The principles of electric circuits, how to draw circuit diagrams, and how to solve circuits
The electric circuit is a device for transferring power from a source at one point to a sink at another, either the small amounts used for signalling, to the large amounts used to produce heat or work. It consists of a closed loop around which the flow of electric charge takes place in response to electrical fields, which exert forces on them. At no point in the circuit can a net electrical charge build up, since this would bring the flow of energy to a halt. Although the electrical potential is different at different parts of the circuit, only very small net charges are necessary to produce these differences, and the circuit as a whole remains electrically neutral, with equal charges of both signs.
It was very difficult to understand the properties of the electric circuit at first, after copious sources of current at low potentials were found around 1800. Only after the energy concept was established was much understood about the electric circuit, through the work of Ohm, Joule and Wheatstone and others. Now it is easy to think of an electron's gaining energy when pushed to a higher potential by chemical or inductive action and moving to some other point, where the energy is given up either by agitating matter as it is drawn to a lower potential, or against a counter-emf in a magnetic field, or indeed by producing chemical action by electrolysis.
Because the things that occur in electric circuits, and electromagnetism at large, are well-removed from daily experience, analogies have been used to give concrete expression to the mysterious goings-on. There is the hydraulic analogy of water given energy by a pump, through being raised to a higher pressure or elevation, flowing through a pipe, and giving up its energy by turning a wheel and doing useful work, or simply chaotically cascading down a beautiful but useless waterfall. In fact, electric potential is sometimes actually called "pressure." Other analogies are the "cutting of lines of force" to generate electromotive force, and the view of "lines of force" as elastic bands. Such analogies are helpful, and under many circumstances true, but cannot be pushed beyond a certain limit. The hydraulic analogy led some early workers to regard insulation as a kind of pipe, and they assumed that a bare wire would leak electricity. Not until this misapprehension was overcome was long-distance telegraphy possible.
In this paper I want to give an elementary discussion of direct-current circuits, emphasizing the role of the circuit schematic or circuit diagram that is an extremely valuable aid to thought and reasoning. The circuit diagram shows lumped elements with certain simple properties connected by conductors, wires, along which the electrical potential is constant. This is an idealization, which is only approximately realized in practice. In most cases the idealization is very close to the truth, which is what makes it useful.
Electrical potential has already been mentioned several times, so it is high time to say what it is. It is energy per unit charge. Energy is measured in joule, J (1 kilowatt-hour is 3 600 000 J, 4.196 J is a small calorie, 1 Btu is 1055 J), while charge is measured in coulomb, C. It takes 6.24 x 1018 electrons to make up a coulomb. Electrical potential, therefore, is measured in J/C, and this unit is given the name volt, V. It is one of the curiosities of energy that it always refers to some reference level. Kinetic energy, energy of motion, is referred to the state of zero velocity. There is no such simple zero for most energy. In electric circuits, the most we can say is that an electron at point b has a certain amount of energy more than an electron at point a. The difference of energy, divided by the charge, gives the difference in electrical potential Eba or Vba. E and V are the usual letters used to denote a difference of electrical potential. The first subscript (if used) shows which point is more positive when the value is positive, and the second subscript the reference point. Instead of "difference of electrical potential," it is usual to speak of "voltage," and we shall do so from now on.
Current is just the net charge in C/s flowing past any point, just like the amount of water in the hydraulic analogy. The usual symbol is I, and the unit C/s is called the ampere, A. When water, or crude oil, flows through a permeable rock like sandstone, the rate of flow is proportional to the pressure gradient between any two points (pressure difference divided by distance). The flow through the pores of the rock is viscous, and the fluid gives up its energy by friction. This heats the rock up slightly, an effect generally ignored but quite valid. When electric current flows through a conductor, the electrons collide with the atoms of the material and give up energy to them. G. S. Ohm (1787-1854) around 1827 conceived that this was exactly like the fluid case, and the current would be proportional to the voltage gradient. For a certain sample of length l and uniform cross section A, a current I would require a voltage V = I (σ l/A), where σ is a constant characteristic of the material called the conductivity. The quantity R = σ l/A is the resistance of the conductor. Its unit, V/A, is called the ohm, Ω.
We now have the three basic quantities in DC circuits: the voltage V in volts, the current I in ampere, and the resistance R in ohm. The units are the practical units, originally defined without reference to other electromagnetic quantities for technical use. The standard ohm was the resistance of a uniform column of mercury 1 meter long, of cross-section 1 mm2, and the volt was the voltage between the terminals of a Daniell cell. The ampere was the current that one Daniell cell would cause to flow in the standard ohm. By only slight changes, these units were redefined in absolute terms in the Giorgi, or MKS, consistent system of units that now makes part of the SI unit system. The relation between them, V = IR, is Ohm's Law.
