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This might be a stupid question but I can't find an explanation.

Voltage measurement is done in parallel. But when we measure the voltage of a stand alone voltage source, say a battery, it looks like we are doing it in series and yet we get the correct reading.

schematic

simulate this circuit – Schematic created using CircuitLab

So we connect the voltmeter like above to measure the voltage of an isolated voltage source. Although it is connected in series, we get the correct reading.

Am I missing something here?

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  • \$\begingroup\$ The battery and voltmeter are connected in parallel in your circuit. \$\endgroup\$
    – HandyHowie
    Dec 23 '15 at 12:01
  • \$\begingroup\$ @HandyHowie How would they look had they were connected in series? \$\endgroup\$
    – Utku
    Dec 23 '15 at 12:02
  • \$\begingroup\$ You can't connect just 2 components in series. \$\endgroup\$
    – HandyHowie
    Dec 23 '15 at 12:03
  • \$\begingroup\$ @HandyHowie That's not true, and this example is a counterexample to your assertion. \$\endgroup\$
    – Shamtam
    Dec 23 '15 at 12:27
  • \$\begingroup\$ @Shamtam OK, I was trying to describe it too simply. Thanks. \$\endgroup\$
    – HandyHowie
    Dec 23 '15 at 12:53
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This situation is a something of a paradox, yes, but you need to understand that 'series' does not mean 'not parallel', and vice versa.

  • Any components which form a loop in which current is allowed to flow are in series.

  • Any loops which share the same potential difference across them are in parallel.

In your example, current flows from the one battery terminal to the other through the voltmeter, which acts like a very large resistance (10M-50M, typically). Current is flowing in a loop, therefore the two components are in series with each other. The voltage across the battery terminals is equal to the voltage across the voltmeter, therefore the two components are also in parallel with each other.

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  • \$\begingroup\$ I'd be a bit more pedantic with your series definition by describing two series components as necessarily sharing the same current through them (for there are no other current paths between them). \$\endgroup\$
    – Shamtam
    Dec 23 '15 at 12:29
  • \$\begingroup\$ This circuit is like the Schrödinger's cat. \$\endgroup\$
    – Hagah
    Dec 23 '15 at 13:11
  • \$\begingroup\$ @Hagah Yes: the battery is both fully charged and flat until you turn the voltmeter on! \$\endgroup\$ Dec 23 '15 at 13:16
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The voltmeter has a high ohm resistor (e.g. 10 MΩ) which is placed in series with the battery. The voltmeter is actually measuring the voltage across this resistor using an ADC (analog to digital converter) and using that voltage which is then displayed.

enter image description here

I replaced the voltmeter in your circuit by the one in the diagram. Right-click on the image and choose View Image to see a larger version.

In this circuit, the input resistance is a little over 11 MΩ. There is a voltage divider made up of R1 through R4 which is connected to a selector switch to select one of several voltage ranges (20 mV, 2V, 20V and 200V) to be input to the ADC (which is contained inside IC1).

In this case, you would set the range switch to 20V since you are measuring a 5V battery (and the next lower range is 2V). This creates a voltage divider with 10MΩ on the top leg, and 1MΩ+110K+1.1K on the bottom. So with a 5V input, the ADC will see

$$5V\times\frac{1MΩ+110K+1.1K}{10MΩ+1MΩ+110K+1.1K} = 5V\times\frac{1.1111MΩ}{11.1111MΩ} = 0.1V$$

When you measure the voltage across the red and black terminals, you are really measuring the voltage across a resistor divider, which is also the voltage of the battery. This is then scaled for the ADC input.

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  • \$\begingroup\$ I like your avatar now. \$\endgroup\$
    – Roh
    Dec 23 '15 at 12:19
  • \$\begingroup\$ This answer is so far off topic that it should be deleted. \$\endgroup\$ Dec 23 '15 at 13:13
  • \$\begingroup\$ @MichaelKaras What? I'm explaining how a digital voltmeter measures voltage across its terminals, which is very on-topic. I combined my diagram with is to makes things a little clearer. It's obvious the OP doesn't understand what the circuit of a digital voltmeter looks like. \$\endgroup\$
    – tcrosley
    Dec 23 '15 at 13:45
  • \$\begingroup\$ I don't see any reference to 'digital' in the OP's question. It's clear to me that he understands how the meter works. His question is why is what appears to be a series circuit connection also a parallel connection. \$\endgroup\$
    – Transistor
    Dec 23 '15 at 21:26
  • \$\begingroup\$ @transistor I show that also -- the10M resistor is in series with battery, and the ADC is in parallel with resistor (thus measuring the voltage). \$\endgroup\$
    – tcrosley
    Dec 24 '15 at 18:57
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Look again at your picture. In this simple case the meter is in parallel with the voltage source from the perspective of voltage measurement.

