Voltage vs Current: What Is the Difference?

Voltage is energy per unit charge between two points, measured in volts. Current is the rate charge flows past a point, measured in amperes. You measure voltage across an element and current through it. Series elements all share one current; parallel branches all share one voltage.

AP Physics: Unit 11 (topics 11.1 Electric Current, 11.3 Resistance, Resistivity, and Ohm's Law, 11.6 Kirchhoff's Loop Rule, 11.7 Kirchhoff's Junction Rule). Current is defined in AP Physics 2 Topic 11.1, where essential knowledge 11.1.A.1 gives I = delta q / delta t and 11.1.A.1.i states that charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force or emf. Potential difference itself is built in Unit 10, Topics 10.4 and 10.5. The two are related by Ohm's law in Topic 11.3, printed on the sheet as I = delta V / R. The sharing rules come from Topic 11.5: 11.5.A.1.i says the current in each element in series must be the same, and 11.5.A.1.ii says the potential difference across each parallel path is the same. The two conservation rules are separate topics: Kirchhoff's loop rule (11.6, a consequence of conservation of energy) governs voltage, and Kirchhoff's junction rule (11.7, a consequence of conservation of electric charge) governs current. Unit 11, Electric Circuits, carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods.

Across versus through: the distinction in one line

Current is a through quantity. It counts how much charge passes one cross-section of a wire each second. The AP Physics 2 sheet prints it as

I = \frac{\Delta q}{\Delta t}

and one ampere is one coulomb per second. Current is something a single point in a circuit has.

Voltage is an across quantity. It is the difference in electric potential between two points, so it counts the joules each coulomb gains or loses in going from one to the other. One volt is one joule per coulomb. A single point in a circuit does not have a voltage in the sense a problem means; it has a voltage relative to something else, which is why the sheet writes $\Delta V$ and not $V$ .

Everything else on this page is a consequence of those two sentences. Current needs one location and a stopwatch. Voltage needs two locations and an energy ledger. That is why an ammeter has to be spliced into the circuit and a voltmeter only has to be touched to two points, and it is why "the voltage through the resistor" and "the current across the resistor" are both grammatically wrong descriptions of a circuit.

Side by side

	Voltage $\Delta V$	Current $I$
What it measures	Energy transferred per unit charge	Charge passing per unit time
Unit	Volt (V), which is J/C	Ampere (A), which is C/s
Defined at	Two points, as a difference	One cross-section
Preposition	Across an element	Through an element
Meter, and how it is wired	Voltmeter, in parallel	Ammeter, in series
Shared in series	No, it divides up	Yes, identical in every element
Shared in parallel	Yes, identical across every branch	No, it splits between branches
Conservation rule	Sums to zero around a closed loop	In equals out at every junction
Which law it follows from	Conservation of energy	Conservation of charge
Ohm's law role	$\Delta V = IR$ , the cause	$I = \Delta V / R$ , the effect

The two most useful rows are the two conservation rows, and they are not symmetric. Current is conserved in the strict sense: charge is neither created nor destroyed at a junction, so whatever arrives leaves. Voltage is not conserved and is not meant to be. It sums to zero around a loop, which is a different statement: the energy a coulomb gains at the battery is exactly the energy it gives up in the rest of the loop, so it returns to where it started with the same potential it had.

The case that separates them: the same two resistors, wired two ways

Take a $30 \ \Omega$ resistor and a $60 \ \Omega$ resistor and a $9.0 \ \mathrm{V}$ ideal battery. Nothing about the components changes between the two arrangements. Only which quantity they are forced to share changes, and every number follows.

	In series	In parallel
Shared quantity	Current, $0.10 \ \mathrm{A}$ in both	Voltage, $9.0 \ \mathrm{V}$ across both
Across the $30 \ \Omega$	$3.0 \ \mathrm{V}$	$9.0 \ \mathrm{V}$
Across the $60 \ \Omega$	$6.0 \ \mathrm{V}$	$9.0 \ \mathrm{V}$
Through the $30 \ \Omega$	$0.10 \ \mathrm{A}$	$0.30 \ \mathrm{A}$
Through the $60 \ \Omega$	$0.10 \ \mathrm{A}$	$0.15 \ \mathrm{A}$
Which resistor gets more of it	The $60 \ \Omega$ gets more voltage	The $30 \ \Omega$ gets more current

Read the last row twice. In series the larger resistance takes the larger share of the voltage. In parallel the smaller resistance takes the larger share of the current. Those look like opposite rules and they are the same rule seen from two sides: $\Delta V = IR$ with $I$ held fixed makes $\Delta V$ rise with $R$ , and $I = \Delta V / R$ with $\Delta V$ held fixed makes $I$ fall as $R$ rises.

