposition is less.
Diagram
Fig. 11.15 – Two wires of the same material: one long and thin, one short and thick, showing that resistance rises with length and falls with cross-sectional area.
Combining i and ii,
\[R = \rho\,\frac{l}{A} \qquad \text{--- (11.31)}\]
where \(\rho\) is a constant of proportionality and it is called the specific resistance or resistivity of the material of the conductor at a given temperature. Here \(R\) is in ohm (\(\Omega\)), \(l\) is in metre (m), and \(A\) is in square metre (m\(^2\)).
From Eq. (11.31), we write
\[\rho = \frac{RA}{l} \qquad \text{--- (11.32)}\]
The SI unit of resistivity is the ohm-metre (\(\Omega\)m).
Resistivity of a conductor is numerically the resistance per unit length and per unit area of cross-section of the material of the conductor. That is, when \(R = 1\ \Omega\), \(A = 1\ \text{m}^2\) and \(l = 1\) m, then \(\rho = 1\ \Omega\)m.
Notice also that Eq. (11.31) already contains the microscopic picture from Section 11.5. There we found \(\rho = m/(ne^2\tau)\) — a formula built entirely from properties of the material, with no mention of length or thickness. That is exactly why \(\rho\) is the material property and \(R\) is the object property.
Conductivity: the reciprocal of resistivity is called the conductivity of a material, \(\sigma = (\rho)^{-1}\).
The SI unit of \(\sigma\) is \((\Omega\text{m})^{-1}\), i.e. siemens/metre (Sm\(^{-1}\)).
A high resistivity means a poor conductor; a high conductivity means a good one. The two quantities carry exactly the same information, stated from opposite ends.
| Material | Resistivity ρ (Ω.m) | Material | Resistivity ρ (Ω.m) |
|---|---|---|---|
| Conductors | Semiconductors | ||
| Silver | 1.59 × 10⁻⁸ | Carbon | 3.5 × 10⁻⁵ |
| Copper | 1.72 × 10⁻⁸ | Germanium | 0.5 |
| Gold | 2.44 × 10⁻⁸ | Silicon | 3 × 10⁴ |
| Aluminium | 2.82 × 10⁻⁸ | Insulators | |
| Tungsten | 5.6 × 10⁻⁸ | Glass | 10¹¹ - 10¹³ |
| Iron | 9.7 × 10⁻⁸ | Mica | 10¹¹ - 10¹⁵ |
| Mercury | 95.8 × 10⁻⁸ | Rubber (hard) | 10¹³ - 10¹⁶ |
| Nichrome (alloy) | 100 × 10⁻⁸ | Teflon | 10¹⁶ |
| Wood (maple) | 3 × 10⁸ |
The range in this table is staggering — from about \(10^{-8}\) \(\Omega\)m for silver to about \(10^{16}\) \(\Omega\)m for Teflon. That is a factor of roughly \(10^{24}\), and few physical quantities vary over so vast a range. It is this enormous gap that makes it possible to build circuits at all: the current can be confined to the conducting paths we choose, because everything around them resists it so overwhelmingly.
Resistivity \(\rho\) is a property of a material, while the resistance \(R\) refers to a particular object. The same distinction runs through the other quantities of this chapter.
The electric field \(\vec{E}\) at a point is specified in a material, where the potential difference is specified across the resistance. Likewise, the current density \(\vec{J}\) is specified in a material, instead of the current \(I\) in the resistor. Then for an isotropic material — one whose properties are the same in all directions —
\[\rho = \frac{E}{J} \qquad \text{or} \qquad \vec{E} = \rho \vec{J} \qquad \text{--- (11.33)}\]
Again, the SI unit of \(\rho\) can be checked from this equation:
\[\frac{\text{unit}(E)}{\text{unit}(J)} = \frac{\text{V/m}}{\text{A/m}^2} = \frac{\text{V}}{\text{A}}\,\text{m} = \Omega.\text{m}\]
In terms of the conductivity \(\sigma\) of a material, from (11.33),
\[\vec{J} = \frac{1}{\rho}\vec{E} = \sigma \vec{E} \qquad \text{--- (11.34)}\]
Ohm’s law came with a condition attached: “provided the physical state of the conductor is unchanged.” Temperature is the physical state that changes most easily, so we must now ask what heating does to a conductor.
Resistivity of a material varies with temperature. It is a property of the material. Fig. 11.16 shows the variation of resistivity of copper as a function of temperature (K).
Diagram
Fig. 11.16 – Resistivity of copper (in units of 10⁻⁸ Ω.m) plotted against temperature in kelvin, showing a straight line over the range shown, with room temperature marked.