The meter is also in series with the voltage source from the perspective of current that wants to flow around the loop. Hopefully the internal impedance of the volt meter is very high so very little current flows.

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    \$\begingroup\$ How could something be connected both in series and in parallel at the same time? I am confused. \$\endgroup\$
    – Utku
    Dec 23 '15 at 12:05
  • \$\begingroup\$ @Utku - Well you will have to accept that in the two component case series and parallel connections look exactly the same in a schematic. \$\endgroup\$ Dec 23 '15 at 12:07
  • \$\begingroup\$ Could you direct me to further reading in order to get a better understanding? \$\endgroup\$
    – Utku
    Dec 23 '15 at 12:08
  • \$\begingroup\$ @Utku - I have no go to guide for you other than to think in terms of what I already wrote. \$\endgroup\$ Dec 23 '15 at 12:11
  • \$\begingroup\$ @Utku Yes, components can be connected in series and parallel at the same time. See CharlieHanson's answer. \$\endgroup\$
    – Shamtam
    Dec 23 '15 at 12:30
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It may help you to think of it a little differently as in your example you have a power source (the battery) and a load rather than, say, two resistors in parallel.

schematic

simulate this circuit – Schematic created using CircuitLab

Here we can clearly see that the two voltages are in parallel. What may not be so clear is that the currents are in anti-parallel - i.e., they are forming a loop and so, as your hunch led you to believe, the current is running in series through the loop.

Another way of looking at it: on the battery the current is exiting the '+' terminal. On the load (voltmeter) the current is entering on the '+' terminal.

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The 'voltmeter' we normally use is really a very sensitive ammeter with a large series resistance.

What the 'voltmeter' is doing is measuring the small current that flows through it (a serial measurement) and converting this current value back into a 'voltage' reading (i.e. the voltage across the terminals = voltage across the 'voltmeter').

By Ohm's law (V=IR) we know that V is directly proportional to I, so its just a matter of how big R (the constant of proportionality) is to convert one to the other.

It's much easier to measure a current directly (even very small ones) than it is to measure voltage directly.

Examples of 'voltmeters' that are really ammeters:

A moving coil multimeter: A low cost moving coil 'voltmeter' is typically a 50uA movement ( 1000 ohm coil). For a ONE VOLT reading (full scale) we need to have a total resistance of 20K. (you'll sometimes see this marked on the meter as 20K/volt) i.e 19k (external) + 1k (meter).

The range selector switch adds suitable parallel/series resistance into the basic meter circuit

A digital multimeter - requires/measures a much lower current something like 1M0 per volt and is closer to an 'ideal' voltmeter which would take no current from the circuit (infinite resistance?).

The resistance of the 'voltmeter' is only really important if the current taken from the circuit significantly interferes with the reading. In the case of a battery (very low internal resistance) this is highly unlikely.

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  • \$\begingroup\$ they why wouldn't hooking a voltmeter in series with a battery and resistor also work. That would tell you the voltage across the voltmeter terminals. I do this and get zero but the voltage drop should be large because of the large resistance of the voltmeter \$\endgroup\$
    – user5419
    Jan 21 '18 at 0:59
  • \$\begingroup\$ @user5419. Given that any practical voltmeter is really an ammeter (with an internal series resistor fitted) that is exactly what you are doing when measuring the voltage, regardless of any external resistor you may or may not add, provided it makes a complete circuit from one battery terminal to the other. Adding an external resistor will simply lower the reading on the voltmeter. If you get a zero reading there is something wrong with your circuit because no current flows through it (i.e an open circuit) \$\endgroup\$ Jan 22 '18 at 17:15
  • \$\begingroup\$ @user5419. Given that any practical voltmeter is really an ammeter (with an internal series resistor fitted) that is exactly what you are doing when measuring the voltage, regardless of any external resistor you may or may not add, provided it makes a complete circuit from one battery terminal to the other. Adding an external resistor will simply lower the reading on the voltmeter. If you get a zero reading there is something wrong with your circuit because no current flows through it (i.e an open circuit) \$\endgroup\$ Jan 22 '18 at 17:16

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