This is the reading skill the pair is really testing. Given a schematic, before computing anything, find the quantity you already know. Two elements end to end with no junction between them: you know their currents are equal. Two elements spanning the same two nodes: you know their potential differences are equal. That one identification decides which equation you write first, and the series vs parallel circuits guide carries the reduction procedure from there.

The two conservation rules, and what each one forbids

The CED files these as separate topics, and the split is exactly the voltage-current split.

[Kirchhoff's junction rule](/ap-physics-2/unit-11-electric-circuits/11-7-kirchhoffs-junction-rule) is about current. Essential knowledge 11.7.A.1 states that it is a consequence of the conservation of electric charge, and 11.7.A.2 states that the total amount of charge entering a junction per unit time must equal the total amount exiting it. What this forbids: current disappearing anywhere. Not in a resistor, not in a bulb, not in a motor.

[Kirchhoff's loop rule](/ap-physics-2/unit-11-electric-circuits/11-6-kirchhoffs-loop-rule) is about voltage. Essential knowledge 11.6.A.2 states that it is a consequence of the conservation of energy, and 11.6.A.3 states that the sum of potential differences across all circuit elements in a single closed loop must equal zero. What this forbids: a coulomb returning to its starting point with more or less energy than it left with.

Put the two together and the answer to "what does a bulb use up" falls out. It is not the current, which is identical on both sides of the filament. It is the energy each coulomb carries, which is why the potential drops across the bulb and does not drop across the ideal wire beside it. A bulb is a place where charge loses potential, not a place where charge goes missing.

That single reframing dissolves most beginner circuit confusion. Charge is a conveyor belt that runs at one speed through the whole loop; voltage is how much cargo each item on the belt is carrying at each point around it.

How each one is measured, and why the wiring differs

The wiring of the two meters is the definition made physical, and the CED states it directly in Topic 11.5.

Ammeters must be connected in series with the element whose current is being measured (11.5.C.1.i), because a through quantity can only be counted by putting the meter in the path so that the same charge goes through it.
Voltmeters must be connected in parallel with the element across which the potential difference is being measured (11.5.C.2.i), because an across quantity is a comparison of two points, so the meter has to touch both.

The ideal resistances follow from the same logic and point in opposite directions: an ideal ammeter has zero resistance so that it does not affect the current it sits in (11.5.C.1.ii), while an ideal voltmeter has infinite resistance so that no charge flows through it (11.5.C.2.ii). Swapping the two does something specific and bad to a circuit, and the ammeter vs voltmeter comparison works through what.

One consequence worth noticing now. You can measure a voltage without breaking anything, by touching two probes to a live circuit. You cannot measure a current without cutting the circuit open and inserting the meter into the gap. That asymmetry is not a quirk of instrument design; it is the across-versus-through distinction showing up on a lab bench.

When it costs a mark

"The current is used up by the bulb." The junction rule forbids it. Whatever current enters a series element leaves it. On a free-response question this error usually shows up as a claim that the bulb nearer the battery is brighter, which is false for identical bulbs in series.

"Voltage flows through the circuit." Nothing flows but charge. Voltage is a difference between two places, and a difference does not travel. This is not pedantry: writing it this way is what leads to putting a voltmeter in series.

Mixing a voltage from one part of the circuit with a resistance from another. Using the full battery voltage with a single resistor's value in a multi-resistor circuit inflates the current. Pair a resistor's own $\Delta V$ with that resistor's own $R$ , or pair the battery's $\Delta V$ with the equivalent resistance of everything it drives.

Assuming the bigger resistor always gets more of everything. It gets more voltage in series and less current in parallel. Which one applies depends only on how it is wired.

Milliamp slips. $250 \ \mathrm{mA}$ is $0.250 \ \mathrm{A}$ . Substituting 250 makes a resistance a thousand times too small, and the answer will still look like a plausible number of ohms, which is what makes this error survive a sanity check.

When they coincide, and why that lulls you

In the simplest circuit there is, a single battery driving a single resistor, the two quantities carry no distinguishing information at all. There is one current, and it is the same everywhere. There is one potential difference, and it appears across the only element there is. Every question about that circuit can be answered by finding either quantity and applying Ohm's law, and nothing ever forces you to say which one is shared, because both are.

This is the circuit most people learn on, and it is the reason the distinction stays invisible for so long. The words "voltage" and "current" behave like two names for the same thing called strength until a second element appears.

They also coincide numerically whenever the resistance happens to be $1 \ \Omega$ , since $I = \Delta V / R$ then returns the same digits. That is a coincidence of units and nothing more, but it makes a worked example look like it proves the two are interchangeable.