It can be seen in Fig. 11.16 that the variation is linear over a certain range of temperatures — the graph is a straight line, not a curve. Such a linear relation can be expressed as
\[\rho = \rho_0[1 + \alpha(T - T_0)] \qquad \text{--- (11.35)}\]
where \(T_0\) is the chosen reference temperature and \(\rho_0\) is the resistivity at the chosen temperature; for example, \(T_0\) can be 0 °C.
The physical reason behind this rise is already familiar from Section 11.5. Raising the temperature makes the ion cores vibrate more violently, so the drifting electrons run into them more often. The average time \(\tau\) between successive collisions therefore falls, and since \(\rho = m/(ne^2\tau)\), a smaller \(\tau\) means a larger \(\rho\). Heat the metal, crowd the collisions, raise the resistivity.
Rearranging Eq. (11.35) to make \(\alpha\) the subject:
\[\alpha = \frac{\rho - \rho_0}{\rho_0(T - T_0)} = \frac{R - R_0}{R_0(T - T_0)} \qquad \text{--- (11.36)}\]
The second form follows because, for a given wire, \(R\) is proportional to \(\rho\) by Eq. (11.31) — so the fractional change in resistance is the same as the fractional change in resistivity.
Here \(\alpha\) is called the temperature coefficient of resistivity. Table (11.1) shows the resistivity of some of the metals.
Read Eq. (11.36) as a recipe for that definition: the numerator is the increase in resistance, dividing by \(R_0\) makes it “per unit original resistance”, and dividing by \((T - T_0)\) makes it “per degree rise”.
From Eq. (11.36),
\[R = R_0[1 + \alpha(T - T_0)] \qquad \text{--- (11.37)}\]
This is the working formula for almost every numerical problem in this section. Here \(R_0\) is the resistance at the reference temperature \(T_0\), and \(R\) is the resistance at temperature \(T\).
For a small difference in temperatures,
\[\alpha = \frac{1}{R_0} \cdot \frac{dR}{dT} \qquad \text{--- (11.38)}\]
When charges flow through a conductor, a potential difference has to be established between the two ends of the conductor. And for a steady flow of charges, this potential difference is required to be maintained across the two ends of the conductor — its terminals — not merely set up once and left to decay.
There is a device that does exactly this, by doing work on the charges and thereby maintaining the potential difference. Such a device is called an emf device, and it provides the emf \(\varepsilon\). The charges move in the conductor owing to the energy provided by the emf device, and the device supplies this energy through the work it does.
You must have used some of these emf devices already. Power cells, batteries, solar cells, fuel cells, and even generators are some familiar examples of emf devices.
Fig. 11.17 shows a circuit with an emf device and a resistor \(R\). Here, the emf device keeps the positive terminal (+) at a higher electric potential than the negative terminal (−).
Diagram
Fig. 11.17 – A circuit with an emf device of internal resistance r connected to an external resistor R, with current I marked.
The emf is represented by an arrow drawn from the negative terminal to the positive terminal of a device such as a voltaic cell. When the circuit is open, there is no net flow of charge carriers within the device.
When it is connected in a circuit, there is a flow of carriers from one terminal to the other terminal inside the emf device. The positive charge carriers move towards the positive terminal, which acts as the cathode inside the emf device.
Now notice how strange this is. Thus the positive charge carriers move from a region of lower potential energy to a region of higher potential energy, which is the cathode inside the emf device. Left to themselves, charges would never do this — they would run downhill, not uphill. Something must be pushing them, and that something is the source of energy inside the device.
Here, the energy source is chemical in nature. In a solar cell, it is the photon energy in the solar radiation.
Now suppose that a charge \(dq\) flows through the cross section of the circuit (Fig. 11.17) in time \(dt\).
It is clear that the same amount of charge \(dq\) flows throughout the circuit, including through the emf device. Inside the device, it enters at the negative terminal (the low potential terminal) and leaves at the positive terminal (the higher potential terminal).
Hence the device must do work \(dw\) on the charge \(dq\), so that it moves in the above manner. Thus we define the emf of the emf device:
\[\varepsilon = \frac{dw}{dq} \qquad \text{--- (11.39)}\]
The SI unit of emf is joule/coulomb (J/C), which is the volt.
In an ideal device, there is no internal resistance to the motion of the charge carriers. The emf of the device is then equal to the potential difference across the two terminals of the device.
In a real emf device, there is an internal resistance to the motion of the charge carriers. This resistance arises because the charges must travel through the material of the device itself — the electrolyte of a cell, for instance — and that material resists their passage just as any conductor would.
If such a device is not connected in a circuit, there is no current through it. In that case the emf is equal to the potential difference across the two terminals of the emf device. This is a useful practical fact: the reading of a voltmeter across a cell on open circuit gives the emf directly, because with no current flowing there is no drop across the internal resistance.
