Theory, worked examples and 27 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
Some books are to be tasted, others to be swallowed, and some few to be chewed and digested. — Francis Bacon
Electric circuit theory and electromagnetic theory are the two fundamental theories on which every branch of electrical engineering is built. Power systems, electric machines, control, electronics, communications and instrumentation all rest on circuit theory, and for this reason a first course in circuits is the natural starting point for anyone entering the subject. Circuit theory is equally valuable to students of the physical sciences in general, because circuits provide a good model for the study of energy systems and because of the applied mathematics, physics and topology that their analysis involves. In electrical engineering we are frequently concerned with communicating or transferring energy from one point to another, and to do this we require an interconnection of electrical devices. Such an interconnection is called an electric circuit, and each component of the circuit is known as an element.
An electric circuit, then, is an interconnection of electrical elements. The simple circuit of Figure 1.1 consists of three basic elements: a battery, a lamp and the connecting wires. A circuit of this kind can exist on its own and has many uses, for example in a torch or a searchlight.
Figure 1.1 A simple electric circuit: a battery, a lamp and connecting wires.
1.1 Charge, current, voltage, power and energy
(a) Charge
Charge is an electrical property of the atomic particles of which matter consists, and it is measured in coulombs (C). All matter is built from atoms, and each atom consists of electrons, protons and neutrons. The charge \(e\) of the electron is negative and of magnitude \(1.602\times10^{-19}\,\text{C}\), while the proton carries a positive charge of the same magnitude. An atom with equal numbers of protons and electrons is therefore electrically neutral.
Three properties of charge deserve emphasis. First, the coulomb is a large unit: one coulomb contains \(1/(1.602\times10^{-19}) = 6.24\times10^{18}\) electronic charges, so that realistic laboratory charges are of the order of pC, nC or \(\mu\)C. Secondly, experiment shows that the only charges occurring in nature are integral multiples of the electronic charge \(-e=-1.602\times10^{-19}\,\text{C}\). Thirdly, the law of conservation of charge states that charge can be neither created nor destroyed, only transferred; the algebraic sum of the charges in an isolated system therefore does not change.
Example 1.1. How much charge is represented by 4600 electrons?
Solution. Each electron carries a charge of \(-1.602\times10^{-19}\,\text{C}\). Hence 4600 electrons carry
Electric current is the time rate of change of charge, measured in amperes (A). When a conducting wire is connected to a battery, which is a source of electromotive force, the charges in the wire are compelled to move: positive charges move in one direction and negative charges in the opposite direction, and this motion constitutes an electric current. By convention the direction of current is taken as the direction of motion of positive charge, that is, opposite to the flow of negative charge, as Figure 1.2 illustrates. The convention was introduced by Benjamin Franklin (1706–1790). Although we now know that the current in a metallic conductor is carried by negatively charged electrons, we shall follow the universally accepted convention that current is the net flow of positive charge.
Figure 1.2 Electric current due to the flow of electrons in a conductor; the conventional current \(I\) flows opposite to the electrons.
Mathematically, the current \(I\), the charge \(Q\) and the time \(t\) are related by
\[\boxed{I=\frac{dQ}{dt}},\qquad Q=\int I\,dt,\]
where the current is measured in amperes, \(1\,\text{A}=1\,\text{C/s}\).
Figure 1.3 Two common types of current: (a) direct current (d.c.), (b) alternating current (a.c.).
A current that remains constant in time is a direct current (d.c.), and by convention the upper-case symbol \(I\) is used for it. A time-varying current is denoted by the lower-case symbol \(i\); its most common form is the alternating current (a.c.), which varies sinusoidally with time, as in Figure 1.3. Alternating current is used in the household to run air conditioners, refrigerators, washing machines and other appliances.
