# Why is a ‘Voltage Drop’ considered when a current is passed through a resistance?

A resistance, by definition offers restriction to the flow of current. Thus when propelling a charge through a resistance, we have to do more work. Therefore shouldn’t the work done per unit charge, i.e, voltage, increase?

• You are thinking about this the wrong way round. It isn't that you drive a current and see what voltage develops. Rather, you establish a voltage and then the resistance determines the level of current that flows (Ohm's Law). Commented Aug 6, 2020 at 12:10
• @Oscar Bravo you certainly can drive a resistor with a current source instead of a voltage source. You shouldn’t think of Ohms law as being “establish V then resistance determines I” instead Ohms law is simply a constraint which enforces a relationship between I and V.
– Dale
Commented Aug 6, 2020 at 13:44
• @Oscar, indeed if V is your independent variable and I is the dependent variable you should be calling the device a conductor and expressing ohm’s law as $I=GV$. Commented Aug 6, 2020 at 14:09
• @Dale Sure if you get yourself a fancy-pants current-driver, you can select a current. But that thing works by adjusting the voltage via feedback until it gets the desired current. Things like batteries and power supplies output a fixed voltage and the current drawn is a consequence of the resistance. Commented Aug 7, 2020 at 12:43
• @Oscar Bravo “that thing works by adjusting the voltage via feedback until it gets the desired current” That is not generally true. A transistor with a Zener diode on the base and a resistor on the emitter gives a simple current source without feedback. I am not sure where you are getting your anti-current-source outlook from. Current sources are every bit as legitimate as voltage sources. Not every source is a battery.
– Dale
Commented Aug 7, 2020 at 20:53

No, the voltage drop is a term for the energy spent in pushing the charge through.

If you do more work, then you push that charge through faster. Energy is still lost to heat etc. as it passes through the resistor. That is what we call the voltage drop

(Remember that voltage is a word for difference in electric potential energy per charge, so we are just talking about a certain energy amount (per charge).)

Think of voltage as the pressure difference in a water hose. If you squeeze it on the middle, then you must open up the faucet a little more to increase the pressure if you still want equally much water to come out. That is the extra work you do. The drop in the pressure as this water passes the constriction represents the voltage drop (on the other side there is less pressure, corresponding to less energy associated with the charge).

shouldn’t the work done per unit charge, i.e, voltage, increase?

The work done by the E field is $$W=\int \vec F \cdot d \vec s$$. And since $$\vec E=\vec F/q$$ we have $$W/q=\int \vec E \cdot d\vec s$$ but $$\Delta V = -\int \vec E \cdot d\vec s$$ so $$\Delta V = -W/q$$

This is sometimes described as the work done against the field rather than the work done by the field. Work done per unit charge is unfortunately too vague to determine the sign of the work. It does not describe whether it is seen from the perspective of the circuit or the perspective of the environment.

Unfortunately, this is a common confusion. I think it is therefore best to simply remember the completely unambiguous $$\Delta V =-\int \vec E \cdot d \vec s$$

A higher resistance requires more energy, at constant current, but this energy is converted into heat.