The moment the distinction turns on is the moment a circuit has two elements and you have to say which quantity they hold in common. Everything before that is the special case.

Where this sits on the AP exam

Current is Topic 11.1, which defines it as the rate charge passes a cross-sectional area and adds the CED's careful point in 11.1.A.2 that current is not a vector even though it has a direction. Potential difference arrives earlier, in Unit 10, and enters circuits through 11.1.A.1.i, which says charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force or emf. Unit 11 carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods.

The relationship between the two is Topic 11.3, Ohm's law, which the sheet prints as $I = \Delta V / R$ rather than the more familiar $V = IR$ . That arrangement is a hint about causation: a potential difference is applied, and a current results. The Ohm's law guide covers all three rearrangements and the power relation $P = I \Delta V$ , which is literally the product of this page's two quantities, joules per coulomb times coulombs per second.

For the resistance side of the picture, and why a material property and an object property are not the same either, see resistance vs resistivity.

The same two resistors, in series and then in parallel

A $30 \ \Omega$ resistor and a $60 \ \Omega$ resistor are connected to a $9.0 \ \mathrm{V}$ ideal battery. Find the current through and the potential difference across each resistor when they are (a) in series and (b) in parallel. (c) Say which quantity was shared in each case.

(a) In series, charge has one path, so the current is the same in both. Add the resistances: $R_{eq} = 30 + 60 = 90 \ \Omega$ .
$I = \Delta V / R_{eq} = 9.0 / 90 = 0.10 \ \mathrm{A}$ , and this same $0.10 \ \mathrm{A}$ is in both resistors.
Potential differences: $\Delta V_{30} = (0.10)(30) = 3.0 \ \mathrm{V}$ and $\Delta V_{60} = (0.10)(60) = 6.0 \ \mathrm{V}$ .
Loop check: $3.0 + 6.0 = 9.0 \ \mathrm{V}$ , matching the battery, as Kirchhoff's loop rule requires.
(b) In parallel, both resistors span the same two nodes, so both have the full $9.0 \ \mathrm{V}$ across them.
Branch currents: $I_{30} = 9.0/30 = 0.30 \ \mathrm{A}$ and $I_{60} = 9.0/60 = 0.15 \ \mathrm{A}$ .
Junction check: the battery must supply $0.30 + 0.15 = 0.45 \ \mathrm{A}$ . Confirm with the equivalent resistance: $1/R_{eq} = 1/30 + 1/60 = 2/60 + 1/60 = 3/60$ , so $R_{eq} = 20 \ \Omega$ and $I = 9.0/20 = 0.45 \ \mathrm{A}$ . The two routes agree.
(c) Series shared the current at $0.10 \ \mathrm{A}$ ; parallel shared the potential difference at $9.0 \ \mathrm{V}$ . Note the reversal in who gets more: in series the $60 \ \Omega$ took twice the voltage, while in parallel the $30 \ \Omega$ took twice the current.

Series: $0.10 \ \mathrm{A}$ through both, with $3.0 \ \mathrm{V}$ and $6.0 \ \mathrm{V}$ across them. Parallel: $9.0 \ \mathrm{V}$ across both, with $0.30 \ \mathrm{A}$ and $0.15 \ \mathrm{A}$ through them. Same components, same battery, and the quantity they share has swapped.

What a bulb actually uses up

A single bulb is connected across a $6.0 \ \mathrm{V}$ ideal battery by ideal wires, and an ammeter placed in the loop reads $0.25 \ \mathrm{A}$ . (a) What is the current on the far side of the bulb? (b) What is the potential difference across the bulb, and across each wire? (c) How much energy does each coulomb deliver to the bulb, and at what rate is energy delivered? (d) State what the bulb consumes.

(a) $0.25 \ \mathrm{A}$ . The bulb and the ammeter are in series, so the same charge passes through both. Kirchhoff's junction rule is a statement of charge conservation, and it leaves no route for current to be consumed.
(b) An ideal wire has negligible resistance (11.5.B.1), so $\Delta V = IR$ gives essentially zero across each wire. The loop rule then forces the whole $6.0 \ \mathrm{V}$ to appear across the bulb.
(c) Each coulomb crosses $6.0 \ \mathrm{V}$ , so each coulomb delivers $6.0 \ \mathrm{J}$ .
The rate is the sheet's power relation: $P = I \Delta V = (0.25 \ \mathrm{A})(6.0 \ \mathrm{V}) = 1.5 \ \mathrm{W}$ .
Check by unit reasoning instead of by formula: $0.25 \ \mathrm{C}$ passes each second and each carries $6.0 \ \mathrm{J}$ , so $(0.25)(6.0) = 1.5 \ \mathrm{J}$ per second. Same number, and it shows that power is literally current times voltage.
(d) The bulb consumes energy at $1.5 \ \mathrm{W}$ . It consumes no charge and no current.