But if a current (\(I\)) flows through an emf device, there is an internal resistance (\(r\)), and the emf (\(\varepsilon\)) differs from the potential difference across its two terminals (\(V\)):
\[V = \varepsilon - (I)(r) \qquad \text{--- (11.40)}\]
The negative sign is due to the fact that the current \(I\) flows through the emf device from the negative terminal to the positive terminal.
Physically, part of the energy given to each coulomb of charge by the cell is spent inside the cell itself, pushing that charge through the internal resistance. Only what is left over appears across the terminals for the external circuit to use.
By the application of Ohm’s law, Eq. (11.9), to the external resistor,
\[V = IR\]
Hence, substituting this into Eq. (11.40),
\[IR = \varepsilon - Ir \qquad \text{--- (11.41)}\]
Collecting the current terms, \(IR + Ir = \varepsilon\), so \(I(R + r) = \varepsilon\), or
\[I = \frac{\varepsilon}{R + r} \qquad \text{--- (11.42)}\]
This is one of the most useful equations in circuit work. It says that the current is decided by the emf divided by the total resistance of the circuit — the external resistance and the internal resistance added together, since the current must pass through both in turn.
Forgetting to include \(r\) in the denominator is among the commonest errors in numerical problems on cells — always check whether the internal resistance has been given before assuming the circuit is ideal.
Thus, the maximum current that can be drawn from the emf device is when \(R = 0\), i.e. when the terminals are joined directly by a wire of negligible resistance:
\[I_{max} = \frac{\varepsilon}{r} \qquad \text{--- (11.43)}\]
This is the maximum allowed current from an emf device (or a cell). Even in the extreme case of a short circuit, the internal resistance is still there, and it alone limits the current. This decides the maximum current rating of a cell or a battery.
A single cell provides a fixed emf. When a larger voltage is needed, cells are combined — and the simplest way is to join them in series.
In a series combination, cells are connected in a single electrical path, such that the positive terminal of one cell is connected to the negative terminal of the next cell, and so on. This is the essential rule: positive to negative, all the way along the chain, so that every cell pushes the charge in the same direction round the circuit.
The terminal voltage of the battery is then equal to the sum of the voltages of the individual cells in series, as shown in Fig. 11.18 (a).
Diagram
Fig. 11.18 (a) – Two 1.5 V cells connected in series, positive terminal of one joined to the negative terminal of the next.
The figure shows two 1.5 V cells in series. This combination provides a total voltage of 3.0 V (\(1.5 \times 2\)).
The equivalent emf of \(n\) number of cells in a series combination is the algebraic sum of their individual emfs. The word algebraic matters: if a cell is accidentally reversed, its emf must be entered with a negative sign, and it will subtract from the total instead of adding to it.
The equivalent internal resistance of \(n\) cells in a series combination is the sum of their individual internal resistances. This follows directly from Section 11.8.2 — the internal resistances lie one after another in a single path, so they add exactly like resistors in series.
Putting both results into the terminal-voltage relation of Eq. (11.40), the terminal voltage of the whole series combination carrying a current \(I\) is
\[V = \sum_i \varepsilon_i - I.\sum_i r_i \qquad \text{--- (11.44)}\]
Here \(\varepsilon_i\) is the emf of the \(i\)-th cell in volt (V), \(r_i\) is its internal resistance in ohm (\(\Omega\)), and \(I\) is the current in the circuit in ampere (A).
(i) The cells connected in series produce a larger resultant voltage. This is the whole purpose of the arrangement — it is the way to obtain a voltage higher than any one cell can supply.
(ii) Cells which are damaged can be easily identified, and hence can be easily replaced. Because there is only one path, a single dead cell interrupts the whole circuit, which makes the fault obvious at once.
Consider two cells which are connected in parallel. Here, the positive terminals of all the cells are connected together, and the negative terminals of all the cells are connected together.
Notice how this differs from the series case: there, positive was joined to negative; here, like is joined to like. In parallel connection, the current is divided among the branches, i.e. into \(I_1\) and \(I_2\), as shown in Fig. 11.18 (b).
Diagram
Fig. 11.18 (b) – Two cells connected in parallel between points B₁ and B₂, with the total current I dividing into I₁ and I₂ through the two cells.
Consider the points \(B_1\) and \(B_2\) having potentials \(V_{B_1}\) and \(V_{B_2}\) respectively. Since both cells are connected between these same two points, both must have the same potential difference \(V = V_{B_1} - V_{B_2}\) across them. This single fact is what drives the whole derivation.
