8
\$\begingroup\$

This question already has an answer here:

Imagine a trivial circuit with battery and one resistor. To measure the "voltage drop" across the resistor, we stick a voltmeter in parallel with it. However, this means the voltmeter is also directly electrically connected to the terminals of the battery. Therefore, why doesn't it simply measure the battery voltage, regardless of the resistor?

schematic

I hypothesized that the answer is because some of the available voltage/pressure/force/power/energy/magic/whatever-it-is from the battery is being drained through the resistor, it will not be passing through the voltmeter and will therefore not be measured by it, but this still doesn't fit, because that means the voltmeter is not measuring the voltage drop; it's measuring the voltage that wasn't dropped. If the battery supplies 5 V and voltmeter reads 4 V, how is 4 V possibly the voltage drop of the resistor, when the available voltage has actually dropped by 1 V and I still seem to have a non-dropped 4 V to play with?

I guess I really do not understand what the voltage drop is supposed to mean. I don't know why the resistor is dropping voltage at all, when it is supposed to resist current. I'm so lost.

|improve this question
\$\endgroup\$

marked as duplicate by PeterJ, Dmitry Grigoryev, Voltage Spike, winny, DoxyLover Sep 19 '17 at 9:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    \$\begingroup\$ Consider KVL for a moment. It must be the case--in the circuit you provided--that the voltage drop across the resistor is the same as the voltage provided by the battery. Effectively the voltmeter is measuring both of those voltages because they are identical. Note also that the ideal voltmeter does not allow any current to pass through it. \$\endgroup\$ – Paul Stiverson Jun 17 '15 at 21:22
  • \$\begingroup\$ @PaulStiverson "It must be the case--in the circuit you provided--that the voltage drop across the resistor is the same as the voltage provided by the battery." Ah-ha! That is news to me. That might answer it. But then how does the voltmeter measure anything when R1 is not present? \$\endgroup\$ – Boann Jun 17 '15 at 21:46
  • 1
    \$\begingroup\$ The battery still has potential difference (this is what we refer to as voltage) even when there is no load. The effects of the load are better explained by the other answers. \$\endgroup\$ – Paul Stiverson Jun 17 '15 at 22:04
  • 1
    \$\begingroup\$ I believe kirchoff might answer that nicely... \$\endgroup\$ – Jorge Aldo Jun 22 '16 at 4:55
15
\$\begingroup\$

[I'm ignoring non-ideal behavior since that doesn't seem to be what you're interested in.]

Your assumption is wrong. Measuring the voltage drop across the resistor does measure the battery voltage. The battery voltage and the resistor voltage are the same in your circuit. The general rules are:

Components in parallel share the same voltage

Components in series share the same current

Ideally, adding or removing the resistor doesn't change the voltmeter's measurement at all. The battery, the resistor, and the meter are all in parallel, so they all share the same voltage. If the battery voltage is 5 V, then the resistor voltage and the meter voltage must also be 5 V.

Voltage is basically a measurement of potential energy due to an electric force field. If you go around the circuit in a loop, you end up back at the same potential, which means you lose whatever energy you gained along the way. (Gravity also works this way.) When you move from negative to positive through a battery, you gain energy. When you move through a resistor, you lose energy. If a battery and a resistor are in parallel and you move around that loop, the energy gained in the battery will equal the energy lost in the resistor. In other words, their voltages are the same! This principle is called Kirchhoff's Voltage Law. More formally, it says that the sum of the voltages around a closed loop must equal zero.

Resistance describes a relationship between the voltage across the resistor and the current through it. So resistors do "resist" the flow of current, but the way they do that is by dissipating energy. It's similar to the way that friction resists the movement of an object.

Now as others have pointed out, in real life a battery is not an ideal voltage source. The voltage of a real battery changes depending on how much current is being drawn and how much charge is left. So in real life, adding a resistor can change the battery voltage. But the battery voltage and the resistor voltage will still be (almost) equal. (The parasitic resistance of the wires is normally very small.)

Hopefully this has clarified things. Please feel free to post follow-up questions if you're still confused.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ I was confused because I had missed the fact that the voltage drop and the supply voltage must balance each other. So because they are the same value, there is no conflict! And the 4 V that I measured was apparently just due to the feebleness of the battery when under load. It makes sense to me now, thank you. \$\endgroup\$ – Boann Jun 17 '15 at 22:30
7
\$\begingroup\$