The current is $0.25 \ \mathrm{A}$ on both sides of the bulb. The bulb has $6.0 \ \mathrm{V}$ across it and dissipates $1.5 \ \mathrm{W}$ . What is used up is the energy each coulomb carries, which is voltage, not the coulombs themselves, which is current.

Charge, energy, and why neither one alone is the amount of electricity

A $9.0 \ \mathrm{V}$ battery drives a steady current of $0.25 \ \mathrm{A}$ through a circuit for $3.0$ minutes. (a) How much charge passes through the battery? (b) How much energy does the battery deliver? (c) Verify the energy a second way using power. (d) Which of the two quantities in the title of this page is the amount of electricity used?

(a) Convert the time: $3.0$ minutes is $180 \ \mathrm{s}$ . Rearrange the sheet's definition $I = \Delta q / \Delta t$ to get $\Delta q = I \Delta t = (0.25 \ \mathrm{A})(180 \ \mathrm{s}) = 45 \ \mathrm{C}$ .
(b) Each coulomb is lifted through $9.0 \ \mathrm{V}$ , so it receives $9.0 \ \mathrm{J}$ . For $45 \ \mathrm{C}$ : $E = q \Delta V = (45 \ \mathrm{C})(9.0 \ \mathrm{V}) = 405 \ \mathrm{J}$ .
(c) By power: $P = I \Delta V = (0.25)(9.0) = 2.25 \ \mathrm{W}$ , and over $180 \ \mathrm{s}$ that is $(2.25)(180) = 405 \ \mathrm{J}$ . The two routes agree exactly, because $q\Delta V$ and $I \Delta V \Delta t$ are the same product rearranged.
(d) Neither. Current is coulombs per second and voltage is joules per coulomb. Only their product, $2.25 \ \mathrm{W}$ , is a rate of energy use, and only the product multiplied by time, $405 \ \mathrm{J}$ , is an amount of energy.
This is why a device rating quotes both, or quotes their product in watts. A $9.0 \ \mathrm{V}$ label alone tells you nothing about how much energy anything uses.

$\Delta q = 45 \ \mathrm{C}$ and $E = 405 \ \mathrm{J}$ , confirmed by $P = 2.25 \ \mathrm{W}$ over $180 \ \mathrm{s}$ . Voltage and current are the two factors of electrical power; neither on its own is an amount of anything delivered.

Frequently asked questions

What is the difference between voltage and current?

Current is the rate at which charge flows past one point in a circuit, measured in amperes, where one ampere is one coulomb per second. Voltage is the difference in electric potential between two points, measured in volts, where one volt is one joule per coulomb. Current is a through quantity that a single cross-section has; voltage is an across quantity that only exists between two locations. That is why an ammeter is wired in series and a voltmeter in parallel.

Is voltage or current more dangerous?

Physics can answer the narrow part of this. What passes through a body is current, and how much current passes is set by the potential difference across the body divided by its resistance, from I = ΔV/R. So a high voltage is dangerous because of the current it can drive, and a high-voltage source that can only supply a tiny current does far less than the same voltage backed by a large one. For anything beyond exam physics, follow lab safety rules rather than a formula.

What stays the same in a series circuit and what stays the same in parallel?

In series the current is the same in every element, because charge has one path and no junction to leave by, and the potential differences divide up so that they add to the source voltage. In parallel the potential difference is the same across every branch, because all branches span the same two nodes, and the current splits between them so that the branch currents add to the total. The larger resistance takes the larger share of the voltage in series and the smaller share of the current in parallel.

Does a bulb use up current?

No. Kirchhoff's junction rule is a consequence of charge conservation, and it means the current leaving a bulb equals the current entering it. What the bulb uses is the energy each coulomb carries, which is why the electric potential drops across the bulb. Two identical bulbs in series are equally bright for exactly this reason: they carry the same current and each takes half the voltage.

Why is Ohm's law written as I = ΔV/R on the AP sheet?

The AP Physics 2 sheet prints it in the form that isolates current, which matches the physics: a potential difference is applied across a resistor and a current results. The delta is a reminder that voltage is a difference between the two ends of the element, not something a single point holds. It is the same relationship as V = IR, rearranged, so use whichever arrangement isolates your unknown.

Can you have voltage without current?

Yes. A battery sitting on a shelf has a potential difference between its terminals and no current at all, because there is no complete path for charge. The same happens across an open switch in a live circuit: the full source voltage appears across the gap while the current everywhere in that loop is zero. The reverse is closer to impossible in ordinary circuits, since driving a current through anything with resistance requires a potential difference across it.