Two other basic components are inductance, L, and capacitance C. Inductance is exhibited by a coil of wire. When a current flows in the wire, it produces a magnetic field in the region of the coil. If this field changes, it induces a voltage in the wire by Faraday's Law. L is the constant in the relation V = L (dI/dt), and its unit (V-sec/A) is called the henry, H. Capacitance is exhibited by two conducting surfaces close together. When charges collect on the surface, an electric field in created betweent the surfaces, or plates. A current results if this electric field changes. C is the constant in the relation I = C (dV/dt), and its unit, C/V, is called the Farad. The henry and the farad are impractically large. Millihenry, mH, and microfarad, μF are often convenient in practice. In DC circuits, V and I are constant, so an inductance acts like a short circuit, and a capacitance like an open circuit.
The term "short circuit" means a connection of small resistance, so that a large current will flow if there is a potential difference. An "open circuit" is a connection of high resistance, usually no connection at all, so that no current flows.
The lumped components R, L and C are available in small packages with two leads, and are connected by wires with negligible R, L and C themselves. What is negligible depends on the situation. In DC circuits, only R is significant, and the usual resistors, available in values from less than an ohm to 10 MΩ, behave very much as expected. Symbols used in circuit diagrams for these components are shown at the right. At the top are the symbols used in power engineering in the past. The resistor (R) symbol recalls a line of spools of resistance wire. The inductor or winding (L) symbol looks like a helix seen from the side, without curved lines. The capacitor (C) symbol recalls the Leyden jar, an early capacitor. These symbols are all easy to draw with a 30-60 triangle. The next line shows the symbols typically used in radio or electronics in the same era. The resistor symbol is a zigzag of 60° lines, and the capacitor is now two lines, one straight and one curved. The inductor symbol has small loops, and is hard to draw. Circuit diagrams of this era showed that conductors crossed with no connection by the semicircular jump, and connected if the lines joined. The standard symbols now used in the US retained the zigzag resistor, but the capacitor has two straight lines and the inductor no loops. The new inductor symbol is somewhat easier to draw, but still inconvenient. These symbols are in ANSI standards. Internationally, the ISO uses only a rectangle to represent a circuit component, whose type is shown by the value or identifying symbol next to it. In both ANSI and ISO standards, joining of two wires is explicitly shown by a dot, and lines that simply cross do not imply any connection. At the bottom of the box a popular way of labeling of the value of a component is shown. The k or M replaces the decimal point for resistors. A decimal point implies microfarad, μF (10-6 F) for capacitors, and millihenry, mH for inductors, while no decimal point means picofarad, pF (10-12 F) or microhenry, μH.
Some miscellaneous circuit symbols are shown in the box at the left. The meaning of many symbols is obvious from their form and use, but many can remain mysterious. There were many variations in these symbols with date and field of application. For a full list, refer to the ANSI or ISO specifications. The ground or earth symbol refers to a common reference potential. It can be the potential of the metal chassis on which a circuit is mounted, the "ground" trace of a printed circuit board, a bus-bar serving the same purpose, or even the potential of a copper rod driven into the earth, from which it took its name. A contact looks like a capacitor, except that the lines are short. The letter inside the circle representing a meter shows what kind of meter it is. V = voltmeter, A = ammeter. A shunt is a low-resistance link used with an ammeter to extend its range. The symbols for switches are easy to understand. SPDT means "single-pole double-throw," and the other designations can be worked out easily. The DPDT switch is shown wired as a reversing switch, which connects pairs of wires in two ways.
Finally, we can look at a circuit. The box at the right shows a simple circuit consisting of a battery and four resistors. The symbol for the battery is the traditional one. The long line represents the copper electrode, and the short line the zinc electrode, of an early battery. The + shows that this end of the battery is at the higher potential, and the voltage is represented by E. There is no significance to how many "cells" are represented in the battery, except that more cells can be used to imply a higher voltage. One cell is usually sufficient. The resistors are identified with the letter R and a subscript. If there were more than one battery, the E's would have subscripts. L and C are used for inductors and capacitors. Every component should have a designation. On an actual circuit, the values of the components are usually given, either beside the component, or in a table. The resistors R1 and R2 are said to be in series since the same current flows through each, and the voltages V1 and V2 across them add. The resistors R3 and R4 are said to be in parallel since there is the same voltage across each of them, and the currents I3 and I4 through them add. The points a and b where wires connect are called nodes, labeled for easy reference. The voltage of node a with respect to node b is Vab, and is the voltage across the parallel resistors. A continuous path can be traced from the + terminal of the battery through the circuit back to the - pole of the battery. It is not usually necessary to mark both + and -.