(c) Voltage
To move the electrons in a conductor in a particular direction requires work, or a transfer of energy. This work is performed by an external electromotive force (e.m.f.), typically provided by a battery. The e.m.f. is also known as the voltage or potential difference. The voltage \(V_{ab}\) between two points \(a\) and \(b\) of a circuit is the energy (or work) needed to move a unit charge from \(a\) to \(b\):
\[\boxed{V_{ab}=\frac{dW}{dq}},\]
where \(W\) is the energy in joules (J) and \(q\) is the charge in coulombs. Voltage is measured in volts (V), named in honour of the Italian physicist Alessandro Volta (1745–1827), who invented the first voltaic battery. It follows from the definition that
Voltage, or potential difference, is thus the energy required to move a unit charge through an element. Figure 1.4 shows the voltage across an element, represented by a rectangular block, connected between the points \(a\) and \(b\). The plus and minus signs define the reference direction, or polarity, of the voltage. The symbol \(V_{ab}\) may be read in two equivalent ways: point \(a\) is at a potential \(V_{ab}\) volts higher than point \(b\), or the potential of point \(a\) with respect to point \(b\) is \(V_{ab}\). It follows that in general
\[V_{ab}=-V_{ba}.\]
Figure 1.4 Polarity of the voltage \(v_{ab}\) across an element.
Figure 1.5 gives two representations of the same voltage. In part (a) point \(a\) is \(+9\,\text{V}\) above point \(b\); in part (b) point \(b\) is \(-9\,\text{V}\) above point \(a\). We may say that in (a) there is a \(9\,\text{V}\) voltage drop from \(a\) to \(b\) or, equivalently, a \(9\,\text{V}\) voltage rise from \(b\) to \(a\). In other words, a voltage drop from \(a\) to \(b\) is equivalent to a voltage rise from \(b\) to \(a\).
Figure 1.5 Two equivalent representations of the same voltage: (a) point \(a\) is \(9\,\text{V}\) above point \(b\); (b) point \(b\) is \(-9\,\text{V}\) above point \(a\).
Current and voltage are the two basic variables of electric circuits. The general term signal is used for an electrical quantity such as a current or a voltage (or even an electromagnetic wave) when it conveys information; engineers prefer to call such variables signals rather than mathematical functions of time because of their importance in communications and related disciplines. As with current, a constant voltage is called a d.c. voltage and is denoted by \(V\), whereas a sinusoidally time-varying voltage is called an a.c. voltage and is denoted by \(v\). A d.c. voltage is commonly produced by a battery, and an a.c. voltage by an electric generator.
(d) Power and energy
Power is the time rate of expending or absorbing energy, measured in watts (W):
\[p=\frac{dw}{dt},\]
where \(p\) is the power in watts, \(w\) the energy in joules and \(t\) the time in seconds. Using the chain rule together with the definitions of voltage and current,
The energy absorbed or supplied by an element between two instants is therefore
\[w=\int p\,dt=\int v\,i\,dt.\]
Energy is the capacity to do work, measured in joules. Electric power utilities measure energy in watt-hours (Wh), where \(1\,\text{Wh}=3600\,\text{J}\).
1.2 Circuit elements and sources
(a) Active and passive elements
An electric circuit is simply an interconnection of elements, and circuit analysis is the process of determining the voltages across, or the currents through, the elements of the circuit. Two kinds of element are found in circuits. An active element is capable of generating energy, while a passive element is not. Resistors, capacitors and inductors are passive elements; generators, batteries and operational amplifiers are typical active elements.
The most important active elements are voltage and current sources, which generally deliver power to the circuit connected to them. Sources are of two kinds, independent and dependent.
(b) Independent sources
An ideal independent source is an active element that provides a specified voltage or current that is completely independent of the other circuit elements. The symbols for independent voltage sources are shown in Figure 1.6. An ideal independent voltage source delivers to the circuit whatever current is necessary to maintain its terminal voltage. Physical sources such as batteries and generators may be regarded as approximations to ideal voltage sources.
Figure 1.6 Symbols for independent voltage sources: (a) used for constant or time-varying voltage, (b) used for constant (d.c.) voltage.
Similarly, an ideal independent current source is an active element that provides a specified current, completely independent of the voltage across the source; it delivers to the circuit whatever voltage is necessary to maintain the designated current. Its symbol, a circle containing an arrow that gives the direction of the current, is shown in Figure 1.7.
