Why does a voltmeter read lower across a load than across a supply?

Question

From the Wikipedia article about series and parallel circuits:

As an example, consider a very simple circuit consisting of four light bulbs and one 6 V battery. If a wire joins the battery to one bulb, to the next bulb, to the next bulb, to the next bulb, then back to the battery, in one continuous loop, the bulbs are said to be in series. If each bulb is wired to the battery in a separate loop, the bulbs are said to be in parallel. If the four light bulbs are connected in series, there is same current through all of them, and the voltage drop is 1.5 V across each bulb, which may not be sufficient to make them glow. If the light bulbs are connected in parallel, the currents through the light bulbs combine to form the current in the battery, while the voltage drop is 6.0 V across each bulb and they all glow.

So with the classic resistor example, why is it that when the voltmeter is put over the resistor it shows a lower voltage? Shouldn't it show the same voltage as the supply?

Answer 1

There is one thing that has been missed out from your 'classic resistor example' - the internal resistance of the battery.

If you simply measure the battery voltage with a voltmeter you get a higher reading due to the fact that there is no (or very little) voltage dropped across the internal battery resistance.

The voltage measured across a load resistor (or bulb) OR the terminals of the battery will be smaller by an amount equal to I * R(int).

The connecting wire has very little resistance and so does not contribute to a significant voltage drop between the terminals of the battery and the resistor so the voltages at these two places are, in practice, the same.

Answer 2

In these circuits, the voltmeter result is the battery voltage. Both 9 V.

^{simulate this circuit – Schematic created using CircuitLab}

In this circuit it is less. 3 V.

^{simulate this circuit}

Don't confuse the two situations

Answer 3

measuring the supply voltage across the power source we found 9v

but if we measure the voltage across the load we found let say 8.03v ie a lower voltage than the source voltage

if we imagine that the upper wire have 6 ohme and the lower wire have 6 ohme the total is 12 ohme

the total courrent is i(t)=V/R= 9/112 =0.0803 A

Vlamp (load) = i(t) X R = 0.0803 x 100 = 8.03 V

the drop of voltage is in the wire =9-8.3=0.97 v

its because of the resistance of the wires.All wires have some resistance. to reduce the drop of voltage you need thicker wires

Answer 4

The reason why there is always a voltage drop in an electric circuit without giving consideration to the internal resistance of the battery or the voltage source is the resistance in the wires used in the circuit connection. Remember the Resistance in any wire which is not a super conductor is given as R= PL/A

Answer 5

A volt/multimeter takes in 2 values and gives you the Value between them. Yes, this behaviour is programmed or hardcoded. You could just aswell display both values on the display its receiving and calculate the drop yourself. which is suprising that it does not give you the the two reference values for added info, you could know so much more about unknown system this way.