For the first cell, the potential difference across its terminals is given by Eq. (11.40):
\[V = V_{B_1} - V_{B_2} = \varepsilon_1 - I_1 r_1 \qquad \text{--- (11.45)}\]
Rearranging to make the branch current the subject,
\[\therefore I_1 = \frac{\varepsilon_1 - V}{r_1} \qquad \text{--- (11.46)}\]
Points \(B_1\) and \(B_2\) are connected exactly similarly to the second cell. Hence, considering the second cell, we write
\[V = V_{B_1} - V_{B_2} = \varepsilon_2 - I_2 r_2\ ; \qquad I_2 = \frac{\varepsilon_2 - V}{r_2} \qquad \text{--- (11.47)}\]
We know that the total current is the sum of the branch currents,
\[I = I_1 + I_2\]
Combining the last three equations,
\[\therefore I = \frac{\varepsilon_1}{r_1} - \frac{V}{r_1} + \frac{\varepsilon_2}{r_2} - \frac{V}{r_2} \qquad \text{--- (11.48)}\]
Grouping the emf terms together and the \(V\) terms together,
\[= \left(\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right)\]
Thus,
\[V\left(\frac{1}{r_1} + \frac{1}{r_2}\right) = \left(\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\right) - I\]
Taking the common denominator \(r_1 r_2\) on both sides,
\[\therefore V\left(\frac{r_1 + r_2}{r_1 r_2}\right) = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} - I\]
and finally, multiplying through by \(\dfrac{r_1 r_2}{r_1 + r_2}\),
\[V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\,\frac{r_1 r_2}{r_1 + r_2} \qquad \text{--- (11.49)}\]
Equation (11.49) has a revealing shape. The first term contains no current at all, while the second is proportional to the current — exactly the form \(V = \varepsilon - Ir\) that a single cell would have.
So suppose we replace the cells by a single cell connected between the points \(B_1\) and \(B_2\), with emf \(\varepsilon_{eq}\) and internal resistance \(r_{eq}\), as in Fig. 11.18 (b). Then
\[V = \varepsilon_{eq} - I r_{eq} \qquad \text{--- (11.50)}\]
Considering Eq. (11.49) and Eq. (11.50), and matching the two terms one by one, we can write
\[\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}\]
\[r_{eq} = \frac{r_1 r_2}{r_1 + r_2}\]
i.e.
\[\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}\]
\[\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \qquad \text{--- (11.51)}\]
The internal resistances have combined exactly like resistors in parallel — which is what we should have expected, since the cells offer the current two alternative paths between the same two points.
For \(n\) number of cells connected in parallel with emfs \(\varepsilon_1, \varepsilon_2, \varepsilon_3, \dots, \varepsilon_n\) and internal resistances \(r_1, r_2, r_3, \dots, r_n\),
\[\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \dots + \frac{1}{r_n} \qquad \text{--- (11.52)}\]
and
\[\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \dots + \frac{\varepsilon_n}{r_n} \qquad \text{--- (11.53)}\]
Substitution of emfs should be done algebraically, by considering the proper ± signs according to polarity. A cell connected the wrong way round enters the sum with a negative emf and works against the others.
Advantages of cells in parallel: For cells connected in parallel in a circuit, the circuit will not break open even if a cell gets damaged or goes open. The other cells still provide a complete path between the same two points, so the current continues to flow — a useful kind of redundancy that a series chain does not have.
Disadvantages of cells in parallel: The voltage developed by the cells in a parallel connection cannot be increased by increasing the number of cells present in the circuit. However many identical cells are added, the equivalent emf stays at the value of one cell.
Taken together, the two arrangements answer two different needs: series raises the emf, parallel lowers the internal resistance. A real battery bank often uses both, in a series-parallel arrangement, to get the voltage and the current it needs.
Electrical cells can be divided into several categories, such as the primary cell, the secondary cell, the fuel cell, and so on. The classification rests on one main question: once the chemical energy inside has been used up, can it be put back?
A primary cell cannot be charged again. It can be used only once, and when its chemicals are exhausted it must be thrown away. Dry cells and alkaline cells are different examples of primary cells.
Primary cells are low in cost and can be used easily, which is why they are so common in torches, remote controls and wall clocks. But these are not suitable for heavy loads — they cannot supply a large current for long. Secondary cells are used for such applications.
The secondary cells are rechargeable and can be reused. The chemical reaction in a secondary cell is reversible: driving a current backwards through the cell reverses the reaction and restores the original chemicals, storing energy again.
The lead acid cell and the fuel cell are some examples of secondary cells. The lead acid battery is used widely in vehicles and in other applications which require high load currents. Solar cells are secondary cells that convert solar energy into electrical energy.
Fuel cell vehicles (FCVs) are electric vehicles that use fuel cells instead of lead acid batteries to power the vehicles. Hydrogen is used as a fuel in fuel cells. The by-product after its burning is water.
This is important in terms of reducing the emission of greenhouse gases produced by traditional gasoline fuelled vehicles. The hydrogen fuel cell vehicles are thus more environment friendly — the exhaust is water rather than carbon dioxide.
When many atoms come together in a metal, the valence electrons become de-localized. These conduction electrons (free electrons) constitute an electric current when a potential difference is applied across the conductor.