Figure 1.7 Symbol for an independent current source.
(c) Dependent sources
An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current in the circuit. Dependent sources are usually drawn with diamond-shaped symbols, as in Figure 1.8. Since the controlling quantity may be a voltage or a current elsewhere in the circuit, and the source itself may be a voltage or a current source, there are four possible types of dependent source:
a voltage-controlled voltage source (VCVS);
a current-controlled voltage source (CCVS);
a voltage-controlled current source (VCCS);
a current-controlled current source (CCCS).
Figure 1.8 Symbols for (a) a dependent voltage source, (b) a dependent current source.
Dependent sources are useful in modelling elements such as transistors, operational amplifiers and integrated circuits. An example of a current-controlled voltage source is shown in Figure 1.9, where the voltage \(10i\) of the source on the right depends on the current \(i\) through element \(C\). Students are sometimes surprised that the value of this source is \(10i\) volts rather than \(10i\) amperes, given that it is controlled by a current. The point to remember is that a voltage source carries the polarity signs \(+\) and \(-\) in its symbol, while a current source carries an arrow, irrespective of the quantity on which it depends.
Figure 1.9 The source on the right-hand side is a current-controlled voltage source.
1.3 Ohm’s law and the passive elements
(a) Ohm’s law and the resistor
Ohm’s law states that the voltage \(v\) across a resistor is directly proportional to the current \(i\) flowing through it, \(v\propto i\). Ohm defined the constant of proportionality to be the resistance\(R\) of the resistor, measured in ohms (\(\Omega\)), so that
\[\boxed{v=Ri}.\]
The resistance is a property of the material and geometry of the element, and it changes if the internal or external conditions of the element are altered, for example if its temperature changes.
Exam tip. Ohm’s law may be applied either to an entire circuit or to a part of it. When it is applied to the entire circuit, the total voltage and the resistance of the entire circuit must be used; when it is applied to a part of the circuit, the voltage across and the resistance of that part must be used.
The two common symbols for a resistor, the zig-zag and the rectangle, are shown below.
(b) Power in a resistor
Power is dissipated in a resistor whenever a current flows through it: the energy of the moving electrons is converted into heat as they collide with the atoms of the resistor material. Combining \(p=vi\) with Ohm’s law gives three equivalent expressions,
The last two forms show that the power dissipated in a given resistor varies as the square of the current or of the voltage.
(c) Limitations of Ohm’s law
Ohm’s law is not universal. It does not apply to non-linear devices such as diodes and voltage regulators, whose current is not proportional to the applied voltage. Nor does it hold for non-metallic conductors, for which the voltage–current relation takes the form \(V=kI^m\), where \(k\) and \(m\) are constants. It also presupposes that the physical conditions of the conductor, in particular its temperature, remain unchanged.
(d) The capacitor
The basic relation describing a capacitor connects the charge \(Q\) on the capacitor to the voltage \(V\) across it,
\[Q=CV.\]
The constant of proportionality \(C\) is the capacitance, whose unit is the farad (F); from the equation above, \(1\,\text{F}=1\,\text{C/V}\). Since current is the rate of flow of charge, \(i=dq/dt\), we differentiate \(q=Cv\) with respect to time (for a fixed capacitance) to obtain
\[\frac{dq}{dt}=C\frac{dv}{dt},\]
and hence
\[\boxed{i=C\frac{dv}{dt}}.\]
The current in a capacitor is therefore proportional to the time rate of change of the voltage across it. The two common symbols for a capacitor are shown below.
The instantaneous power delivered to a capacitor is
\[p=vi=vC\frac{dv}{dt},\]
and the energy \(U\) stored in it is the time integral of this power,
If the capacitor voltage is zero at the start of the integration, the integral gives
\[\boxed{U=\tfrac12Cv^2}.\]
Unlike a resistor, in which the energy is lost as heat, an ideal capacitor does not dissipate energy. The energy is stored in the electric field associated with the separated charge, and it is recovered when the charge flows back out of the capacitor.
