*"Provided its temperature is constant, the current flowing through an electrical conductor is directly proportional to the potential difference [voltage] across its ends."*

Ohm's Law

*"The algebraic sum of the currents entering and leaving a junction is zero."*

Kirchhoff's First Law

*"The algebraic sum of potential differences and emfs around any closed circuit is zero."*

Kirchhoff's Second Law

*"In a linear resistive network [one to which Ohm's Law applies] containing more than one source of emf, the resultant current in any branch is the algebraic sum of the currents that would be produced by each emf acting on its own while the other emfs are replaced by their respective internal resistances."*

Superposition Theorem

*"Any active network can be replaced at any pair of terminals by an equivalent emf in series with an equivalent resistance."*

Thevenin's Theorem

*"Any active network can be replaced at any pair of terminals by an equivalent current source in parallel with an equivalent resistance."*

Norton's Theorem

A battery or other power supply has a certain electromotive force (emf, denoted by E). This is a measure of the ability to drive an electric current through an electrical circuit. The SI unit of emf is the volt (V).

The symbol for a battery consists of two parallel lines - the line denoting the negative terminal being shorter - perpendicular to and straddling the wire. There is no wire between the terminals.

A resistor is a component that resists the flow of current. Resistance (R) is given by the equation R=V/I, where V denotes the potential difference (pd), also measured in volts, between the ends of the resistor. The SI unit of resistance is the ohm (Ω).

A resistor is denoted by a rectangle straddling the wire. The wire is hidden between the ends of the resistor.

An ohmic conductor is one whose resistance is constant. Other components, for example lamps, warm up in use and therefore generate a higher resistance. Higher temperature is in turn caused by a higher current.

The sum of the voltage drops or potential differences across the components is equal to the emf of the power supply. However, all power supplies have an internal resistance, denoted by r. Because the current also flows through this resistance, the voltage that can be used by the circuit is equal to E-Ir.

For a combination of resistors, one can calculate the equivalent single resistance R_{e}. If the resistors are in series, the same current will flow through them all. For resistor n, therefore, V_{n}=IR_{n}. The pd across the series is equal to the sum of the pds across the individual resistors, i.e. V=ΣV_{n}. Therefore, IR_{e}=Σ(IR_{n}). Factorising and cancelling gives R_{e}=ΣR_{n}.

In a parallel combination, the same pd is dropped across each resistor. For resistor n, therefore, V=I_{n}R_{n}. The current splits before passing through the resistors and recombines on the far side, which means I=ΣI_{n}. This leads to V/R_{e}=Σ(V/R_{n}). Factorising and cancelling gives 1/R_{e}=Σ(1/R_{n}).

Resistors may be used either to restrict the amount of current that can flow or in a potential divider. A potential divider consists of a power supply, two resistors - whose resistances are R_{1} and R_{2} - and a branch that leaves the circuit between the resistors. The potential is V_{in} at the positive terminal, 0V at the negative terminal and V_{out} at the branch. Current I_{out} flows along the branch.

Because the voltage drop across R_{2} is equal to V_{out}-0, V_{out}=IR_{2}. Similarly, V_{in}=I(R_{1}+R_{2}). Substituting for I in the first equation gives V_{out}=V_{in}R_{2}/(R_{1}+R_{2}). This only applies if I_{out} is negligible compared to I.

As with gravitational potential, one can arbitrarily define the zero of electrical potential, with all other potentials measured relative to that point. For our potential divider, the zero is the most negative potential. The zero of potential may be called the earth, ground, chassis or common. Its symbol may be an inverted triangle joined to the bottom of a vertical line; alternatively, the triangle may be replaced by a series of horizontal lines that also form an inverted triangle, or the symbol may simply consist of a horizontal line labelled "0V".

In the Thevenin equivalent, a circuit is replaced between two terminals - which we shall call A and B - by emf E_{t} in series with resistance R_{t}. The Norton equivalent uses current source I_{n} in parallel with resistance R_{n}. Incidentally, a current source is denoted by two overlapping circles, both centred on the wire; the wire is hidden by the symbol.

If each arrangement forms a black box, how are the unknown quantities determined?

First consider the Thevenin equivalent. Measure the open-circuit voltage V_{oc} by connecting a high impedance meter between A and B. Because no current can flow through R_{t}, there is no voltage drop across R_{t} and E_{t}=V_{oc}.

Next, measure the short-circuit current I_{sc} by connecting an ammeter of negligible resistance across AB. Because the only resistance is R_{t}, I_{sc}=E_{t}/R_{t}. Combining this with the previous equation leads to R_{t}=V_{oc}/I_{sc}.

Now consider the Norton equivalent. For a particular pd, current is inversely proportional to resistance (I ∝ 1/R). Measuring the open-circuit voltage therefore leads to V_{oc}=I_{n}R_{n}, given that the current passes exclusively through R_{n}.

When the short-circuit current is measured, then because there is negligible resistance across AB, there is negligible current through R_{n} and I_{n}=I_{sc}. Combining this with the previous equation leads to R_{n}=V_{oc}/I_{sc}.

It will be noticed that R_{n}=R_{t}. This is because the combined resistance of a part of a circuit has a definite value, regardless of which equivalent is being considered.

It is not a good idea to short-circuit the terminals of a black box, as one can confirm by putting a thick copper wire between the terminals of a car battery. If, instead, a resistor R is connected across AB and the voltage V_{R} across that resistor measured, it can be shown that:

R_{t}=R(V_{oc}-V_{R})/V_{R}; and

I_{n}=V_{R}V_{oc}/R(V_{oc}-V_{R}).

When the parameters of the equivalent circuits have been determined, they can be used to study the effects of attaching different loads across the terminals.