If \(q^{+}\) and \(q^{-}\) are the positive and negative charges crossing a plane in the forward direction in time \(t\), the net charge is \(q = q^{+} - q^{-}\), and for a steady flow the electric current is \[I = \frac{q}{t} \qquad \text{(11.1)}\] SI unit: ampere (A); charge in coulomb (C), time in second (s).
For a current varying with time, the instantaneous current is \[I(t) = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} \qquad \text{(11.2)}\]
Lightning currents may be as high as 10,000 A; household currents are of the order of a few amperes; currents in semiconductor devices are of the order of mA, µA or nA.
In an electrolyte both positive and negative charges conduct; in a metal only the free electrons conduct. A steady electric field, maintained by cells and batteries, keeps a steady current flowing.
Sign convention: the direction of current is drawn in the direction in which positive charges would move, even though in a metal the electrons actually move opposite to that direction.
Without a field, the free electrons move randomly with no net motion. Under an applied field they drift with drift speed \(V_d\), opposite to \(\vec{E}\).
Drift speed in copper is of the order of \(10^{-4}\) m/s to \(10^{-5}\) m/s, while the random speed is of the order of \(10^{6}\) m/s.
Total charge in length \(L\) of a wire: \(q = nALe\) (11.3); time to cross it: \(t = L/V_d\) (11.4). Hence \[I = n A V_d e \qquad \text{(11.5)}\] where \(n\) = number of free electrons per unit volume (m\(^{-3}\)), \(A\) = area of cross-section (m\(^2\)), \(e\) = electronic charge (C).
\[V_d = \frac{I}{nAe} = \frac{J}{ne} \qquad \text{(11.6)}\]
Current density \(J = \dfrac{I}{A}\) (11.7), SI unit A/m\(^2\); in vector form \(\vec{J} = (ne)\vec{V}_d\) (11.8). For electrons \(ne\) is negative, so \(\vec{J}\) and \(\vec{V}_d\) point in opposite directions.
Ohm’s law: the current \(I\) through a conductor is directly proportional to the potential difference \(V\) applied across its ends, provided the physical state of the conductor is unchanged.
\[V = IR \qquad \text{or} \qquad R = \frac{V}{I} \qquad \text{(11.9)}\] \(R\) is the resistance, SI unit ohm (\(\Omega\)), where \(1\ \Omega = 1\ \text{volt}/1\ \text{ampere}\).
Conductance \(C = \dfrac{1}{R}\) (11.10); unit siemens or \((\Omega)^{-1}\).
Electrons collide with the ion cores (not with each other), and each collision destroys the drift gained before it. Acceleration of an electron: \[\vec{a} = \frac{e\vec{E}}{m} \qquad \text{(11.11)}\]
With \(\tau\) as the average time between two successive collisions, \[V_d = a\tau = \frac{eE\tau}{m} \qquad \text{(11.12)}\] \[V_d = \frac{J}{ne} = \frac{eE\tau}{m} \qquad \text{(11.13)}\] \[E = \left(\frac{m}{e^2 n \tau}\right) J \qquad \text{(11.14)}\]
Hence \(E = \rho J\), with resistivity \[\rho = \frac{m}{ne^2\tau} \qquad \text{(11.15)}\] For a given material \(m\), \(n\), \(e^2\) and \(\tau\) are constant, so \(\rho\) is constant and independent of the applied field \(\vec{E}\) — which is Ohm’s law.
Ohmic (linear) devices: \(V\)–\(I\) curve is a straight line through the origin; resistance is constant.
Non-ohmic devices: \(I\)–\(V\) curve is not a straight line; resistance depends on \(V\) or \(I\). Examples: liquid electrolytes, vacuum tubes, junction diodes, thermistors.
For a non-linear device, the resistance at a particular potential difference \(V\) is \[R = \lim_{\Delta I \to 0} \frac{\Delta V}{\Delta I} = \frac{dV}{dI} \qquad \text{(11.16)}\]
Potential difference: \(V = \dfrac{W}{Q}\), so \(W = VQ\) (11.17). Loss of potential energy of charge \(Q\) falling through \(V\): \(\Delta U = QV = I\,\Delta t\,V\) (11.18).
Power, the time rate of transfer of energy: \[P = \frac{dU}{dt} = I.V \qquad \text{(11.19), (11.20)}\] SI unit: watt (W).
Using Ohm’s law, \[P = \frac{V^2}{R} = I^2 R \qquad \text{(11.21)}\]
Energy is dissipated as heat because electrons hand over the energy gained from the field to the ion cores at each collision, increasing the lattice vibrations. This heating makes a bulb filament glow.
Commercial unit of energy: 1 kWh = 1 unit of electrical energy.