(e) The inductor
The voltage across an inductor is directly proportional to the time rate of change of the current through it,
\[\boxed{v=L\frac{di}{dt}}.\]
This property arises from the ability of the inductor to store energy in the magnetic field that surrounds it; the stored magnetic energy can be returned to the circuit by driving an electric current. The constant of proportionality \(L\) is the inductance, whose unit is the henry (H). Why this property appears in coils of wire is a matter of the intimate relationship between electricity and magnetism, which lies outside the scope of this chapter; here we simply accept that the voltage across an inductor is proportional to the rate of change of its current. The symbol for an inductor is shown below.
As with the capacitor, the inductor relation may be written in integral form to give \(i\) in terms of \(v\), and the close kinship between the two elements then becomes evident:
The lower limit \(-\infty\) means that the inductor current at time \(t\) depends on the entire past history of the inductor voltage (and, likewise, the capacitor voltage on the entire past history of its current). In practice we start the integration from a known current \(i_0\) at a known instant, such as \(t=0\), and keep track of the changes from there:
\[\boxed{i(t)=\frac1L\int_0^{t}v\,dt'+i_0}.\]
The instantaneous power delivered to an inductor is
\[p=vi=iL\frac{di}{dt},\]
and the energy \(U\) stored in its magnetic field is
If the current is zero at the start of the integration,
\[\boxed{U=\tfrac12Li^2}.\]
Unlike a resistor, an ideal inductor does not dissipate energy: the energy stored in its magnetic field can be fully recovered when the field collapses and the energy is converted back into an electric current in the wire.
1.4 Nodes, branches and loops
A branch represents a single element, such as a voltage source or a resistor; in other words, a branch is any two-terminal element. The circuit of Figure 1.10 has five branches: the \(10\,\text{V}\) voltage source, the \(2\,\text{A}\) current source and the three resistors.
A node is a point of connection between two or more branches. A node is usually indicated by a dot in a circuit diagram. If a short circuit (a connecting wire) joins two nodes, the two constitute a single node. The circuit of Figure 1.10 has three nodes, \(a\), \(b\) and \(c\). The three points that form node \(b\) are joined by perfectly conducting wires and therefore constitute a single point, and the same is true of the four points that form node \(c\). That the circuit has only three nodes is made clear by redrawing it as in Figure 1.11. The two circuits are identical; in Figure 1.10 the nodes \(b\) and \(c\) are merely spread out along perfect conductors for clarity.
Figure 1.10 Nodes, branches and loops.Figure 1.11 The three-node circuit of Figure 1.10 redrawn.
A loop is any closed path in a circuit. More precisely, a loop is a closed path formed by starting at a node, passing through a set of nodes and returning to the starting node without passing through any node more than once. A loop is said to be independent if it contains at least one branch that is not part of any other independent loop. Independent loops lead to independent sets of equations. It is also possible to form an independent set of loops in which one of the loops does not contain such a branch. In Figure 1.11, the loop \(abca\) containing the \(2\,\Omega\) resistor is independent; a second loop, containing the \(3\,\Omega\) resistor and the current source, is also independent; and the third loop may be taken as the one formed by the \(2\,\Omega\) resistor in parallel with the \(3\,\Omega\) resistor. These three loops form an independent set.
A network with \(b\) branches, \(n\) nodes and \(l\) independent loops satisfies the fundamental theorem of network topology,
\[\boxed{b=l+n-1}.\]
For the circuit of Figure 1.10, \(b=5\) and \(n=3\), so that \(l=5-3+1=3\), in agreement with the three independent loops identified above.
1.5 Kirchhoff’s laws
(a) Energy sources: ideal and practical
There are basically two types of energy source, the voltage source and the current source, and each may be classified as ideal or practical. An ideal voltage source maintains its terminal voltage whatever current it delivers, which is equivalent to saying that its internal resistance is zero; a practical voltage source behaves as an ideal source in series with a small internal resistance, so that its terminal voltage falls as the load current increases. Conversely, an ideal current source delivers its specified current whatever the voltage across it, corresponding to an infinite internal resistance, while a practical current source behaves as an ideal source in parallel with a large internal resistance.
