If an electric circuit has a well-defined output terminal, the circuit connected to this terminal (or its input impedance) is the load. (The term 'load' may also refer to the power consumed by a circuit; that topic is not discussed here.)

Load affects the performance of circuits that output voltages or currents, such as sensors, voltage sources, and amplifiers. A household's power outlets provide an easy example: they are a voltage source, outputting 120 V AC for example (in USA), with the household's appliances collectively making up the load. When a power-hungry appliance switches on, it dramatically reduces the load impedance, causing the output voltage to drop. This drop is easily observed; for instance, turning on a vacuum cleaner dims the lights.

A more technical approach

(Two sidenotes on generality, for advanced readers: This discussion will disregard nonlinearity. It will also use simple resistances, but they can be readily generalized to impedances for AC analysis.)

When discussing the effect of load on a circuit, it is helpful to disregard the circuit's actual design and consider only the Thévenin equivalent. (The Norton equivalent works just as well, but this discussion will use the Thévenin form.) The Thévenin equivalent of a circuit looks like this:

The circuit is represented by an ideal voltage source Vs in series with an internal resistance Rs.

With no load (open-circuited terminals), all of $V_S$ falls across the output; the output voltage is $V_S$. However, the circuit will behave differently if a load is added. We would like to ignore the details of the load circuit, as we did for the power supply, and represent it as simply as possible. If we use an input resistance to represent the load, the complete circuit looks like this:

Whereas the voltage source by itself was an open circuit, adding the load makes a closed circuit and allows current to flow. This current places a voltage drop across $R_S$, so the voltage at the output terminal is no longer $V_S$. The output voltage can be determined by the voltage division rule:
$V_{OUT} = V_S \cdot \frac{R_{L}}{R_{L} + R_S}$