Because we are living in a non-ideal world, where batteries and wires have non-zero resistance and the actual circuit is as following:

schematic

simulate this circuit – Schematic created using CircuitLab

Voltage measured in such a circuit would be \$V=5\frac{R_1}{R_{wire}+R_{battery}+R_1}\$ which is less than 5V. The non-ideality of the voltmeter is not taken in account here.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ I'm sorry, I don't understand... My complaint is not that the voltmeter measurement is slightly inaccurate. My problem is that it seems be measuring completely the wrong thing, because it is connected to the battery just as strongly, regardless of whether R1 is there or not. \$\endgroup\$ – Boann Jun 17 '15 at 21:12
  • 4
    \$\begingroup\$ If R1 is not there, there won't be any current in the circuit. And since there is no current - there won't be any voltage drop on R_wire and R_battery, so you will measure the full 5V. But if R1 is there, the current will flow and produce voltage drop on these, so you will measure voltage lower than 5V. See the update about the calculation. \$\endgroup\$ – Eugene Sh. Jun 17 '15 at 21:15
  • \$\begingroup\$ So the voltmeter is not measuring the voltage drop of R1 at all, but the addition of R1 causes it to suddenly measure a reduced voltage from the battery due to a voltage drop through R_battery and R_wire, but for some reason this is referred to as measuring the voltage drop of R1, even though it is actually measuring the remaining voltage and not the voltage drop, and it is really measuring the rest of the circuit and not R1 ?? \$\endgroup\$ – Boann Jun 17 '15 at 21:31
  • \$\begingroup\$ It is measuring the voltage across R1 if connected close enough to it. Or a voltage drop on it. Or a potential difference between it's leads. Call it whatever you want. Yes, you can say that it is measuring the voltage across the whole circuit minus R1 as well. \$\endgroup\$ – Eugene Sh. Jun 17 '15 at 21:33
  • \$\begingroup\$ Also, do not ignore the voltmeter's input impedance. Replace VM with an impedance that equals the voltmeter's input impedance. (IOW, the voltmeter's input impedance is in parallel with resistor R1.) For example, the (default) input impedance of an Agilent 34401A digital multimeter is 10 megohms when making DC Volts measurements. A Simpson 260-6XLP has 20,000 ohms/volt input impedance in DC Volts measurement mode. See also: allaboutcircuits.com/textbook/direct-current/chpt-8/… \$\endgroup\$ – Jim Fischer Jun 22 '16 at 5:44
1
\$\begingroup\$

Lets simplify, the resistor or load in the circuit is consuming voltage. If you were to remove (R1) creating a open circuit essentially you would be measuring battery voltage. closing the circuit is putting the load to work. in the first example r1 the meter should read 5v, showing the resister is using full battery voltage. (KVL) voltage drop is an excellent way to identify undesirable loss in a circuit. if in example 1, the meter reads 2.5 volts and a known good battery is 5v would indicate unwanted resistance in the circuit. (bad connections, long distance etc.)this is known due to KVL, and could be verified by doing a voltage drop from battery source to load, and after load to battery ground. the total readings will equate 5v and identify the unwanted resistance. in multi load circuits it can identify malfunctioning loads and voltage use from individual loads.

|improve this answer
\$\endgroup\$
  • 1
    \$\begingroup\$ +1 for a nice answer, however, please try to avoid acronyms like KVL (Kirchhoff's Voltage Law). It makes things harder for newbies. \$\endgroup\$ – jose.angel.jimenez Dec 14 '15 at 20:04
1
\$\begingroup\$

If you have an ideal battery you got 5v when you measure the voltage across the battery even if you put a resistor of 100 Ω

Because it is ideal battery i.e. no internal resistance

enter image description here

But in your case you measured 4v instead of 5v which mean your battery have an internal resistor And by apply a kirchouve law you find 4v with neglecting the wire resistance which is in mili ohms

VR=(V*R)/(R internal+R)

VR=(5*100)/(25+100)

VR=4v

enter image description here

|improve this answer
\$\endgroup\$
0
\$\begingroup\$

in your example, measuring the voltage drop across the resistor may not exactly match the results of measuring the voltage of the battery terminals. think of very long leads between the battery terminals and resistive load which are also adding resistance in the mix. in that scenario choosing where exactly to measure becomes practical. Assuming a resistor is simply directly tied in parallel with the battery, there will be practically no voltage drop as the resistance will be extremely low. that scenario also provides a few extra complications. the battery is not an 'ideal' power source, and you are just essentially measuring the voltage of a battery as a resistor drains it down.

|improve this answer
\$\endgroup\$
-1
\$\begingroup\$

Old post, but my two cents. In order to measure the voltage drop across ANY load, be it resistor, whatever, you must have a SECOND load. In the OP's circuit, any difference in voltage measured with and without the resistor (no open circuit of course!) is the voltage drop of the all the other 'loads' in the circuit, except the resistor! Which is what many of the posts are saying, in a round-about way. Put a second resistor in series with the first! Then you are measuring the voltage drop across the resistor.

|improve this answer
\$\endgroup\$
  • 1
    \$\begingroup\$ This answer seems very confused about how to take a voltage measurement. "In order to measure the voltage drop across ANY load, be it resistor, whatever, you must have a SECOND load." This is not true. \$\endgroup\$ – Transistor Jul 13 '16 at 20:46

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