(b) Kirchhoff’s voltage law
Ohm’s law by itself is not sufficient to analyse circuits. When it is combined with Kirchhoff’s two laws, however, we have a powerful set of tools for analysing a large variety of electric circuits. These laws were introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887) and are known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). The current law is based on the conservation of charge, which requires that the algebraic sum of the charges within a system cannot change; the voltage law expresses the conservation of energy.
Kirchhoff’s voltage law states that the algebraic sum of all the voltages around any closed loop of a circuit is zero. Mathematically,
\[\boxed{\sum_{k=1}^{n}v_k=0},\]
where \(n\) is the number of voltages in the loop (or the number of branches in the loop) and \(v_k\) is the \(k\)th voltage.
Figure 1.12 A single-loop circuit illustrating KVL.
To illustrate KVL, consider the circuit of Figure 1.12. The sign given to each voltage is the polarity of the terminal encountered first as we travel round the loop. We may start with any branch and go round the loop either clockwise or anticlockwise. Suppose we start with the voltage source \(v_1\) and go clockwise round the loop as shown; the voltages are then \(-v_1\), \(+v_2\), \(+v_3\), \(-v_4\) and \(+v_5\), in that order. For example, on reaching branch 3 we meet the positive terminal first, so we write \(+v_3\); for branch 4 we meet the negative terminal first, so we write \(-v_4\). KVL therefore gives
\[-v_1+v_2+v_3-v_4+v_5=0,\]
which on rearrangement becomes
\[v_2+v_3+v_5=v_1+v_4.\]
This may be read as
\[\text{sum of voltage drops} = \text{sum of voltage rises},\]
which is an alternative form of KVL.
Example 1.2. For the circuit shown, find the voltages \(v_1\) and \(v_2\).
Solution. Let a current \(i\) flow clockwise round the loop, as shown below.
This current enters the \(2\,\Omega\) resistor at the \(+\) terminal of \(v_1\) but enters the \(3\,\Omega\) resistor at the \(-\) terminal of \(v_2\). Ohm’s law therefore gives
\[v_1=2i,\qquad v_2=-3i.\]
Applying KVL clockwise round the loop, starting at the lower terminal of the source,
Example 1.3. Determine \(v_0\) and \(i\) in the circuit shown.
Solution. The current \(i\) flows clockwise round the single loop, as indicated below.
Applying KVL clockwise, starting at the lower terminal of the \(12\,\text{V}\) source, we meet in turn the rise of the \(12\,\text{V}\) source, the drop \(4i\) across the \(4\,\Omega\) resistor, the dependent source \(2v_0\) (entered at its \(+\) terminal), the \(4\,\text{V}\) source (entered at its \(-\) terminal) and the drop \(6i\) across the \(6\,\Omega\) resistor:
\[-12+4i+2v_0-4+6i=0.\]
The current \(i\) enters the \(6\,\Omega\) resistor at the \(-\) terminal of \(v_0\), so Ohm’s law gives \(v_0=-6i\). Substituting \(2v_0=-12i\),
The negative sign of \(i\) shows that the current actually circulates anticlockwise.
(c) Kirchhoff’s current law
Kirchhoff’s current law states that the algebraic sum of the currents entering a node (or a closed boundary) is zero. Mathematically,
\[\boxed{\sum_{k=1}^{m}i_k=0},\]
where \(m\) is the number of branches connected to the node and \(i_k\) is the \(k\)th current entering (or leaving) the node. Currents entering a node may be counted as positive and currents leaving it as negative, or vice versa.
