3
$\begingroup$

I am struggling to reconcile the definition of voltage as work done moving a unit of electric charge between two points in an electric field and voltage in Ohm's law.

In the work definition, everything is more or less clear, but in Ohm's law we are measuring voltage at a point, so work done should be very small since there is almost no distance. Could somebody point me at right reading direction?

$\endgroup$
  • 3
    $\begingroup$ Why do you say that in Ohm's law we are measuring voltage at one point? It is always a quantity involving (usually well separated) two points. $\endgroup$ – GiorgioP Mar 28 at 12:57
  • $\begingroup$ @GiorgioP, I heard it, but second point is not too clear. Some say it is groud, some say it is negative terminal (what does it even mean for alternating current). Could you give me an example of two points involved? $\endgroup$ – user1700890 Mar 28 at 13:18
  • 1
    $\begingroup$ I added an Ohm's law tag $\endgroup$ – Dale Mar 28 at 13:39
5
$\begingroup$

While one can specify a voltage at a point, it must always be with respect to another point, sometimes referred to as common, ground or reference point where the voltage is arbitrarily assigned a value of zero. The zero voltage point is typically chosen at the negative terminal of the voltage source.

For the simple circuit below, I have labeled points A, B, C, and D. I have assigned the voltage at point D (the negative terminal of the battery) to be zero volts. The voltages at points A, B and C are then measured with respect to point D.

If I want to apply Ohms Law to resistor $R_2$, the applicable voltage is the difference in voltage between points B and C, or

$$I=\frac{(V_{B}-V_{C})}{R_2}$$

If I want to apply Ohm’s law to the series combination of the three resistors, it is then

$$I=\frac{(V_{A}-V_{D})}{(R_{1}+R_{2}+R_{3})}$$

$$I=\frac{V}{(R_{1}+R_{2}+R_{3})}$$

So when you see Ohm’s law written as

$$I=\frac{V}{R}$$

$V$ in the equation is meant to be the difference in potential (voltage) across the resistor.

I hope this helps.

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Check your formula for the current in the diagram. I think you have a typo, either in the resistor subscript or the voltage/potential subscripts. $\endgroup$ – Bill N Mar 29 at 14:03
  • 1
    $\begingroup$ @Bill N Ugh yes. Thought I fixed it but then pasted the wrong version. Will try again. Thanks $\endgroup$ – Bob D Mar 29 at 14:06
3
$\begingroup$

in Ohm's law we are measuring voltage at a point so work done should be very small since there is almost no distance

This is the key mistake. In Ohm’s law the voltage is the voltage across the resistor. In other words, the voltage in Ohm’s law is measured at two points. One point is always the point where the current is entering the resistor and the other point is always the point where the current is leaving the resistor.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Sometimes notation may induce misunderstandings. It is true that in many formulas voltage appears as $\Delta V$, making clear that it refers to the difference of voltage between two points. However, arrived at Ohnm's law, it is usual to find it expressed as $$ I=\frac{V}{R},$$ thus inducing to think that only voltage at one point is present. Actually in that formula $V$ is the difference of voltage across the resistor. Therefore the numerator on the right hand side of the equation should be written more consistently as $\Delta V$.

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

definition of voltage as work done moving unit of electric charge between two points in electric field

That's almost okay, but what's missing is the concept of electrical potential. And technically, you should say "work per unit charge" to get the units correct (volt = joule per coulomb). That's often not discussed in a lower level (high school) first course, but is the basis for voltage. Electrical potential is defined for a point in space and is the potential energy per unit charge added to a system of charge(s) by adding a "test" charge at the point in space (bringing it in from infinitely far away).

Imagine a space with only $5$ nC charge. The is no force on the charge and no potential energy defined. But there is an electric field everywhere. If we add a $1$ C charge, starting it infinitely far away, and move it to a distance $3$ m from the $5$ nC charge, the E-field of the $5$ nC charge does $-15$ J of work on the test charge. Therefore, the potential energy of the system changes from zero to 15 J (by definition of potential energy).

If I calculate the change in potential energy per charge added I get $$\phi=\frac{15~\mathrm{J}}{1~\mathrm{C}} = 15~\mathrm{V}.$$ A volt is a joule per coulomb. That's the potential at the point $3$ m from the $5$ nC charge *due to the * $5$ nC charge only. If I take my test charge, again starting at infinity, and move it to $5$ m, I will find that the potential energy added is $9.0$ J, so the potential at that point is $9.0$ V.

Now voltage is the difference in potential between two points in a system, $$V=\Delta\phi$$so voltage ends up being the difference in [work per unit charge done by the electric field] in moving a test charge from [infinity to point A] and [infinity to point B], which is also the net work per unit charge done by the field while moving the test from B to A. The voltage between $3$ m and $5$ m from the $5$ nC charge is $6.0$ V with $3$ m being the higher potential point. The voltage across that gap is $6.0$ V.

An electrical circuit is a complex set of charges and materials which affect how space responds to those charges, but the net result is an approximation model known as Ohm's Law. As @BobD details, the voltage, V, in Ohm's Law is a difference in potential between two points in the circuit, usually across some resistor or set of resistors.

Most importantly, even if the reference potential ($V_D$) is not zero, the voltages across the resistors will not change, and the current in the circuit will not change.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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