Resistors limit the current in a branch of a circuit. Two main commercial types: carbon resistors and wire wound resistors. Carbon resistor values are marked by a four-band colour code standardized by the EIA: first two bands give the digits, the third is the decimal multiplier, and the fourth (set apart by a gap) gives the tolerance.
Mnemonic for the colour sequence: “B. B. Roy in Great Britain has Very Good Wife” — B B R O Y G B V G W.
A rheostat is an adjustable resistor used to vary the current or resistance in a circuit — for adjusting potential difference, changing light intensity, controlling motor speed, and volume control in hi-fi equipment.
Series combination: same current through each resistor; supply voltage divides. \[R_s = R_1 + R_2 + R_3 + \dots + R_n = \sum_{i=1}^{i=n} R_i \qquad \text{(11.26)}\]
Parallel combination: same voltage across each resistor; total current divides. \[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} = \sum_{i=1}^{n}\frac{1}{R_i} \qquad \text{(11.30)}\] For two resistors, \(R_p = \dfrac{R_1 R_2}{R_1 + R_2}\).
Resistance of a conductor of uniform cross-section: \(R \propto l\) and \(R \propto \dfrac{1}{A}\), hence \[R = \rho\,\frac{l}{A} \qquad \text{(11.31)} \qquad \text{and} \qquad \rho = \frac{RA}{l} \qquad \text{(11.32)}\]
Resistivity \(\rho\) is a property of the material; resistance \(R\) is a property of a particular object. SI unit of \(\rho\): ohm-metre (\(\Omega\)m). Numerically, \(\rho = 1\ \Omega\)m when \(R = 1\ \Omega\), \(A = 1\ \text{m}^2\), \(l = 1\) m.
Conductivity \(\sigma = (\rho)^{-1}\); SI unit \((\Omega\text{m})^{-1}\), i.e. siemens/metre (Sm\(^{-1}\)).
For an isotropic material, \[\rho = \frac{E}{J} \quad \text{or} \quad \vec{E} = \rho\vec{J} \qquad \text{(11.33)} \qquad\text{and}\qquad \vec{J} = \frac{1}{\rho}\vec{E} = \sigma\vec{E} \qquad \text{(11.34)}\]
Resistivities range from about \(10^{-8}\) \(\Omega\)m for conductors (silver \(1.59 \times 10^{-8}\), copper \(1.72 \times 10^{-8}\), nichrome \(100 \times 10^{-8}\) \(\Omega\)m) through semiconductors to about \(10^{16}\) \(\Omega\)m for insulators such as Teflon.
Resistivity of a metal varies linearly with temperature over a certain range: \[\rho = \rho_0[1 + \alpha(T - T_0)] \qquad \text{(11.35)}\] where \(T_0\) is the chosen reference temperature and \(\rho_0\) the resistivity at that temperature.
\[\alpha = \frac{\rho - \rho_0}{\rho_0(T - T_0)} = \frac{R - R_0}{R_0(T - T_0)} \qquad \text{(11.36)}\] \[R = R_0[1 + \alpha(T - T_0)] \qquad \text{(11.37)}\] \[\alpha = \frac{1}{R_0} \cdot \frac{dR}{dT} \qquad \text{(11.38)}\]
Temperature coefficient of resistance \(\alpha\) = increase in resistance per unit original resistance at the reference temperature, per degree rise in temperature. Unit: °C\(^{-1}\) or K\(^{-1}\). Since only the temperature difference enters, the Celsius and Kelvin scales may be used interchangeably.
Physically, heating increases lattice vibrations, reducing \(\tau\) and hence raising \(\rho\).
Superconductivity: in some metals and alloys the resistivity drops suddenly to zero at a critical temperature \(T_c\) — for example, mercury at 4.2 K. Superconducting magnets produce fields of a few tesla and are used in research quality NMR spectrometers; their coils must be kept below \(T_c\).
An emf device does work on the charges to maintain the potential difference needed for a steady current. Examples: cells, batteries, solar cells, fuel cells, generators.
\[\varepsilon = \frac{dw}{dq} \qquad \text{(11.39)}\] SI unit of emf: joule/coulomb (J/C), i.e. volt.
For a real cell of internal resistance \(r\) delivering current \(I\), the terminal potential difference is \[V = \varepsilon - Ir \qquad \text{(11.40)}\]
With external resistance \(R\): \(IR = \varepsilon - Ir\) (11.41), so \[I = \frac{\varepsilon}{R + r} \qquad \text{(11.42)}\]
Maximum current, when \(R = 0\): \[I_{max} = \frac{\varepsilon}{r} \qquad \text{(11.43)}\] This decides the maximum current rating of a cell or battery.
Cells in series: positive terminal of one joined to the negative terminal of the next. Equivalent emf = algebraic sum of individual emfs; equivalent internal resistance = sum of individual internal resistances. \[V = \sum_i \varepsilon_i - I.\sum_i r_i \qquad \text{(11.44)}\] Advantages: larger resultant voltage; a damaged cell is easily identified and replaced.
