2
\$\begingroup\$

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.

enter image description here

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?

\$\endgroup\$
4
  • \$\begingroup\$ If the resistor is across the supply then the voltage across the resistor IS the supply voltage. \$\endgroup\$
    – Andy aka
    Commented Dec 5, 2013 at 10:18
  • \$\begingroup\$ Which classic resistor example? \$\endgroup\$ Commented Dec 5, 2013 at 10:21
  • \$\begingroup\$ A simple resistor in a circuit like this: sub.allaboutcircuits.com/images/00444.png \$\endgroup\$ Commented Dec 5, 2013 at 10:22
  • \$\begingroup\$ If you are measuring a lower voltage at the load than the supply then its because of the resistance of your wires. All wires have some resistance. If the diffence in voltage is significant you need thicker wires. \$\endgroup\$ Commented Jun 4, 2015 at 13:02

5 Answers 5

5
\$\begingroup\$

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

enter image description here

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.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Dang, +1 for having better psychic powers of intuition. \$\endgroup\$ Commented Dec 5, 2013 at 11:57
  • \$\begingroup\$ What I really don't understand is how the voltmeter is able to show the voltage drop across the resistor. Look at this circuit: imgur.com/xw9PtM7 if I measure across R1 then I will read a value less than 12v, right? But what is I measure across LAMP1, that should show 12v (I think) yet it is connected to the supply in the same way the voltmeter would be. Why doesn't it run at less than 12v? \$\endgroup\$ Commented Dec 6, 2013 at 2:10
  • \$\begingroup\$ @CameronBall - Wrong. The voltage drop across the resistor and the bulb are the same and are equal to the supply voltage. Due to the 'load' of the resistor and bulb in parallel the battery voltage will drop slightly from its 'no load' value. The connecting wires have no significant effect unless the load current is extremely high (e.g. if the battery was short circuited - the wire becomes the load resistor). If the voltages were different it would contradict Kirchoff's and Ohm's law. \$\endgroup\$ Commented Dec 6, 2013 at 11:44
3
\$\begingroup\$

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

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit it is less. 3 V.

schematic

simulate this circuit

Don't confuse the two situations

\$\endgroup\$
4
  • \$\begingroup\$ +1 for giving a nice explanation of series and parallel loading \$\endgroup\$ Commented Dec 5, 2013 at 11:59
  • \$\begingroup\$ OK so what about this circuit: imgur.com/xw9PtM7 if I measure over R1 then I will get a value less than 12v, so what voltage is LAMP1 running at? Is it 12v or is it the same as R1? \$\endgroup\$ Commented Dec 6, 2013 at 1:58
  • \$\begingroup\$ @Cameron, In your diagram, the voltage across V1, R1 and Lamp1 are all the same. (we disregard tiny differences due to non-zero resistance of copper wire) \$\endgroup\$ Commented Dec 6, 2013 at 9:36
  • \$\begingroup\$ Except light bulbs have quite a bit of internal resistance, making all light bulb examples from school books pretty unrealistic... \$\endgroup\$
    – Lundin
    Commented Jun 26, 2018 at 11:46
0
\$\begingroup\$

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

enter image description here

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

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

\$\endgroup\$
0
\$\begingroup\$

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

\$\endgroup\$
1
  • 2
    \$\begingroup\$ What is "PL/A"? Do you mean \$ \rho L/A \$ where \$ \rho \$ is the resistivity, \$ L \$ is the length and \$ A \$ is the cross-sectional area? It's not obvious in your answer. Remember you can edit your answer at any time to improve it. \$\endgroup\$
    – Transistor
    Commented Apr 4, 2021 at 20:51
-5
\$\begingroup\$

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.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ Eh. No. It doesn't work that way. A voltmeter can only read a difference between two points. \$\endgroup\$
    – JRE
    Commented Jun 26, 2018 at 11:09
  • 1
    \$\begingroup\$ To take in two values a third point would have to be referenced. \$\endgroup\$
    – Transistor
    Commented Jun 26, 2018 at 11:22
  • \$\begingroup\$ This behaviour is not "programmed" or "hardcoded"; it's a consequence of a fundamental property of physics, specifically the gauge symmetry of the electromagnetic potential, rendering it literally physically impossible to measure any absolute voltages. Or in other words, absolute voltages don't exist, only voltage differences do. \$\endgroup\$
    – Hearth
    Commented Mar 20, 2020 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.