For concreteness, let us assume that E = 3.0 V, R1 = 1000 Ω, R2 = 220 Ω, R3 = 470 Ω and R4 = 330 Ω. We wish to "solve" the circuit, which means to find all currents and voltage drops. Let us start with the parallel resistors. If Vab is the voltage across them, then I3 = Vab/470, and I4 = Vab/330. Solving for Vab, we find Vab = (I3 + I4) [1 / (1/470 + 1/330)] = 193.9 I1. This is the same as the voltage across a single resistor of resistance 193.9 Ω, which may be considered the equivalent resistance of the parallel combination. The same current, I1, flows through the resistors in series. Therefore, V1 + V2 = 1000I1 + 220I1 = (1000 + 220) I1 = 1220I1. This is the same as the voltage drop across a single resistor of resistance 1220 Ω, which may be considered the equivalent resistance of the series combination. Now we can combine the resistance of 1220 Ω with the resistance 193.9 Ω, which carries the same current, to find a total equivalent resistance of 1414 Ω. The voltage across this equivalent resistance is 3.0 V, so the current is I1 = 3.0 / 1414 = 2.12 mA. The voltage across R1 is 1000 x 0.00212 = 2.12 V, across R2, 220 x 0.00212 = 0.47 V, and Vab = 193.9 x 0.00212 = 0.41 V. These voltages add up to 3.0 V, which they must. Then, I3 = 0.41 / 470 = 0.872 mA, I4 = 0.41 / 330 = 1.244 mA. These currents add up to 2.12 mA, as they must. The circuit has now been completely solved.
The method demonstrated above can be used for a class of circuits called series-parallel circuits in which the resistors are arranged in series or parallel combinations. The way we found the equivalent resistances can be generalized to make the work easier. The equivalent resistance of resistors in parallel is given by the reciprocal of the sum of the reciprocals of each resistance. The equivalent resistance of resistors in series is the sum of the individual resistances. To solve a series-parallel circuit, we just combine resistors until we wind up with one resistor across the battery, then work backwards to the individual resistors.
A general method of solving any circuit can be based on principles we have already used in the series-parallel circuit. These principles are: (1) the sum of the voltages around any closed loop is zero, and (2) the sum of the currents flowing into any node is zero. These are called Kirchhoff's Laws. The first expresses the conservation of energy, and the second the conservation of electric charge. To use these principles, we select a sufficient number of unknown independent currents to specify the current in any wire of the circuit, then use Kirchhoff's Laws to write the same number of linear equations involving the currents, which can then be solved to find the unkown currents. It is always possible to do this, and there are some clever methods to ease selecting the currents and writing down the equations. The details will be left to the references. In practice, one seldom goes further than two or three unknowns for a hand solution. People are too error-prone to carry out the analysis in any more complex problem, so the work is best left to computers. In any case, resist the temptation to use determinants to solve the equations.
For linear circuits, there are some useful theorems that can ease the solution in many cases. The first is the Superposition Theorem: in a linear circuit, the current at any point is the sum of the currents due to each voltage source taken separately, the others replaced by their internal impedances. Another is Thévenin's Theorem: with respect to any two terminals, a linear network can be replaced by a voltage source in series with a resistance; the voltage is equal to the open-circuit voltage between the terminals, and the resistance is equal to the ratio of the open-circuit voltage to the short-circuit current at the terminals. Norton's Theorem is similar, except that the equivalent circuit is a current source in parallel with a resistance, where the current source supplies the short-circuit current at the terminals, and the resistor is the ratio of the open-circuit voltage to the short-circuit current (the same as in Thévenin's Theorem). The Reciprocity Theorem says that if a voltage source E produces a current I at some point in the network, then the source and the current can be interchanged (the internal resistance of the source stays where it was). Of these theorems, the first two are by far the most useful. The Wye-Delta equivalent circuits are often useful; see the references.
