My question is inspired by this: Why do farads multiplied by ohms produce a result that has a unit of seconds?

I want to ask what kilogram is doing in this equation:

$$\mathrm{V = \frac{kg \cdot m^2}{A \cdot s^3}}$$


6 Answers 6


Voltage is used to measure the potential difference between one point and a reference point. The reference point when its ground voltage becomes a node or a point voltage. Now potential is defined as the "Work done in bringing a point charge from infinity to the given location". The Work done is Force multiplied by displacement and The force involves mass of the particle (unit +ve charge), which in SI units is in kg.

  • 3
    \$\begingroup\$ This explains why, whereas the accepted answer is just a bunch of formulae that only really make sense if you already understand why. \$\endgroup\$ Commented Nov 15, 2014 at 0:05
  • \$\begingroup\$ @DavidRicherby I agree too. This is a much better explanation than the raw formulas. I have updated my vote. \$\endgroup\$
    – Kamil
    Commented Jan 16, 2022 at 22:03

It's part of the Force. Well, just force actually. Which is mass times acceleration.


\$\mathrm{J = N\cdot m}\$

\$\mathrm{N = \dfrac{kg\cdot m}{s^2}}\$

\$\mathrm{C = A\cdot s}\$

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    \$\begingroup\$ "Use the Force, Kamil!" \$\endgroup\$
    – Kroltan
    Commented Nov 14, 2014 at 0:22

Let \$P\$ be power in watts, \$I\$ be current in amps, \$W\$ be work in Joules,

\$A\$ Acceleration in meters per \$\text{second}^2\$ \$D\$ distance in meters, \$M\$ Mass in kg.

\$T\$ Time in seconds, \$F\$ Force in newtons and \$V\$ voltage in volts.

We know \$ P = V \cdot I\$ so \$V = \dfrac{P}{I}\$.

Basic physics should tell you Power is Work divided by time \$P = \dfrac{W}{T}\$.

Work is Force times distance \$W = F \cdot D\$

Force is mass times Acceleration \$F = M \cdot A\$.

Putting all this together we see.

\$ V = \dfrac{P}{I} = \dfrac{W}{I \cdot T} = \dfrac{F \cdot D}{I \cdot T} = \dfrac{M \cdot A \cdot D}{I \cdot T} = \dfrac{M \cdot D \cdot D}{I \cdot T \cdot T^2} = \dfrac{M \cdot D^2}{I \cdot T^3}\$

Using standard SI units the volt is therefore \$\dfrac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{A} \cdot \mathrm{s}^3}\$


Imagine a uniform electric field, pointing to the right. Consider two points A,B, where B is one meter to the right of A. Suppose the difference in potential between A and B is one volt.

At point A, put an object with a mass of one kilogram which has a positive charge of one coulomb (one amp-second). The object will be pushed towards B by the electric field. As it moves towards B, it will experience a force sufficient to accelerate it at a rate of one meter per second per second.

In other words, one volt is the potential difference which, over a distance of one meter, will cause an object with charge one coulomb and mass one kilogram to accelerate at a rate of one meter per second per second.

If the object's mass were two kilograms, it would only accelerate at one-half meter per second per second.


As other have mentioned, voltage is a measure of energy per unit charge. But in the physics of E&M fields, voltage is also used to calculate the electric field. Specifically: $$-E = \nabla V$$

In English, this means that the gradient of the voltage gives the electric field. (If you don't know vector calculus, think of it as the vector equivalent of E = dV/dx.) The units of the electric field are Newtons/Coulomb (force / charge), and mass is part of the definition of force. That's why the kilogram is there -- the energy in an electric circuit is transferred through a force field! You can see this more clearly if you separate out the units: $$Physics: Voltage = Efield \cdot distance \space$$ $$Units: V = \frac{N}{C} \cdot m$$ $$V = N \cdot \frac{1}{C} \cdot m$$ $$V = \frac{kg \cdot m}{s^2} \cdot \frac{1}{A \cdot s} \cdot m$$ $$V = \frac{kg \cdot m^2}{A \cdot s^3}$$


As an interesting side light. It may soon be that the kilogram is defined in terms of the volt. (Well more than just the volt.) See the watt balance.

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    \$\begingroup\$ "Any laboratory which had invested the (very considerable) time and money in a working watt balance would be able to measure masses to the same accuracy as they currently measure the Planck constant." Infinite awesomeness achieved. \$\endgroup\$ Commented Nov 13, 2014 at 19:00
  • \$\begingroup\$ @IgnacioVazquez-Abrams, I read a nice physics today article about it few months ago, still here.nymanz.org/Nonlinear/documents/… \$\endgroup\$ Commented Nov 14, 2014 at 0:01
  • \$\begingroup\$ although I find this an interesting find and a valuable comment, it doesn't even try to answer the question asked... George, please try to answer what kg is doing in V definition, otherwise this answer would be eligible to be flagged as "not an answer". \$\endgroup\$
    – user20088
    Commented Nov 15, 2014 at 13:48
  • \$\begingroup\$ @vaxquis, flag away :^) I thought everyone else answered the question. I guess if you read between the lines, the implication is that there are a limited number of fundamental constants. And a force can be mass * acceleration (little g) or B x I * L (a current in a B field for some length.) \$\endgroup\$ Commented Nov 15, 2014 at 20:29
  • \$\begingroup\$ little g is actually the earth's gravitational constant, not acceleration. Acceleration is a. \$\endgroup\$
    – user20088
    Commented Nov 15, 2014 at 21:51

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