The circuit in Fig. 1.1 is used to highlight some of the key aspects of a circuit.
The questions that can be asked about the circuit in Fig. 1 are listed below.
To answer these questions, we need to know what the units for voltage, current, power and energy are.
The study of electric circuits is fundamental to study of electrical engineering. Even though the origin of circuit theory lies in electromagnetic field theory, it is easier to introduce electrical engineering via circuit theory, since it provides simple and sufficiently accurate solutions to practical electric systems. Circuit theory aims to predict and define the electrical behaviour of circuits, both qualitatively and quantitatively and its domain of application is very broad. Hence it is an area of study in its own right.
The study of circuit theory usually begins with definitions and axioms. Here we make a start with description of concepts in this chapter essentially at a superficial level. The purpose is to show the link between the electromagnetic field theory and the circuit theory.
Go to top of the page| QUANTITY | UNIT | SYMBOL |
| Length | Metre | m |
| Mass | Kilogram | kg |
| Time | Second | s |
| Current | Ampere | A |
| Temperature | Degree Kelvin | K |
| Luminous intensity | Candela | cd |
Nowadays the International System of Units, known as SI system or the
metric system, is used widely and the six basic units of this system are
shown in Table 1.1. All other units are derived from these. For example, when
mass of m kilograms is accelerated at rate a in
m/s2, the force behind is defined in newton and it is equal to
m × a. The basic units
used most in this book are current and time.
One of the advantages of a metric system is that a system of decimal prefixes
can be used with any unit, so that appropriate measures can be used as is
warranted. The set of prefixes available is presented in Table 1.2.
| NAME OF PREFIX | SYMBOL | MULTIPLYING FACTOR |
| tera | T | 1012 |
| giga | G | 109 |
| mega | M | 106 |
| kilo | k | 103 |
| milli | m | 10-3 |
| micro | m | 10-6 |
| nano | n | 10-9 |
| pico | p | 10-12 |
| femto | f | 10-15 |
| atto | a | 10-18 |
All electrical phenomena are manifestations of electric charge. The atomic
particles of matter that carry charges are protons and electrons. Protons
carry positive charges and electrons carry negative charges. The charge an
electron carries is small and it is -1.602 x 10-19 coulomb(C).
This means that 1 C contains 6.24 x 1018 electrons. In addition,
this also means that charge exists in discrete quantities. Usually we use
either the literal symbol Q or
q(t) when a reference to
charge contained in a given quantity of matter is made. The conservation of
charge is due to the basic physical postulate that a charge cannot be created
or destroyed and that it can only be transferred. The transfer of charge
leads to charge flow or current.
By nature, charge is bipolar, in the sensethat there are both positive
charges and negative charges. Separation of charges gives rise to voltage, or
the electro-motive force, whereas charge in motion gives rise to current.
The SI unit for current is ampere, and current is the rate of charge flow or the rate of change in charge. An ampere is equal to one coulomb per second. We can state that current i(t) to be:
From the above definition, we can find out the charge that passes through a surface in the time interval - ∞to t. If the current through at any instant be i(t), then
with the assumption that q(-∞) = 0. In practice, an ammeter is used to measure current. In a laboratory, a current probe may be used in order to see the waveform of current on an oscilloscope.
Go to top of the pageA copper wire carrying 10 A has a diameter of 1.5 mm. The flow of current is only due to electrons. The free electron concentration in copper is 1029 electrons per cubic metre. Find the average velocity of electrons in the wire.
The solution is obtained as follows. The current passing through the wire is known. First determine the charge passing through the wire per second. Then calculate the number of electrons that pass through the wire per second. Based on the number of electrons and the free electron concentration in copper, determine the volume of copper needed to hold the required number of electrons. Divide the volume by the area of the wire to get the average velocity of electrons in the wire.
Go to top of the pageThe law of conservation of energy postulates that energy cannot be
created, but only transformed.Chemical energy, kinetic energy and atomic
energy can be transformed into electrical energy. The reason for popularity
of electric energy is its ease of transportability.
Any electric circuit obeys the law of conservation of energy. It means that
the algebraic sum of power in a circuit is zero at any instant. If power
associated with an element in a circuit is p, then
Examples that reflect equation (1.3) are provided in chapter 3, when Kirchoff’s laws are presented.
Go to top of the pageIf work is done on charge, the ratio of energy expended to charge is voltage. For example, work is done on charges by chemical reactions within a battery and the result is a voltage across its terminals. If we use literal w to indicate the work done on charge q, then
where w is in joules, q is in coulombs and v is in volts. The SI unit for voltage is the volt, represented by literal V. In practice, a voltmeter is used to measure the voltage between two points.
Go to top of the pageFrom equation (1.4) for voltage, we get that
If the time taken for work dw be dt, then
The rate of change of energy with respect to time is power. It is also known that current is the rate of charge, as defined by equation (1.1). Hence power p is:
Electrical power is the product of voltage and current. The unit for power is watt(W).
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The charge, i(t) flowing into the circuit, shown as a box in Fig. 1.2, is expressed to be:
Find
a. the total charge that flows into the circuit over the period
b. the peak current and the instant at which it occurs.
Let the total charge flowing into the circuit be Q. Then
As shown above, it is necessary to make use of the formula for integration by parts. It is known that
Alternatively, Euler’s identity can be used to obtain Q.
To find the peak current, differentiate current and equate the derivative to zero. Find the instant at which the derivative to zero
The peak value of current is 0.322 A and it occurs at t = 0.785 s. The plots of current and charge are shown below.
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The current, i(t) flowing into the circuit, shown as a box in Fig. 1.2, is expressed to be:

Sketch the waveform of i(t) and q(t).


Assuming that q(0) = 0, we get that

The plots of i(t) and q(t) are presented by Fig. 1.3.
It is seen that the current has a discontinuity at t = 0 s and its value jumps att = 0 s. Such a function is called as unit step function and is denoted by u(t). Unit-step function has a value of unity for positive values of t and is zero for negative values of t. It is seen that the charge increases linearly with t when t is positive and is zero for negative values of t. Such a function is known as unit ramp function and is denoted by r(t). In other words,

The expression for the unit ramp function shows how unit step function can be used for defining other functions.
Go to top of the pageThe plots of voltage and supplied to a circuit are shown in Fig. 1.4. Show the plots of instantaneous power and the energy absorbed by the circuit.

From Fig. 1.4, the expressions for voltage and current are as follows:

Based on equation (1.7), we get that

Energy is the integral of power, as can be understood from equation (1.7). Hence

The plots of power and energy are shown in Fig. 1.5.
Go to top of the pageThis chapter has presented the fundamental units of the SI system. The other basic electrical parameters such as voltage, charge, power and energy have also been introduced.
Go to top of the pageE1.1: Current, i(t) flowing into a circuit is expressed to be:

E1.2: Energy, w(t) absorbed by a circuit is expressed to be:

a. Find the expression for power during the interval 0 ≤ t ≤ 3 s.
b. Given that current, i(t) is expressed as:

find an expression for voltage v(t).
c. Sketch the waveforms of i(t) and v(t).
E1.3: A laser generates 100 -mJ laser pulses of 25 ns duration.
a. What is the peak instantaneous power of the laser /
b. If 50 pulses are generated per second, what is the average power output of the laser ?
E1.4: A 12 V battery supplies 5 A for 1 hour and the charge supplied by the battery during this period amounts to one-tenth the AH rating of the battery. What is the AH rating of the battery?