Cells in parallel: like terminals joined together; the current divides among the branches. For two cells, \[\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \frac{r_1 r_2}{r_1 + r_2}, \qquad \frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \qquad \text{(11.51)}\]
For \(n\) cells in parallel, \[\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \dots + \frac{1}{r_n} \qquad \text{(11.52)}\] \[\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \dots + \frac{\varepsilon_n}{r_n} \qquad \text{(11.53)}\] Emfs must be substituted algebraically, with proper ± signs according to polarity.
Advantage of parallel cells: the circuit does not break open even if one cell is damaged. Disadvantage: the voltage cannot be increased by adding more cells.
Primary cells cannot be recharged; the chemical reaction is irreversible. Examples: dry cells, alkaline cells. Low cost and easy to use, but unsuitable for heavy loads.
Secondary cells are rechargeable; the chemical reaction is reversible. Examples: lead acid cell, fuel cell. Lead acid batteries are used where high load currents are needed. Solar cells convert solar energy into electrical energy.
Fuel cell vehicles (FCVs) use hydrogen as fuel; the by-product is water, so they emit no greenhouse gases and are more environment friendly than gasoline vehicles.
A metal wire lying on a table contains an enormous number of free electrons moving at speeds of the order of 10⁶ m/s, yet it carries no current. Explain why.
The drift speed of electrons in a copper wire is only about 10⁻⁴ m/s. Why, then, does a bulb light up almost immediately when the switch is closed?
Why is the direction of conventional current taken opposite to the direction in which electrons actually move in a metal? Does adopting the opposite convention change any physics?
State Ohm’s law. Why is the condition “provided the physical state of the conductor is unchanged” an essential part of the statement and not an optional extra?
If the electric field inside a conductor exerts a constant force on each free electron, why does the electron not accelerate indefinitely? What limits its drift speed?
Using \(\rho = \dfrac{m}{ne^2\tau}\), explain why the resistivity of a metal increases when its temperature is raised.
Two wires of the same material, one thick and one thin, are at the same temperature. Which has the greater resistance, and which has the greater resistivity? Justify your answer.
The power dissipated in a resistor can be written as \(P = I^2R\) and also as \(P = V^2/R\). The first suggests that power increases with resistance and the second that it decreases with resistance. Explain how both can be correct.
The same current passes through the copper connecting cord and through the filament of an electric bulb, yet only the filament glows. Explain why, referring to the resistivity values in Table 11.1.
Why are the heating elements of appliances such as electric irons made of nichrome rather than copper?
Distinguish between resistance and resistivity. Which of the two would change if a given wire were drawn out to twice its original length at the same temperature?
Why must the tolerance band of a four-band carbon resistor be separated by a gap from the other three bands? What error would result if a resistor were read from the wrong end?
Explain why the equivalent resistance of a parallel combination is always smaller than the smallest resistance in the combination, while that of a series combination is always larger than the largest.
Why are domestic appliances connected in parallel across the supply rather than in series? Give two distinct reasons.
Distinguish between the emf and the terminal potential difference of a cell. Under what condition are the two equal?
The headlights of a car dim briefly when the engine is started. Explain this observation using the relation \(V = \varepsilon - Ir\).
What happens to the equivalent emf and the equivalent internal resistance when several identical cells are connected (a) in series, and (b) in parallel? In which situation would each arrangement be preferred?
Why can a small dry cell not start a car engine, even though a car battery has an emf only a few times larger?
Distinguish between a primary cell and a secondary cell, giving one example of each. Why are primary cells unsuitable for heavy loads?
A current is started in a superconducting ring cooled below its critical temperature. Explain why it continues to flow without any source connected to the ring.
A junction diode is deliberately non-ohmic. Why is this property essential for its use in converting alternating current into direct current? How is the resistance of such a device defined at a given operating point?
Electric power is transmitted over long distances at very high voltage and low current. Explain the reasoning behind this choice using the expression for power dissipated in the transmission line.
Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
Distinguish between ohmic and non-ohmic substances; explain with the help of an example.
DC current flows in a metal piece of non-uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?
A steady current of 1.5 A flows through a conductor for 3 minutes. Calculate the net charge that crosses any cross-section of the conductor and the number of electrons corresponding to this charge. (Charge on an electron \(= 1.6 \times 10^{-19}\) C)
An aluminium wire of cross-sectional area \(1.5 \times 10^{-6}\) m² carries a current of 2.4 A. Calculate the current density in the wire.