To prove KCL, suppose a set of currents flows into a node. The algebraic sum of these currents is
\[i_T(t)=i_1(t)+i_2(t)+i_3(t)+\cdots.\]
Integrating both sides with respect to time gives
\[q_T(t)=q_1(t)+q_2(t)+q_3(t)+\cdots,\]
where \(q_k=\int i_k\,dt\). The law of conservation of charge requires that the algebraic sum of the charges at the node must not change; that is, the node stores no net charge. Hence \(q_T(t)=0\) at all times, and therefore \(i_T(t)=0\), which confirms the validity of KCL.
Figure 1.13 Currents at a node illustrating KCL.
Consider the node of Figure 1.13. The currents \(i_1\), \(i_3\) and \(i_4\) enter the node, while \(i_2\) and \(i_5\) leave it. KCL gives
\[i_1-i_2+i_3+i_4-i_5=0,\]
and rearranging,
\[i_1+i_3+i_4=i_2+i_5.\]
This is an alternative form of KCL: the sum of the currents entering a node is equal to the sum of the currents leaving it.
Example 1.4. Find the current \(i_0\) and the voltage \(v_0\) in the circuit shown.
Solution. The dependent source drives \(0.5i_0\) and the independent source \(3\,\text{A}\) into node \(a\), while the current \(i_0\) leaves node \(a\) through the \(4\,\Omega\) resistor. Applying KCL at node \(a\),
For the \(4\,\Omega\) resistor, Ohm’s law then gives
\[v_0=4i_0=24\,\text{V}.\]
Example 1.5. Find \(v_0\) and \(i_0\) in the circuit shown.
Solution. All four elements are connected between the same pair of nodes, so they share the voltage \(v_0\). Let \(I\) be the current flowing downwards in the \(8\,\Omega\) resistor. Then
Applying KCL at the upper node, the \(9\,\text{A}\) entering from the source equals the sum of the currents leaving through the \(2\,\Omega\) resistor, the dependent source and the \(8\,\Omega\) resistor:
The need to combine resistors in series or in parallel arises so often that it deserves special attention. The combination is most easily carried out by taking two resistors at a time. With this in mind, consider the single-loop circuit of Figure 1.14. The two resistors are in series, since the same current \(i\) flows through both of them. Applying Ohm’s law to each resistor gives
\[v_1=iR_1,\qquad v_2=iR_2,\]
and applying KVL clockwise round the loop gives
\[-v+v_1+v_2=0.\]
Combining these equations,
\[v=v_1+v_2=i(R_1+R_2)=iR_{eq},\]
so that
\[i=\frac{v}{R_1+R_2},\qquad R_{eq}=R_1+R_2.\]
Figure 1.14 A single-loop circuit with two resistors in series.
The two series resistors may therefore be replaced by a single equivalent resistor \(R_{eq}=R_1+R_2\) without changing the current drawn from the source, as shown in Figure 1.15.
Figure 1.15 Two resistors in series and their equivalent resistance \(R_{eq}=R_1+R_2\).
The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. For \(N\) resistors in series,
\[\boxed{R_{eq}=R_1+R_2+\cdots+R_N}.\]
(b) Resistors in parallel
Consider next the circuit of Figure 1.16, in which two resistors are connected in parallel and therefore have the same voltage \(v\) across them. From Ohm’s law,
The voltage-division rule states that the total voltage applied across a series connection of resistors is divided among the resistors in proportion to their resistances. The largest voltage drop therefore appears across the largest resistance, and the smallest across the smallest. To derive the rule, consider again two resistors in series, as in Figure 1.17. Ohm’s law gives
The total current divides between two parallel resistors in inverse proportion to their resistances: the larger share flows through the smaller resistance.
Exam tip. In the voltage-division rule each resistor’s own resistance appears in the numerator, whereas in the current-division rule it is the resistance of the other branch that appears in the numerator.
(e) Equivalent resistance: worked examples
Example 1.6. Find \(R_{eq}\) for the circuit shown.
Solution. To find \(R_{eq}\) we combine resistors in series and in parallel, working from the end farthest from the terminals, as summarised in the sequence of reduced circuits below.
