The basic circuit is shown in the first circuit diagram. The output voltage across AB is given by :

Note that the input voltage (V) in this case supplied by the battery is constant. The current flowing through
both resistors is the same (series circuit) and so the output voltage across one of them
depends simply on the two resistance values and the input voltage.

(V = IR_{1}
+ IR_{2} and V_{2} = IR_{2} and so V_{2}/V =
R_{2}/[R_{1}+R_{2}])

If we now
attempt to actually MEASURE the output voltage things may change.

(a) Firstly
consider using a digital voltmeter with very high (if not virtually infinite) resistance (R_{V}). The
resistor R_{2} and the voltmeter are connected in parallel and so their combined
resistance (R) is given by the equation;

1/R = 1/R_{2} +1/R_{V}

but as we
have said, R_{V} is huge – almost infinite and so 1/R_{V} is virtually 0 and can be ignored.

This means that 1/R = 1/R_{2} and so R = R_{2}.

The output voltage (V_{o}) measured by
the meter really is that across R_{2}, in other words V_{2}.

(b) A moving coil meter. These meters have a much lower resistance than a digital meter, usually
some tens of kW. This means that the combined resistance of R_{2} and R_{V} is affected
by the resistance of the voltmeter and is actually lower than R_{2}. (Connecting two
resistors in parallel gives a resulting resistance lower than either resistor).

The
proportion of the input voltage (V) dropped across R_{2} therefore falls and so the
output voltage (V_{o}) is less than that measured with a digital
meter.

The LDR is a component that has a resistance that changes when light falls on it. As the intensity of
the light is increased so the resistance of the LDR falls.

If the LDR is connected as part
of a potential divider as shown in the diagram then as the light level is increased its
resistance falls and the proportion of the input voltage dropped across it will also
fall.

So in the light V_{2} is low and in the dark V_{2} is
high.

Something very similar happens if R_{2} is replaced by a
thermistor. As the temperature of the thermistor rises its resistance falls and so the voltage
dropped across it falls.

When the thermistor is hot V_{2} is low and when the
thermistor is cold V_{2} is high.

Of course both
these examples have considered R_{2} being replaced by another component. If
R_{1} is replaced then if the voltage across this component rises the output voltage
across R_{2} will fall. (The total voltage across both the resistor and the other
component in the circuit must always stay the same and be equal to the supply voltage of the
battery.)