A copper wire of diameter 1.0 mm carries a current of 2.0 A. If copper contains \(8.5 \times 10^{28}\) free electrons per cubic metre, calculate the drift speed of the electrons. (Charge on an electron \(= 1.6 \times 10^{-19}\) C)
A potential difference of 6 V applied across a conductor produces a current of 0.4 A through it. Calculate the resistance of the conductor and its conductance.
A nichrome wire has a resistivity of \(100 \times 10^{-8}\) Ω m. Calculate the resistance of a 2.0 m length of this wire whose cross-sectional area is \(0.5 \times 10^{-6}\) m².
A wire has a resistance of 8 Ω. It is drawn out uniformly so that its length becomes twice the original length, the volume remaining unchanged. Calculate its new resistance.
Two resistors of 6 Ω and 12 Ω are connected in parallel, and this combination is joined in series with a 6 Ω resistor across a 12 V supply of negligible internal resistance. Calculate the equivalent resistance of the network and the current drawn from the supply.
Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series across a 60 V supply. Find the current in the circuit and the potential difference across each resistor.
Four resistors, each of 8 Ω, are connected in parallel across a 24 V supply of negligible internal resistance. Calculate the equivalent resistance, the total current drawn, and the current through each resistor.
An electric kettle rated 1500 W operates on a 230 V supply. Calculate the current drawn by it, its resistance while working, and the electrical energy in kilowatt-hour consumed if it is used for 45 minutes each day for 30 days.
A resistor of 25 Ω carries a current of 2.0 A for 5 minutes. Calculate the heat produced in the resistor in joule.
A tungsten filament has a resistance of 20 Ω at 0 °C. Its temperature coefficient of resistance is \(4.5 \times 10^{-3}\)/°C. Calculate its resistance at 120 °C.
The resistance of a metal wire is 15 Ω at 20 °C and 18 Ω at 120 °C. Determine the temperature coefficient of resistance of the material of the wire.
A cell of emf 2.0 V and internal resistance 0.6 Ω is connected to an external resistance of 5.4 Ω. Calculate the current in the circuit, the terminal potential difference of the cell, and the power dissipated in the external resistor.
A battery of emf 9 V has an internal resistance of 0.5 Ω. Calculate the maximum current that can be drawn from it.
Four cells, each of emf 1.5 V and internal resistance 0.25 Ω, are connected in series across an external resistance of 5.0 Ω. Calculate the equivalent emf, the equivalent internal resistance, the current in the circuit, and the terminal voltage of the combination.
Two cells of emf 1.5 V and 2.0 V, with internal resistances 0.3 Ω and 0.6 Ω respectively, are connected in parallel with their like terminals joined together. Calculate the equivalent emf and the equivalent internal resistance of the combination.
A carbon resistor carries the bands Brown, Green, Orange, Silver. Write down its resistance and tolerance, and state the maximum and minimum values its resistance could actually have.
A constantan wire of resistivity \(5.0 \times 10^{-7}\) Ω m has a diameter of 0.80 mm. Calculate the length of this wire required to make a resistance of 10 Ω.
A silver wire of length 2.0 m and cross-sectional area \(0.8 \times 10^{-6}\) m² has a resistivity of \(1.59 \times 10^{-8}\) Ω m. Calculate its resistance and its conductivity.
An electric field of 0.10 V/m is maintained inside a copper conductor of conductivity \(5.8 \times 10^{7}\) Sm⁻¹. Calculate the current density in the conductor.
Two bulbs rated 60 W and 100 W, both at 230 V, are connected in series across a 230 V supply. Assuming their resistances remain unchanged, calculate the resistance of each bulb and the current in the circuit.
In the given circuit diagram, two resistors of 8 Ω and 6 Ω are connected in series to a 5 V supply.
Diagram
Fig. 11.21 – An 8 Ω resistor and a 6 Ω resistor connected in series across a 5 V supply.
What is the resistance of one of the rails of a railway track 20 km long at 20 °C? The cross section area of the rail is 25 cm² and the rail is made of steel having resistivity at 20 °C as 6 × 10⁻⁸ Ω m.
A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive the starting motor of a car?
A battery of emf 12 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A,
The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is 8.47 × 10²², calculate the drift velocity of the electrons through the copper wire. (Charge on an electron \(e = 1.6 \times 10^{-19}\) C)
Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination.
Two resistors 1 kΩ and 2 kΩ are connected in parallel combination.
A silver wire has a resistance of 4.2 Ω at 27 °C and a resistance of 5.4 Ω at 100 °C. Determine the temperature coefficient of resistance.
A 6 m long wire has a diameter of 0.5 mm. Its resistance is 50 Ω. Find the resistivity and the conductivity.
Find the value of resistances for the following colour codes.
Find the colour code for the following values of resistor, having tolerance ± 10%:
A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?
The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal.).
Diagram
Fig. 11.20 – Five dry cells of 1.5 V each connected in a single row, with some cells reversed in polarity relative to the others.