The \(6\,\Omega\) and \(3\,\Omega\) resistors are in parallel, so their equivalent resistance is
The symbol \(\parallel\) denotes a parallel combination. The \(1\,\Omega\) and \(5\,\Omega\) resistors are in series, and their equivalent resistance is \(1+5=6\,\Omega\). The two \(2\,\Omega\) resistors are now in series, giving
\[2\,\Omega+2\,\Omega=4\,\Omega.\]
This \(4\,\Omega\) resistance is in parallel with the \(6\,\Omega\) resistance:
Finally, the \(4\,\Omega\), \(2.4\,\Omega\) and \(8\,\Omega\) resistances are in series, so the equivalent resistance of the circuit is
\[R_{eq}=4+2.4+8=14.4\,\Omega.\]
Example 1.7. Calculate the equivalent resistance \(R_{ab}\) of the circuit shown.
Solution. The \(3\,\Omega\) and \(6\,\Omega\) resistors are in parallel because they are connected to the same two nodes, \(c\) and \(b\). Their combined resistance is
The \(1\,\Omega\) and \(5\,\Omega\) resistors on the right are in series, giving \(1+5=6\,\Omega\) between \(d\) and \(b\). The circuit is thus reduced to the one below.
The \(3\,\Omega\) and \(6\,\Omega\) resistances between \(d\) and \(b\) are in parallel, giving \(2\,\Omega\), and this is in series with the \(1\,\Omega\) resistor between \(c\) and \(d\), giving \(1+2=3\,\Omega\). The circuit becomes the one below.
The \(2\,\Omega\) and \(3\,\Omega\) resistances between \(c\) and \(b\) are in parallel:
This \(1.2\,\Omega\) is in series with the \(10\,\Omega\) resistor, so that
\[R_{ab}=10+1.2=11.2\,\Omega.\]
Example 1.8. In the circuit shown, the resistor in series with the \(12\,\text{V}\) source is \(3\,\Omega\). Find the current through each element (a) using the current-division rule, (b) using KVL.
Solution. (a) The two \(7\,\Omega\) resistors are in series and form a \(14\,\Omega\) branch, which is in parallel with the \(6\,\Omega\) resistor. The resistance seen by the source is therefore
\[3+\frac{6\times14}{6+14}=3+4.2=7.2\,\Omega,\]
and the source current is
\[I=\frac{12}{7.2}=\frac{5}{3}\,\text{A}.\]
By the current-division rule, the current in the \(14\,\Omega\) branch (through both \(7\,\Omega\) resistors) and the current in the \(6\,\Omega\) resistor are
The superposition theorem states that if a number of voltage or current sources act simultaneously in a linear network, the resultant current in any branch is the algebraic sum of the currents that would be produced in it by each source acting alone, all the other independent sources being replaced by their internal resistances. The theorem follows from the linearity of the network equations. It is applied in three steps:
Consider one independent source at a time, with all the other independent sources turned off. This means that every other voltage source is replaced by \(0\,\text{V}\) (a short circuit) and every other current source by \(0\,\text{A}\) (an open circuit).
Repeat step 1 for each of the other independent sources.
Find the total response by adding algebraically the contributions of all the independent sources.
Dependent sources are never turned off in this procedure, since they are controlled by the circuit variables themselves.
Exam tip. Superposition applies to currents and voltages, which are linear in the sources, but not directly to power, which is quadratic in them: the power in a resistor must be computed from the total current, not by adding the powers due to each source.
(b) Worked examples
Example 1.9. In the circuit of Figure 1.19, two batteries \(B_1=28\,\text{V}\) and \(B_2=7\,\text{V}\), of negligible internal resistance, feed the resistors \(R_1=4\,\Omega\), \(R_2=2\,\Omega\) and \(R_3=1\,\Omega\). Use the superposition theorem to find the current in, and the voltage across, each resistor, and verify the result by nodal analysis.
Figure 1.19 A two-source circuit for the application of superposition.
Solution. Consider first the source \(B_1\) acting alone, with \(B_2\) replaced by a short circuit (its internal resistance being zero), as in Figure 1.20.