To illustrate how the theorems can be useful, and to obtain a result that is important in its own right, suppose we wish to connect a 2 Ω load, represented by R3 in diagram (a) in the box at the right, across two 1.5 V D cells in parallel, hoping that the two cells will share the load equally. However, suppose that one cell is fresh, with a voltage E1 = 1.5 V and an internal resistance R1 of 1 Ω. The other cell is old, however, and its voltage is only E2 = 1.4 V, but its internal resistance is also 1 Ω (for simplicity). This is not a series-parallel circuit. It can be solved by Kirchhoff's Laws, but we will use the Superposition Theorem instead. Circuits (b) and (c) show one electromotive force (emf) set to zero. Each of these circuits is a series-parallel circuit, and can easily be solved. Superimposing these circuits means to add the currents. The voltages at the nodes in the two circuits are different, and are not the same as those in the actual circuit. Once the currents are known, however, we can calculate the actual voltages.
If you solve the circuits, you find that E1 contributes a current of I' = 0.30 A in the load, and E2 a current of I" = .28 A, for a total I = 0.58 A. The load voltage is 1.16 V. However, the current in E1 is 0.34 A, and the current in E2 is 0.24 A. The cells do not share the load equally, the weaker cell taking less load. This is probably acceptable in the present case. Now work the problem again, assuming that the internal resistances are 0.1 Ω instead of 1 Ω. You will find that the load current is 0.342 + 0.365 = 0.707 A. This is higher, as expected, because of the lower internal resistances. However, something remarkable lurks in the problem. Cell 1 supplies 0.85 A, more than is supplied to the load. The difference, about 0.14 A, is forced backwards through the weak cell! If we removed the weak cell, the load current would be 0.714 A, a little higher, but more importantly, the good cell would be relieved of supplying 0.14 A. We see that one must take care in connecting sources of emf in parallel, and in general some series resistance is necessary to equalize the load. Lead-acid cells, car batteries, have very low internal resistances, so connecting them in parallel is risky, unless all are at the same state of charge and in equally good condition.
We can also solve the circuit by using Thévenin's theorem. Circuit (d) is the open-circuit case, and circuit (e) is the short-circuit case. We can see without calculation that the open-circuit voltage is the average of the emf's of the two cells, since the internal resistances are equal. Hence, Eoc = 1.45 V = ET. The short-circuit current is 2.9 A if the internal resistances are 1 Ω, so RT = 0.5 Ω This is just the same as calculating the equivalent resistance of the circuit with the emf's reduced to zero. Two 1 Ω resistances in parallel are equivalent to 0.5 Ω. The Thévenin equivalent is shown in (f) feeding the load, and the current is seen to be 1.45 V / 2.5 Ω = 0.58 A, as we found using Superposition. The case of 0.1 Ω internal resistance is just as easy, and we find the current 1.45 V / 2.05 Ω = 0.707 A. Note that the Thévenin equivalent gives no hint of the problem with the cells in the second case. The equivalent is only equivalent as far as the external circuit goes, and tells nothing about conditions in the supply network.
Now that you can find the currents and voltages, you can determine the power furnished by the voltage source and the power dissipated by the resistors. Since volts are joules per coulomb, and amperes are coulombs per second, VI is joules per second, or watt, W. EI is the power furnished by the battery; note that the current comes out of the + terminal. VI is the power dissipated in a resistor; note that the current goes into the + terminal. One is an energy source, the other an energy sink. Since V = IR, the power P = I2R. The power dissipated in resistance is proportional to the square of the current. We also have P = E2/R, but this is less useful. P = VI is called Joule's Law, after J. P. Joule, who established its truth experimentally in the 1840's.
In our series-parallel circuit above, the power furnished is 3.0 V x 2.12 mA = 6.36 mW. The powers dissipated in each resistor are 4.49, 1.00, .36 and 0.51 mW, which adds up to 6.36 mW. These are very small powers, and will not heat a common 1/4 W resistor enough to be noticeable. If we connected the 1000 Ω resistor across 10 V, it would dissipate 100 mW, and would become warm. If we connected it across 100 V, the dissipation would be 10 W, at least for a short while before it destroyed itself. Electrical things generally fail from overheating, so it is best to watch the power dissipation closely.
In electronics, milliamperes and kiloohms are the usual thing, so they are directly used in Ohm's Law and Joule's Law, together with volts and milliwatts.
Many books in electrical engineering include circuits as introductory material. Texts on circuits alone have become rather rare, and their emphasis has changed. Older texts were often quite good, with a great deal of practical information on things like batteries, conductors and other practical apparatus. Books for schools are very bad, if they exist at all. Perhaps an encyclopedia would be a good place to start if you are taking this up for the first time.
Composed by J. B. Calvert
Created 7 February 2001
Last revised 29 May 2004