Figure 1.20 The circuit of Figure 1.19 with \(B_1\) acting alone.
The resistors \(R_2\) and \(R_3\) are now in parallel, \(2\parallel1=\tfrac23\,\Omega\), and this combination is in series with \(R_1\). The current delivered by \(B_1\) is
flowing through \(R_3\) away from \(B_2\), and by current division
\[I_1''=\frac{2}{2+4}\times3=1\,\text{A}\ \text{(towards } B_1 \text{ in } R_1),\qquad I_2''=\frac{4}{2+4}\times3=2\,\text{A}\ \text{(downwards in } R_2).\]
By the superposition theorem the net currents are found by adding these contributions with due regard to their directions. In \(R_1\) the two contributions are opposed, so
\[I_1=I_1'-I_1''=6-1=5\,\text{A},\]
and the voltage drop across \(R_1\) is \(5\times4=20\,\text{V}\). In \(R_2\) both contributions flow downwards, so
\[I_2=I_2'+I_2''=2+2=4\,\text{A},\]
and the voltage drop across \(R_2\) is \(4\times2=8\,\text{V}\). In \(R_3\) the contributions are again opposed, so
\[I_3=I_3'-I_3''=4-3=1\,\text{A},\]
flowing towards \(B_2\), and the voltage drop across \(R_3\) is \(1\times1=1\,\text{V}\). These voltages are marked in Figure 1.22.
Figure 1.22 The resulting voltage drops: \(20\,\text{V}\) across \(R_1\), \(8\,\text{V}\) across \(R_2\) and \(1\,\text{V}\) across \(R_3\).
As a check, we analyse the same circuit by the node method, Figure 1.23. Take the lower wire as the reference node and let the potential of the upper node \(A\) be \(V\).
Figure 1.23 The circuit of Figure 1.19 prepared for nodal analysis.
Counting all currents as leaving node \(A\), KCL gives
and the voltage drops are \(5\times4=20\,\text{V}\) across \(R_1\), \(4\times2=8\,\text{V}\) across \(R_2\) and \(1\times1=1\,\text{V}\) across \(R_3\), exactly as found by superposition.
Example 1.10. Use the superposition theorem to find \(v\) in the circuit shown.
Solution. Let \(v=v_1+v_2\), where \(v_1\) and \(v_2\) are the contributions of the \(6\,\text{V}\) voltage source and the \(3\,\text{A}\) current source respectively. Both partial circuits are shown below.
To obtain \(v_1\) we set the current source to zero (an open circuit), as in part (a). The \(8\,\Omega\) and \(4\,\Omega\) resistors are then in series with the \(6\,\text{V}\) source, and KVL round the loop gives
To obtain \(v_2\) we set the voltage source to zero (a short circuit), as in part (b). The \(8\,\Omega\) and \(4\,\Omega\) resistors are now in parallel across the \(3\,\text{A}\) source, and by current division the current in the \(4\,\Omega\) resistor is
\[i_3=\frac{8}{4+8}\times3=2\,\text{A},\]
so that \(v_2=4i_3=8\,\text{V}\). Adding the two contributions,
\[v=v_1+v_2=2+8=10\,\text{V}.\]
As a check, KCL at the upper node of the original circuit gives \((v-6)/8+v/4=3\), that is \(3v-6=24\), or \(v=10\,\text{V}\).
1.8 Thévenin’s theorem
(a) Statement of the theorem
It often happens in practice that one element of a circuit, the load, is variable while the rest of the circuit is fixed. A household socket, for example, may be connected to many different appliances. Each time the load is changed the whole circuit would have to be analysed afresh, unless the fixed part can be replaced once and for all by something simpler. Thévenin’s theorem provides exactly such a replacement.
Thévenin’s theorem. A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source \(V_{\text{Th}}\) in series with a resistor \(R_{\text{Th}}\), where \(V_{\text{Th}}\) is the open-circuit voltage at the terminals and \(R_{\text{Th}}\) is the input (or equivalent) resistance at the terminals when all the independent sources are turned off.
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